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← Grade 10 Optional Math

Exam practice

Every question from the end of every unit, with hints and answers to open. After you check an answer, mark how you did, and come back to the ones you didn’t get.

254 questions

Unit 7 · Trigonometry

Exercise 7.1

Question 2

a.

cos⁡15°=3+122 \cos 15° = \dfrac{\sqrt{3} + 1}{2\sqrt{2}}

Answer

cos⁡(45°−30°)=12⋅32+12⋅12=3+122 \cos(45° - 30°) = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3} + 1}{2\sqrt{2}}

b.

sin⁡75°=3+122 \sin 75° = \dfrac{\sqrt{3} + 1}{2\sqrt{2}}

Answer

sin⁡(45°+30°)=12⋅32+12⋅12=3+122 \sin(45° + 30°) = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3} + 1}{2\sqrt{2}}
(the same as cos⁡15° \cos 15° , since 75°=90°−15° 75° = 90° - 15° ).

c.

tan⁡75°=2+3 \tan 75° = 2 + \sqrt{3}

Answer

tan⁡(45°+30°)=1+131−13=3+13−1=(3+1)23−1=4+232=2+3 \tan(45° + 30°) = \frac{1 + \frac{1}{\sqrt{3}}}{1 - \frac{1}{\sqrt{3}}} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} = \frac{(\sqrt{3} + 1)^2}{3 - 1} = \frac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3}

d.

cos⁡105°=1−322 \cos 105° = \dfrac{1 - \sqrt{3}}{2\sqrt{2}}

Answer

cos⁡(60°+45°)=12⋅12−32⋅12=1−322 \cos(60° + 45°) = \frac{1}{2} \cdot \frac{1}{\sqrt{2}} - \frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{2}} = \frac{1 - \sqrt{3}}{2\sqrt{2}}
Negative, as it should be: 105° is past 90°.

e.

tan⁡105°=−(2+3) \tan 105° = -(2 + \sqrt{3})

Answer

tan⁡(60°+45°)=3+11−3=−3+13−1=−(2+3) \tan(60° + 45°) = \frac{\sqrt{3} + 1}{1 - \sqrt{3}} = -\frac{\sqrt{3} + 1}{\sqrt{3} - 1} = -(2 + \sqrt{3})
using c.

f.

cot⁡15°=2+3 \cot 15° = 2 + \sqrt{3}

Answer

cot⁡15°=tan⁡(90°−15°)=tan⁡75°=2+3 \cot 15° = \tan(90° - 15°) = \tan 75° = 2 + \sqrt{3}
by c. Or directly: cot⁡15°=1/tan⁡(45°−30°)=(3+1)/(3−1)=2+3 \cot 15° = 1/\tan(45° - 30°) = (\sqrt{3} + 1)/(\sqrt{3} - 1) = 2 + \sqrt{3} .

Question 3

a.

sin (A + B)

Answer

35⋅1213+45⋅513=36+2065=5665 \tfrac{3}{5} \cdot \tfrac{12}{13} + \tfrac{4}{5} \cdot \tfrac{5}{13} = \tfrac{36 + 20}{65} = \tfrac{56}{65}

b.

cos (A + B)

Answer

45⋅1213−35⋅513=48−1565=3365 \tfrac{4}{5} \cdot \tfrac{12}{13} - \tfrac{3}{5} \cdot \tfrac{5}{13} = \tfrac{48 - 15}{65} = \tfrac{33}{65}
Check: 562+332=4225=652 56^2 + 33^2 = 4225 = 65^2 ✓.

c.

sin (A − B)

Answer

36−2065=1665 \tfrac{36 - 20}{65} = \tfrac{16}{65}

d.

cos (A − B)

Answer

48+1565=6365 \tfrac{48 + 15}{65} = \tfrac{63}{65}

e.

tan (A + B)

Answer

sin⁡(A+B)cos⁡(A+B)=56/6533/65=5633 \frac{\sin(A + B)}{\cos(A + B)} = \frac{56/65}{33/65} = \frac{56}{33}

f.

cot⁡(A−B) \cot(A - B)

Answer

cos⁡(A−B)sin⁡(A−B)=63/6516/65=6316 \frac{\cos(A - B)}{\sin(A - B)} = \frac{63/65}{16/65} = \frac{63}{16}

Question 4

a.

sin (A + B) +  sin (A − B) = 2 sin A cos B

Answer

( sin A cos B +  cos A sin B) + ( sin A cos B −  cos A sin B) = 2 sin A cos B.

b.

sin (A + B) −  sin (A − B) = 2 cos A sin B

Answer

( sin A cos B +  cos A sin B) − ( sin A cos B −  cos A sin B) = 2 cos A sin B.

c.

cos (A + B) +  cos (A − B) = 2 cos A cos B

Answer

( cos A cos B −  sin A sin B) + ( cos A cos B +  sin A sin B) = 2 cos A cos B.

d.

cos (A + B) −  cos (A − B) = −2 sin A sin B

Answer

( cos A cos B −  sin A sin B) − ( cos A cos B +  sin A sin B) = −2 sin A sin B. The minus sign matters: with A=B=45° A = B = 45° , cos⁡90°−cos⁡0°=−1=−2⋅12 \cos 90° - \cos 0° = -1 = -2 \cdot \frac{1}{2} .

Question 5

a.

sin⁡75°−sin⁡15°=12 \sin 75° - \sin 15° = \dfrac{1}{\sqrt{2}}

Answer

3+122−3−122=222=12 \frac{\sqrt{3} + 1}{2\sqrt{2}} - \frac{\sqrt{3} - 1}{2\sqrt{2}} = \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}}

b.

cos⁡15°−sin⁡15°=12 \cos 15° - \sin 15° = \dfrac{1}{\sqrt{2}}

Answer

3+122−3−122=12 \frac{\sqrt{3} + 1}{2\sqrt{2}} - \frac{\sqrt{3} - 1}{2\sqrt{2}} = \frac{1}{\sqrt{2}}

c.

sin⁡105°−cos⁡75°=12 \sin 105° - \cos 75° = \dfrac{1}{\sqrt{2}}

Answer

This is Example 3: 3+122−3−122=12 \dfrac{\sqrt{3} + 1}{2\sqrt{2}} - \dfrac{\sqrt{3} - 1}{2\sqrt{2}} = \dfrac{1}{\sqrt{2}} .

d.

cos⁡105°+cos⁡15°=12 \cos 105° + \cos 15° = \dfrac{1}{\sqrt{2}}

Answer

1−322+3+122=222=12 \frac{1 - \sqrt{3}}{2\sqrt{2}} + \frac{\sqrt{3} + 1}{2\sqrt{2}} = \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}}

e.

tan⁡15°+cot⁡15°=4 \tan 15° + \cot 15° = 4

Answer

tan⁡15°=tan⁡(45°−30°)=(3−1)/(3+1)=2−3 \tan 15° = \tan(45° - 30°) = (\sqrt{3} - 1)/(\sqrt{3} + 1) = 2 - \sqrt{3} , and cot⁡15°=2+3 \cot 15° = 2 + \sqrt{3} (Question 2 f). Their sum is 4.

f.

cos⁡18°−sin⁡18°=2sin⁡27° \cos 18° - \sin 18° = \sqrt{2}\sin 27°

Answer

LHS=2(12cos⁡18°−12sin⁡18°)=2(sin⁡45°cos⁡18°−cos⁡45°sin⁡18°)=2sin⁡(45°−18°)=2sin⁡27° \begin{aligned} \text{LHS} &= \sqrt{2}\left(\tfrac{1}{\sqrt{2}}\cos 18° - \tfrac{1}{\sqrt{2}}\sin 18°\right) = \sqrt{2}(\sin 45° \cos 18° - \cos 45° \sin 18°) \\ &= \sqrt{2}\sin(45° - 18°) = \sqrt{2}\sin 27° \end{aligned}

g.

sin⁡35°+cos⁡35°=2cos⁡10° \sin 35° + \cos 35° = \sqrt{2}\cos 10°

Answer

LHS=2(cos⁡45°cos⁡35°+sin⁡45°sin⁡35°)=2cos⁡(45°−35°)=2cos⁡10° \text{LHS} = \sqrt{2}(\cos 45° \cos 35° + \sin 45° \sin 35°) = \sqrt{2}\cos(45° - 35°) = \sqrt{2}\cos 10°

h.

3cos⁡20°+sin⁡20°=2sin⁡80° \sqrt{3}\cos 20° + \sin 20° = 2\sin 80°

Answer

LHS=2(32cos⁡20°+12sin⁡20°)=2(sin⁡60°cos⁡20°+cos⁡60°sin⁡20°)=2sin⁡80° \text{LHS} = 2\left(\tfrac{\sqrt{3}}{2}\cos 20° + \tfrac{1}{2}\sin 20°\right) = 2(\sin 60° \cos 20° + \cos 60° \sin 20°) = 2\sin 80°

Question 6

a.

sin⁡(25°+α)cos⁡(65°−α)+cos⁡(25°+α)sin⁡(65°−α)=1 \sin(25° + \alpha)\cos(65° - \alpha) + \cos(25° + \alpha)\sin(65° - \alpha) = 1

Answer

It is sin⁡{(25°+α)+(65°−α)}=sin⁡90°=1 \sin\{(25° + \alpha) + (65° - \alpha)\} = \sin 90° = 1 .

b.

cos⁡(25°+α)cos⁡(65°+α)−sin⁡(25°+α)sin⁡(65°+α)=−sin⁡2α \cos(25° + \alpha)\cos(65° + \alpha) - \sin(25° + \alpha)\sin(65° + \alpha) = -\sin 2\alpha

Answer

It is cos⁡{(25°+α)+(65°+α)}=cos⁡(90°+2α)=−sin⁡2α \cos\{(25° + \alpha) + (65° + \alpha)\} = \cos(90° + 2\alpha) = -\sin 2\alpha .

c.

cos⁡(60°+β)−cos⁡β+cos⁡(60°−β)=0 \cos(60° + \beta) - \cos\beta + \cos(60° - \beta) = 0

Answer

cos⁡(60°+β)+cos⁡(60°−β)=2cos⁡60°cos⁡β=cos⁡β \cos(60° + \beta) + \cos(60° - \beta) = 2\cos 60° \cos\beta = \cos\beta (Question 4 c), so the left side is cos β −  cos β = 0.

d.

tan⁡(45°+ϕ)−tan⁡ϕ1+tan⁡(45°+ϕ)tan⁡ϕ=1 \dfrac{\tan(45° + \phi) - \tan\phi}{1 + \tan(45° + \phi)\tan\phi} = 1

Answer

It is tan⁡{(45°+ϕ)−ϕ}=tan⁡45°=1 \tan\{(45° + \phi) - \phi\} = \tan 45° = 1 .

Question 7

a.

tan A +  tan B +  tan A tan B = 1

Answer

tan⁡(A+B)=tan⁡45°=1 \tan(A + B) = \tan 45° = 1 : ( tan A +  tan B)/(1 −  tan A tan B) = 1, so tan A +  tan B = 1 −  tan A tan B, which rearranges to the result.

b.

(1 +  tan A)(1 +  tan B) = 2

Answer

(1 +  tan A)(1 +  tan B) = 1 + ( tan A +  tan B +  tan A tan B) = 1 + 1 = 2, by a.

Question 8

a.

If tan X = 2/3 and tan Y = 1/5, prove that X+Y=45° X + Y = 45° .

Answer

tan⁡(X+Y)=23+151−23⋅15=13/1513/15=1 \tan(X + Y) = \frac{\frac{2}{3} + \frac{1}{5}}{1 - \frac{2}{3} \cdot \frac{1}{5}} = \frac{13/15}{13/15} = 1
Both angles are acute, so X+Y=45° X + Y = 45° .

b.

If tan A = m and tan B = 1/m (m > 0), prove that A+B=(π/2)c A + B = (\pi/2)^c .

Answer

tan A tan B = 1, so the bottom of tan (A + B) = ( tan A +  tan B)/(1 −  tan A tan B) is 0: tan (A + B) is not defined, and A+B=90°=(π/2)c A + B = 90° = (\pi/2)^c . Or: tan⁡B=cot⁡A=tan⁡(90°−A) \tan B = \cot A = \tan(90° - A) , so B=90°−A B = 90° - A .

Question 9

a.

cos⁡35°−sin⁡35°cos⁡35°+sin⁡35°=tan⁡10° \dfrac{\cos 35° - \sin 35°}{\cos 35° + \sin 35°} = \tan 10°

Answer

Divide by cos⁡35° \cos 35° : 1−tan⁡35°1+tan⁡35°=tan⁡(45°−35°)=tan⁡10° \dfrac{1 - \tan 35°}{1 + \tan 35°} = \tan(45° - 35°) = \tan 10° .

b.

cos⁡17°+sin⁡17°cos⁡17°−sin⁡17°=tan⁡62° \dfrac{\cos 17° + \sin 17°}{\cos 17° - \sin 17°} = \tan 62°

Answer

1+tan⁡17°1−tan⁡17°=tan⁡(45°+17°)=tan⁡62° \dfrac{1 + \tan 17°}{1 - \tan 17°} = \tan(45° + 17°) = \tan 62° .

c.

cos⁡8°+sin⁡8°cos⁡8°−sin⁡8°=tan⁡53° \dfrac{\cos 8° + \sin 8°}{\cos 8° - \sin 8°} = \tan 53°

Answer

1+tan⁡8°1−tan⁡8°=tan⁡(45°+8°)=tan⁡53° \dfrac{1 + \tan 8°}{1 - \tan 8°} = \tan(45° + 8°) = \tan 53° . (It also equals cot⁡37° \cot 37° .)

d.

cos⁡10°−sin⁡10°cos⁡10°+sin⁡10°=tan⁡35° \dfrac{\cos 10° - \sin 10°}{\cos 10° + \sin 10°} = \tan 35°

Answer

1−tan⁡10°1+tan⁡10°=tan⁡(45°−10°)=tan⁡35° \dfrac{1 - \tan 10°}{1 + \tan 10°} = \tan(45° - 10°) = \tan 35° .

Question 10

a.

tan⁡5A−tan⁡4A1+tan⁡5Atan⁡4A=tan⁡A \dfrac{\tan 5A - \tan 4A}{1 + \tan 5A \tan 4A} = \tan A

Answer

It is tan (5A − 4A) =  tan A.

b.

tan⁡7A−tan⁡4A1+tan⁡7Atan⁡4A=tan⁡3A \dfrac{\tan 7A - \tan 4A}{1 + \tan 7A \tan 4A} = \tan 3A

Answer

It is tan (7A − 4A) =  tan 3A.

c.

tan⁡(A+B)−tan⁡B1+tan⁡(A+B)tan⁡B=tan⁡A \dfrac{\tan(A + B) - \tan B}{1 + \tan(A + B)\tan B} = \tan A

Answer

It is tan {(A + B) − B} =  tan A.

d.

tan⁡(A+B)tan⁡(A−B)=tan⁡2A−tan⁡2B1−tan⁡2Atan⁡2B \tan(A + B)\tan(A - B) = \dfrac{\tan^2 A - \tan^2 B}{1 - \tan^2 A \tan^2 B}

Answer

tan⁡A+tan⁡B1−tan⁡Atan⁡B⋅tan⁡A−tan⁡B1+tan⁡Atan⁡B=tan⁡2A−tan⁡2B1−tan⁡2Atan⁡2B \frac{\tan A + \tan B}{1 - \tan A \tan B} \cdot \frac{\tan A - \tan B}{1 + \tan A \tan B} = \frac{\tan^2 A - \tan^2 B}{1 - \tan^2 A \tan^2 B}
(the top and the bottom are each (p+q)(p−q)=p2−q2 (p + q)(p - q) = p^2 - q^2 ).

e.

sin⁡(A+B)sin⁡Asin⁡B=cot⁡B+cot⁡A \dfrac{\sin(A + B)}{\sin A \sin B} = \cot B + \cot A

Answer

sin⁡Acos⁡B+cos⁡Asin⁡Bsin⁡Asin⁡B=cos⁡Bsin⁡B+cos⁡Asin⁡A=cot⁡B+cot⁡A \frac{\sin A \cos B + \cos A \sin B}{\sin A \sin B} = \frac{\cos B}{\sin B} + \frac{\cos A}{\sin A} = \cot B + \cot A

f.

cos⁡(A+B)sin⁡Asin⁡B=cot⁡Acot⁡B−1 \dfrac{\cos(A + B)}{\sin A \sin B} = \cot A \cot B - 1

Answer

cos⁡Acos⁡B−sin⁡Asin⁡Bsin⁡Asin⁡B=cot⁡Acot⁡B−1 \frac{\cos A \cos B - \sin A \sin B}{\sin A \sin B} = \cot A \cot B - 1

Question 11

a.

tan⁡25°+tan⁡20°+tan⁡25°tan⁡20°=1 \tan 25° + \tan 20° + \tan 25° \tan 20° = 1

Answer

25°+20°=45° 25° + 20° = 45° : as in Example 9, tan⁡25°+tan⁡20°=1−tan⁡25°tan⁡20° \tan 25° + \tan 20° = 1 - \tan 25° \tan 20° .

b.

1−tan⁡27°−tan⁡18°−tan⁡27°tan⁡18°=0 1 - \tan 27° - \tan 18° - \tan 27° \tan 18° = 0

Answer

27°+18°=45° 27° + 18° = 45° , so tan⁡27°+tan⁡18°=1−tan⁡27°tan⁡18° \tan 27° + \tan 18° = 1 - \tan 27° \tan 18° . Then 1−(tan⁡27°+tan⁡18°)−tan⁡27°tan⁡18°=1−1+tan⁡27°tan⁡18°−tan⁡27°tan⁡18°=0 1 - (\tan 27° + \tan 18°) - \tan 27° \tan 18° = 1 - 1 + \tan 27° \tan 18° - \tan 27° \tan 18° = 0 .

c.

tan⁡65°−tan⁡20°=1+tan⁡65°tan⁡20° \tan 65° - \tan 20° = 1 + \tan 65° \tan 20°

Answer

65°−20°=45° 65° - 20° = 45° : tan⁡45°=tan⁡65°−tan⁡20°1+tan⁡65°tan⁡20°=1 \tan 45° = \dfrac{\tan 65° - \tan 20°}{1 + \tan 65° \tan 20°} = 1 , so tan⁡65°−tan⁡20°=1+tan⁡65°tan⁡20° \tan 65° - \tan 20° = 1 + \tan 65° \tan 20° .

d.

1+cot⁡18°+cot⁡27°=cot⁡18°cot⁡27° 1 + \cot 18° + \cot 27° = \cot 18° \cot 27°

Answer

cot⁡45°=cot⁡(18°+27°)=cot⁡18°cot⁡27°−1cot⁡27°+cot⁡18°=1 \cot 45° = \cot(18° + 27°) = \frac{\cot 18° \cot 27° - 1}{\cot 27° + \cot 18°} = 1
so cot⁡18°cot⁡27°−1=cot⁡18°+cot⁡27° \cot 18° \cot 27° - 1 = \cot 18° + \cot 27° , which rearranges to the result.

Question 12

a.

tan⁡50°+tan⁡60°+tan⁡70°=tan⁡50°tan⁡60°tan⁡70° \tan 50° + \tan 60° + \tan 70° = \tan 50° \tan 60° \tan 70°

Answer

50°+70°=180°−60° 50° + 70° = 180° - 60° , so tan⁡(50°+70°)=tan⁡(180°−60°)=−tan⁡60° \tan(50° + 70°) = \tan(180° - 60°) = -\tan 60° :
tan⁡50°+tan⁡70°1−tan⁡50°tan⁡70°=−tan⁡60° \frac{\tan 50° + \tan 70°}{1 - \tan 50° \tan 70°} = -\tan 60°
tan⁡50°+tan⁡70°=−tan⁡60°+tan⁡50°tan⁡60°tan⁡70° \tan 50° + \tan 70° = -\tan 60° + \tan 50° \tan 60° \tan 70° , which rearranges to the result. (The three angles add up to 180°, as in a triangle: see 7.6.)

b.

tan 10A −  tan 6A −  tan 4A =  tan 10A tan 6A tan 4A

Answer

10A = 6A + 4A, so tan⁡10A=tan⁡6A+tan⁡4A1−tan⁡6Atan⁡4A \tan 10A = \dfrac{\tan 6A + \tan 4A}{1 - \tan 6A \tan 4A} . Cross-multiply: tan 10A −  tan 10A tan 6A tan 4A =  tan 6A +  tan 4A, which rearranges to the result.

c.

tan⁡20°tan⁡30°+tan⁡30°tan⁡40°+tan⁡40°tan⁡20°=1 \tan 20° \tan 30° + \tan 30° \tan 40° + \tan 40° \tan 20° = 1

Answer

20°+40°=90°−30° 20° + 40° = 90° - 30° , so tan⁡(20°+40°)=cot⁡30°=1/tan⁡30° \tan(20° + 40°) = \cot 30° = 1/\tan 30° :
tan⁡20°+tan⁡40°1−tan⁡20°tan⁡40°=1tan⁡30° \frac{\tan 20° + \tan 40°}{1 - \tan 20° \tan 40°} = \frac{1}{\tan 30°}
tan⁡30°(tan⁡20°+tan⁡40°)=1−tan⁡20°tan⁡40° \tan 30°(\tan 20° + \tan 40°) = 1 - \tan 20° \tan 40° , which rearranges to the result.

d.

cot⁡5Acot⁡4A−cot⁡9Acot⁡5A−cot⁡9Acot⁡4A=1 \cot 5A \cot 4A - \cot 9A \cot 5A - \cot 9A \cot 4A = 1

Answer

9A = 5A + 4A: cot⁡9A=cot⁡5Acot⁡4A−1cot⁡4A+cot⁡5A \cot 9A = \dfrac{\cot 5A \cot 4A - 1}{\cot 4A + \cot 5A} . Cross-multiply: cot⁡9Acot⁡4A+cot⁡9Acot⁡5A=cot⁡5Acot⁡4A−1 \cot 9A \cot 4A + \cot 9A \cot 5A = \cot 5A \cot 4A - 1 , which rearranges to the result.

Question 13

a.

tan⁡50°−tan⁡40°=2tan⁡10° \tan 50° - \tan 40° = 2\tan 10°

Answer

tan⁡10°=tan⁡(50°−40°)=tan⁡50°−tan⁡40°1+tan⁡50°tan⁡40° \tan 10° = \tan(50° - 40°) = \dfrac{\tan 50° - \tan 40°}{1 + \tan 50° \tan 40°} . Since 50°+40°=90° 50° + 40° = 90° , tan⁡40°=cot⁡50° \tan 40° = \cot 50° and tan⁡50°tan⁡40°=1 \tan 50° \tan 40° = 1 . So tan⁡10°=(tan⁡50°−tan⁡40°)/2 \tan 10° = (\tan 50° - \tan 40°)/2 .

b.

tan⁡70°=2tan⁡50°+tan⁡20° \tan 70° = 2\tan 50° + \tan 20°

Answer

tan⁡50°=tan⁡(70°−20°)=tan⁡70°−tan⁡20°1+tan⁡70°tan⁡20°=tan⁡70°−tan⁡20°2 \tan 50° = \tan(70° - 20°) = \dfrac{\tan 70° - \tan 20°}{1 + \tan 70° \tan 20°} = \dfrac{\tan 70° - \tan 20°}{2} , since tan⁡70°tan⁡20°=1 \tan 70° \tan 20° = 1 . So 2tan⁡50°=tan⁡70°−tan⁡20° 2\tan 50° = \tan 70° - \tan 20° . Check: 2(1.1918)+0.3640=2.7475=tan⁡70° 2(1.1918) + 0.3640 = 2.7475 = \tan 70° ✓.

Question 14

a.

sin⁡(β+γ)sin⁡(β−γ)=tan⁡β+tan⁡γtan⁡β−tan⁡γ \dfrac{\sin(\beta + \gamma)}{\sin(\beta - \gamma)} = \dfrac{\tan\beta + \tan\gamma}{\tan\beta - \tan\gamma}

Answer

sin⁡βcos⁡γ+cos⁡βsin⁡γsin⁡βcos⁡γ−cos⁡βsin⁡γ \frac{\sin\beta\cos\gamma + \cos\beta\sin\gamma}{\sin\beta\cos\gamma - \cos\beta\sin\gamma}
Divide the top and the bottom by cos⁡βcos⁡γ \cos\beta\cos\gamma : tan⁡β+tan⁡γtan⁡β−tan⁡γ \dfrac{\tan\beta + \tan\gamma}{\tan\beta - \tan\gamma} .

b.

sin⁡(A+B)sin⁡(A−B)cos⁡2Acos⁡2B=tan⁡2A−tan⁡2B \dfrac{\sin(A + B)\sin(A - B)}{\cos^2 A \cos^2 B} = \tan^2 A - \tan^2 B

Answer

sin⁡(A+B)sin⁡(A−B)=sin⁡2Acos⁡2B−cos⁡2Asin⁡2B \sin(A + B)\sin(A - B) = \sin^2 A \cos^2 B - \cos^2 A \sin^2 B
Divide by cos⁡2Acos⁡2B \cos^2 A \cos^2 B : tan⁡2A−tan⁡2B \tan^2 A - \tan^2 B .

c.

sin⁡(A+B)+sin⁡(A−B)cos⁡(A+B)+cos⁡(A−B)=tan⁡A \dfrac{\sin(A + B) + \sin(A - B)}{\cos(A + B) + \cos(A - B)} = \tan A

Answer

By Question 4 a and c: 2sin⁡Acos⁡B2cos⁡Acos⁡B=tan⁡A \dfrac{2\sin A \cos B}{2\cos A \cos B} = \tan A .

Question 15

a.

sin⁡(A−B)cos⁡Acos⁡B+sin⁡(B−C)cos⁡Bcos⁡C+sin⁡(C−A)cos⁡Ccos⁡A=0 \dfrac{\sin(A - B)}{\cos A \cos B} + \dfrac{\sin(B - C)}{\cos B \cos C} + \dfrac{\sin(C - A)}{\cos C \cos A} = 0

Answer

sin⁡(A−B)cos⁡Acos⁡B=sin⁡Acos⁡B−cos⁡Asin⁡Bcos⁡Acos⁡B=tan⁡A−tan⁡B \frac{\sin(A - B)}{\cos A \cos B} = \frac{\sin A \cos B - \cos A \sin B}{\cos A \cos B} = \tan A - \tan B
In the same way the others are tan B −  tan C and tan C −  tan A. The three add up to 0.

b.

sin (A + B + C) =  cos A cos B cos C( tan A +  tan B +  tan C −  tan A tan B tan C)

Answer

sin⁡{(A+B)+C}=sin⁡(A+B)cos⁡C+cos⁡(A+B)sin⁡C=sin⁡Acos⁡Bcos⁡C+cos⁡Asin⁡Bcos⁡C+cos⁡Acos⁡Bsin⁡C−sin⁡Asin⁡Bsin⁡C \begin{aligned} \sin\{(A + B) + C\} &= \sin(A + B)\cos C + \cos(A + B)\sin C \\ &= \sin A \cos B \cos C + \cos A \sin B \cos C + \cos A \cos B \sin C - \sin A \sin B \sin C \end{aligned}
Take out cos A cos B cos C: each term divided by it gives tan A, tan B, tan C and tan A tan B tan C.

c.

cos (A + B + C) =  cos A cos B cos C(1 −  tan A tan B −  tan B tan C −  tan C tan A)

Answer

cos⁡{(A+B)+C}=cos⁡(A+B)cos⁡C−sin⁡(A+B)sin⁡C=cos⁡Acos⁡Bcos⁡C−sin⁡Asin⁡Bcos⁡C−sin⁡Acos⁡Bsin⁡C−cos⁡Asin⁡Bsin⁡C \begin{aligned} \cos\{(A + B) + C\} &= \cos(A + B)\cos C - \sin(A + B)\sin C \\ &= \cos A \cos B \cos C - \sin A \sin B \cos C - \sin A \cos B \sin C - \cos A \sin B \sin C \end{aligned}
Take out cos A cos B cos C: 1 −  tan A tan B −  tan A tan C −  tan B tan C.

d.

tan⁡(A+B+C)=tan⁡A+tan⁡B+tan⁡C−tan⁡Atan⁡Btan⁡C1−tan⁡Atan⁡B−tan⁡Btan⁡C−tan⁡Ctan⁡A \tan(A + B + C) = \dfrac{\tan A + \tan B + \tan C - \tan A \tan B \tan C}{1 - \tan A \tan B - \tan B \tan C - \tan C \tan A}

Answer

Divide b by c: the cos A cos B cos C cancels.

Exercise 7.2

Question 2

a.

If sin A = 4/5, find the values of sin 2A, cos 2A and tan 2A.

Answer

cos A = 3/5 (A acute). sin⁡2A=2⋅45⋅35=2425 \sin 2A = 2 \cdot \frac{4}{5} \cdot \frac{3}{5} = \frac{24}{25} ; cos⁡2A=925−1625=−725 \cos 2A = \frac{9}{25} - \frac{16}{25} = -\frac{7}{25} ; tan⁡2A=−247 \tan 2A = -\frac{24}{7} . (A is more than 45°, so 2A is more than 90°: its cosine is negative.)

b.

If cos A = 12/13, find the values of sin 2A, cos 2A and tan 2A.

Answer

sin A = 5/13. sin⁡2A=2⋅513⋅1213=120169 \sin 2A = 2 \cdot \frac{5}{13} \cdot \frac{12}{13} = \frac{120}{169} ; cos⁡2A=144−25169=119169 \cos 2A = \frac{144 - 25}{169} = \frac{119}{169} ; tan⁡2A=120119 \tan 2A = \frac{120}{119} .

c.

If cos θ = 4/5, find the values of sin 3θ, cos 3θ and tan 3θ.

Answer

sin θ = 3/5.
sin⁡3θ=3⋅35−4⋅27125=225−108125=117125,cos⁡3θ=4⋅64125−3⋅45=256−300125=−44125 \sin 3\theta = 3 \cdot \frac{3}{5} - 4 \cdot \frac{27}{125} = \frac{225 - 108}{125} = \frac{117}{125}, \qquad \cos 3\theta = 4 \cdot \frac{64}{125} - 3 \cdot \frac{4}{5} = \frac{256 - 300}{125} = -\frac{44}{125}
tan 3θ = −117/44.

d.

If tan y = 1/2, find the value of tan 3y.

Answer

tan⁡3y=3⋅12−181−3⋅14=11/81/4=112 \tan 3y = \frac{3 \cdot \frac{1}{2} - \frac{1}{8}}{1 - 3 \cdot \frac{1}{4}} = \frac{11/8}{1/4} = \frac{11}{2}

Question 3

a.

If cos 2A = 7/25, find the value of sin A.

Answer

sin⁡2A=(1−cos⁡2A)/2=(18/25)/2=9/25 \sin^2 A = (1 - \cos 2A)/2 = (18/25)/2 = 9/25 , so sin A = 3/5 (A acute).

b.

If cos 2A = 13/36, find the value of cos A.

Answer

cos⁡2A=(1+cos⁡2A)/2=(49/36)/2=49/72 \cos^2 A = (1 + \cos 2A)/2 = (49/36)/2 = 49/72 , so cos⁡A=7/72=7/(62)=72/12 \cos A = 7/\sqrt{72} = 7/(6\sqrt{2}) = 7\sqrt{2}/12 .

c.

If tan 2A = 120/119, find the value of tan A.

Answer

With t =  tan A: 2t/(1−t2)=120/119 2t/(1 - t^2) = 120/119 , so 238t=120−120t2 238t = 120 - 120t^2 , 60t2+119t−60=0 60t^2 + 119t - 60 = 0 .
t=−119±1192+4⋅60⋅60120=−119±169120 t = \frac{-119 \pm \sqrt{119^2 + 4 \cdot 60 \cdot 60}}{120} = \frac{-119 \pm 169}{120}
tan A = 5/12 or −12/5. Both are right: 2A could be about 45.2° or that plus 180°.

Question 4

a.

1+cos⁡2Asin⁡2A=cot⁡A \dfrac{1 + \cos 2A}{\sin 2A} = \cot A

Answer

2cos⁡2A2sin⁡Acos⁡A=cos⁡Asin⁡A=cot⁡A \dfrac{2\cos^2 A}{2\sin A \cos A} = \dfrac{\cos A}{\sin A} = \cot A .

b.

1−cos⁡2Asin⁡2A=tan⁡A \dfrac{1 - \cos 2A}{\sin 2A} = \tan A

Answer

2sin⁡2A2sin⁡Acos⁡A=tan⁡A \dfrac{2\sin^2 A}{2\sin A \cos A} = \tan A .

c.

1−cos⁡2A1+cos⁡2A=tan⁡2A \dfrac{1 - \cos 2A}{1 + \cos 2A} = \tan^2 A

Answer

2sin⁡2A2cos⁡2A=tan⁡2A \dfrac{2\sin^2 A}{2\cos^2 A} = \tan^2 A .

d.

1−tan⁡θ1+tan⁡θ=1−sin⁡2θcos⁡2θ \dfrac{1 - \tan\theta}{1 + \tan\theta} = \dfrac{1 - \sin 2\theta}{\cos 2\theta}

Answer

1−sin⁡2θcos⁡2θ=(cos⁡θ−sin⁡θ)2(cos⁡θ−sin⁡θ)(cos⁡θ+sin⁡θ)=cos⁡θ−sin⁡θcos⁡θ+sin⁡θ=1−tan⁡θ1+tan⁡θ \frac{1 - \sin 2\theta}{\cos 2\theta} = \frac{(\cos\theta - \sin\theta)^2}{(\cos\theta - \sin\theta)(\cos\theta + \sin\theta)} = \frac{\cos\theta - \sin\theta}{\cos\theta + \sin\theta} = \frac{1 - \tan\theta}{1 + \tan\theta}
using 1−sin⁡2θ=cos⁡2θ+sin⁡2θ−2sin⁡θcos⁡θ 1 - \sin 2\theta = \cos^2\theta + \sin^2\theta - 2\sin\theta\cos\theta .

e.

cos⁡2α1+sin⁡2α=1−tan⁡α1+tan⁡α \dfrac{\cos 2\alpha}{1 + \sin 2\alpha} = \dfrac{1 - \tan\alpha}{1 + \tan\alpha}

Answer

(cos⁡α−sin⁡α)(cos⁡α+sin⁡α)(cos⁡α+sin⁡α)2=cos⁡α−sin⁡αcos⁡α+sin⁡α=1−tan⁡α1+tan⁡α \frac{(\cos\alpha - \sin\alpha)(\cos\alpha + \sin\alpha)}{(\cos\alpha + \sin\alpha)^2} = \frac{\cos\alpha - \sin\alpha}{\cos\alpha + \sin\alpha} = \frac{1 - \tan\alpha}{1 + \tan\alpha}

f.

1+sin⁡2θ−cos⁡2θ1+sin⁡2θ+cos⁡2θ=tan⁡θ \dfrac{1 + \sin 2\theta - \cos 2\theta}{1 + \sin 2\theta + \cos 2\theta} = \tan\theta

Answer

2sin⁡2θ+2sin⁡θcos⁡θ2cos⁡2θ+2sin⁡θcos⁡θ=2sin⁡θ(sin⁡θ+cos⁡θ)2cos⁡θ(cos⁡θ+sin⁡θ)=tan⁡θ \frac{2\sin^2\theta + 2\sin\theta\cos\theta}{2\cos^2\theta + 2\sin\theta\cos\theta} = \frac{2\sin\theta(\sin\theta + \cos\theta)}{2\cos\theta(\cos\theta + \sin\theta)} = \tan\theta

g.

sin⁡θ+sin⁡2θ1+cos⁡θ+cos⁡2θ=tan⁡θ \dfrac{\sin\theta + \sin 2\theta}{1 + \cos\theta + \cos 2\theta} = \tan\theta

Answer

It is the upside-down of Example 5: sin⁡θ(1+2cos⁡θ)cos⁡θ(1+2cos⁡θ)=tan⁡θ \dfrac{\sin\theta(1 + 2\cos\theta)}{\cos\theta(1 + 2\cos\theta)} = \tan\theta .

h.

tan⁡A+cot⁡A=2cosec⁡2A \tan A + \cot A = 2\operatorname{cosec} 2A

Answer

sin⁡Acos⁡A+cos⁡Asin⁡A=sin⁡2A+cos⁡2Asin⁡Acos⁡A=22sin⁡Acos⁡A=2sin⁡2A=2cosec⁡2A \dfrac{\sin A}{\cos A} + \dfrac{\cos A}{\sin A} = \dfrac{\sin^2 A + \cos^2 A}{\sin A \cos A} = \dfrac{2}{2\sin A \cos A} = \dfrac{2}{\sin 2A} = 2\operatorname{cosec} 2A .

i.

cot⁡A−tan⁡A=2cot⁡2A \cot A - \tan A = 2\cot 2A

Answer

cos⁡2A−sin⁡2Asin⁡Acos⁡A=2cos⁡2Asin⁡2A=2cot⁡2A \dfrac{\cos^2 A - \sin^2 A}{\sin A \cos A} = \dfrac{2\cos 2A}{\sin 2A} = 2\cot 2A .

Question 5

a.

tan⁡(45°+A)=cos⁡2A1−sin⁡2A \tan(45° + A) = \dfrac{\cos 2A}{1 - \sin 2A}

Answer

cos⁡2A1−sin⁡2A=(cos⁡A+sin⁡A)(cos⁡A−sin⁡A)(cos⁡A−sin⁡A)2=cos⁡A+sin⁡Acos⁡A−sin⁡A=1+tan⁡A1−tan⁡A=tan⁡(45°+A) \frac{\cos 2A}{1 - \sin 2A} = \frac{(\cos A + \sin A)(\cos A - \sin A)}{(\cos A - \sin A)^2} = \frac{\cos A + \sin A}{\cos A - \sin A} = \frac{1 + \tan A}{1 - \tan A} = \tan(45° + A)

b.

1−sin⁡2θcos⁡2θ=cos⁡θ−sin⁡θcos⁡θ+sin⁡θ \dfrac{1 - \sin 2\theta}{\cos 2\theta} = \dfrac{\cos\theta - \sin\theta}{\cos\theta + \sin\theta}

Answer

This is the middle step of Question 4 d.

c.

2sin⁡2(45°−A)=1−sin⁡2A 2\sin^2(45° - A) = 1 - \sin 2A

Answer

2sin⁡2x=1−cos⁡2x 2\sin^2 x = 1 - \cos 2x with x=45°−A x = 45° - A : 1−cos⁡(90°−2A)=1−sin⁡2A 1 - \cos(90° - 2A) = 1 - \sin 2A .

d.

tan⁡(45°+θ)=sec⁡2θ+tan⁡2θ \tan(45° + \theta) = \sec 2\theta + \tan 2\theta

Answer

sec⁡2θ+tan⁡2θ=1+sin⁡2θcos⁡2θ=(cos⁡θ+sin⁡θ)2(cos⁡θ+sin⁡θ)(cos⁡θ−sin⁡θ)=1+tan⁡θ1−tan⁡θ=tan⁡(45°+θ) \sec 2\theta + \tan 2\theta = \frac{1 + \sin 2\theta}{\cos 2\theta} = \frac{(\cos\theta + \sin\theta)^2}{(\cos\theta + \sin\theta)(\cos\theta - \sin\theta)} = \frac{1 + \tan\theta}{1 - \tan\theta} = \tan(45° + \theta)

e.

2cos⁡2(45°−A)=1+sin⁡2A 2\cos^2(45° - A) = 1 + \sin 2A

Answer

2cos⁡2x=1+cos⁡2x 2\cos^2 x = 1 + \cos 2x with x=45°−A x = 45° - A : 1+cos⁡(90°−2A)=1+sin⁡2A 1 + \cos(90° - 2A) = 1 + \sin 2A .

f.

cos⁡2(45°−θ)−sin⁡2(45°−θ)=sin⁡2θ \cos^2(45° - \theta) - \sin^2(45° - \theta) = \sin 2\theta

Answer

It is cos⁡2(45°−θ)=cos⁡(90°−2θ)=sin⁡2θ \cos 2(45° - \theta) = \cos(90° - 2\theta) = \sin 2\theta .

g.

sin⁡5Asin⁡A−cos⁡5Acos⁡A=4cos⁡2A \dfrac{\sin 5A}{\sin A} - \dfrac{\cos 5A}{\cos A} = 4\cos 2A

Answer

sin⁡5Acos⁡A−cos⁡5Asin⁡Asin⁡Acos⁡A=sin⁡4A12sin⁡2A=2⋅2sin⁡2Acos⁡2Asin⁡2A=4cos⁡2A \frac{\sin 5A \cos A - \cos 5A \sin A}{\sin A \cos A} = \frac{\sin 4A}{\frac{1}{2}\sin 2A} = \frac{2 \cdot 2\sin 2A \cos 2A}{\sin 2A} = 4\cos 2A

h.

1−tan⁡2(45°−A)1+tan⁡2(45°−A)=sin⁡2A \dfrac{1 - \tan^2(45° - A)}{1 + \tan^2(45° - A)} = \sin 2A

Answer

It is cos⁡2(45°−A)=cos⁡(90°−2A)=sin⁡2A \cos 2(45° - A) = \cos(90° - 2A) = \sin 2A .

Question 6

a.

If cos⁡θ=12(a+1a) \cos\theta = \frac{1}{2}\left(a + \frac{1}{a}\right) , prove that cos⁡2θ=12(a2+1a2) \cos 2\theta = \frac{1}{2}\left(a^2 + \frac{1}{a^2}\right) .

Answer

cos⁡2θ=2cos⁡2θ−1=2⋅14(a2+2+1a2)−1=12(a2+1a2) \cos 2\theta = 2\cos^2\theta - 1 = 2 \cdot \frac{1}{4}\left(a^2 + 2 + \frac{1}{a^2}\right) - 1 = \frac{1}{2}\left(a^2 + \frac{1}{a^2}\right)
(For a real angle, 12(a+1/a) \frac{1}{2}(a + 1/a) can only be a cosine when a = 1 or −1; the algebra works for any a.)

b.

If sin⁡A=12(a+1a) \sin A = \frac{1}{2}\left(a + \frac{1}{a}\right) , prove that cos⁡2A=−12(a2+1a2) \cos 2A = -\frac{1}{2}\left(a^2 + \frac{1}{a^2}\right) .

Answer

cos⁡2A=1−2sin⁡2A=1−12(a2+2+1a2)=−12(a2+1a2) \cos 2A = 1 - 2\sin^2 A = 1 - \frac{1}{2}\left(a^2 + 2 + \frac{1}{a^2}\right) = -\frac{1}{2}\left(a^2 + \frac{1}{a^2}\right)

c.

If sin⁡β=12(a+1a) \sin\beta = \frac{1}{2}\left(a + \frac{1}{a}\right) , prove that sin⁡3β=−12(a3+1a3) \sin 3\beta = -\frac{1}{2}\left(a^3 + \frac{1}{a^3}\right) .

Answer

sin⁡3β=3sin⁡β−4sin⁡3β=32(a+1a)−12(a+1a)3=32(a+1a)−12(a3+3a+3a+1a3)=−12(a3+1a3) \begin{aligned} \sin 3\beta &= 3\sin\beta - 4\sin^3\beta = \frac{3}{2}\left(a + \frac{1}{a}\right) - \frac{1}{2}\left(a + \frac{1}{a}\right)^3 \\ &= \frac{3}{2}\left(a + \frac{1}{a}\right) - \frac{1}{2}\left(a^3 + 3a + \frac{3}{a} + \frac{1}{a^3}\right) = -\frac{1}{2}\left(a^3 + \frac{1}{a^3}\right) \end{aligned}

d.

If cos⁡β=12(a+1a) \cos\beta = \frac{1}{2}\left(a + \frac{1}{a}\right) , prove that cos⁡3β=12(a3+1a3) \cos 3\beta = \frac{1}{2}\left(a^3 + \frac{1}{a^3}\right) .

Answer

cos⁡3β=4cos⁡3β−3cos⁡β=12(a+1a)3−32(a+1a)=12(a3+1a3) \cos 3\beta = 4\cos^3\beta - 3\cos\beta = \frac{1}{2}\left(a + \frac{1}{a}\right)^3 - \frac{3}{2}\left(a + \frac{1}{a}\right) = \frac{1}{2}\left(a^3 + \frac{1}{a^3}\right)

Question 7

a.

(sin⁡2A+sin⁡2B)2+(cos⁡2A−cos⁡2B)2=4sin⁡2(A+B) (\sin 2A + \sin 2B)^2 + (\cos 2A - \cos 2B)^2 = 4\sin^2(A + B)

Answer

Expanding, sin⁡2+cos⁡2 \sin^2 + \cos^2 twice gives 2:
2+2(sin⁡2Asin⁡2B−cos⁡2Acos⁡2B)=2−2cos⁡(2A+2B)=2⋅2sin⁡2(A+B) 2 + 2(\sin 2A \sin 2B - \cos 2A \cos 2B) = 2 - 2\cos(2A + 2B) = 2 \cdot 2\sin^2(A + B)

b.

(sin⁡2A−sin⁡2B)2+(cos⁡2A+cos⁡2B)2=4cos⁡2(A+B) (\sin 2A - \sin 2B)^2 + (\cos 2A + \cos 2B)^2 = 4\cos^2(A + B)

Answer

2+2(cos⁡2Acos⁡2B−sin⁡2Asin⁡2B)=2+2cos⁡(2A+2B)=4cos⁡2(A+B) 2 + 2(\cos 2A \cos 2B - \sin 2A \sin 2B) = 2 + 2\cos(2A + 2B) = 4\cos^2(A + B)

c.

(1+sin⁡2A+cos⁡2A)2=4cos⁡2A(1+sin⁡2A) (1 + \sin 2A + \cos 2A)^2 = 4\cos^2 A(1 + \sin 2A)

Answer

1+cos⁡2A+sin⁡2A=2cos⁡2A+2sin⁡Acos⁡A=2cos⁡A(cos⁡A+sin⁡A) 1 + \cos 2A + \sin 2A = 2\cos^2 A + 2\sin A \cos A = 2\cos A(\cos A + \sin A) . Squared: 4cos⁡2A(cos⁡A+sin⁡A)2=4cos⁡2A(1+sin⁡2A) 4\cos^2 A(\cos A + \sin A)^2 = 4\cos^2 A(1 + \sin 2A) .

Question 8

a.

sin⁡2A−cos⁡2Acos⁡2B=sin⁡2B−cos⁡2Bcos⁡2A \sin^2 A - \cos^2 A \cos 2B = \sin^2 B - \cos^2 B \cos 2A

Answer

LHS=sin⁡2A−cos⁡2A(1−2sin⁡2B)=sin⁡2A−cos⁡2A+2cos⁡2Asin⁡2B \text{LHS} = \sin^2 A - \cos^2 A(1 - 2\sin^2 B) = \sin^2 A - \cos^2 A + 2\cos^2 A \sin^2 B . Write sin⁡2A=1−cos⁡2A \sin^2 A = 1 - \cos^2 A : LHS=1−2cos⁡2A+2cos⁡2Asin⁡2B=1−2cos⁡2Acos⁡2B \text{LHS} = 1 - 2\cos^2 A + 2\cos^2 A\sin^2 B = 1 - 2\cos^2 A \cos^2 B .
In the same way RHS=1−2cos⁡2Bcos⁡2A \text{RHS} = 1 - 2\cos^2 B \cos^2 A . Equal.

b.

cos⁡2α+sin⁡2αcos⁡2β=cos⁡2β+sin⁡2βcos⁡2α \cos^2\alpha + \sin^2\alpha\cos 2\beta = \cos^2\beta + \sin^2\beta\cos 2\alpha

Answer

LHS=cos⁡2α+sin⁡2α(1−2sin⁡2β)=1−2sin⁡2αsin⁡2β \text{LHS} = \cos^2\alpha + \sin^2\alpha(1 - 2\sin^2\beta) = 1 - 2\sin^2\alpha\sin^2\beta , and the right side is the same with α and β swapped, which doesn’t change it.

Question 9

a.

(2 cos θ + 1)(2 cos θ − 1) = 2 cos 2θ + 1

Answer

4cos⁡2θ−1=2(2cos⁡2θ−1)+1=2cos⁡2θ+1 4\cos^2\theta - 1 = 2(2\cos^2\theta - 1) + 1 = 2\cos 2\theta + 1 .

b.

(2 cos θ + 1)(2 cos θ − 1)(2 cos 2θ − 1) = 2 cos 4θ + 1

Answer

By a, the first two give 2 cos 2θ + 1; then a again, with 2θ for θ: (2 cos 2θ + 1)(2 cos 2θ − 1) = 2 cos 4θ + 1.

Question 10

a.

cos⁡3A+sin⁡3Acos⁡A+sin⁡A=1−12sin⁡2A \dfrac{\cos^3 A + \sin^3 A}{\cos A + \sin A} = 1 - \dfrac{1}{2}\sin 2A

Answer

p3+q3=(p+q)(p2−pq+q2) p^3 + q^3 = (p + q)(p^2 - pq + q^2) : the quotient is cos⁡2A−sin⁡Acos⁡A+sin⁡2A=1−12sin⁡2A \cos^2 A - \sin A \cos A + \sin^2 A = 1 - \frac{1}{2}\sin 2A .

b.

cos⁡6θ−sin⁡6θ=cos⁡2θ(1−14sin⁡22θ) \cos^6\theta - \sin^6\theta = \cos 2\theta\left(1 - \frac{1}{4}\sin^2 2\theta\right)

Answer

p3−q3=(p−q)(p2+pq+q2) p^3 - q^3 = (p - q)(p^2 + pq + q^2) with p=cos⁡2θ p = \cos^2\theta , q=sin⁡2θ q = \sin^2\theta :
(cos⁡2θ−sin⁡2θ){(cos⁡2θ+sin⁡2θ)2−sin⁡2θcos⁡2θ}=cos⁡2θ(1−14sin⁡22θ) (\cos^2\theta - \sin^2\theta)\{(\cos^2\theta + \sin^2\theta)^2 - \sin^2\theta\cos^2\theta\} = \cos 2\theta\left(1 - \tfrac{1}{4}\sin^2 2\theta\right)

c.

4(sin⁡6θ+cos⁡6θ)=4−3sin⁡22θ 4(\sin^6\theta + \cos^6\theta) = 4 - 3\sin^2 2\theta

Answer

From Example 7, sin⁡6θ+cos⁡6θ=1−34sin⁡22θ \sin^6\theta + \cos^6\theta = 1 - \frac{3}{4}\sin^2 2\theta . Times 4.

d.

cos⁡6θ−sin⁡6θ=14(cos⁡32θ+3cos⁡2θ) \cos^6\theta - \sin^6\theta = \frac{1}{4}(\cos^3 2\theta + 3\cos 2\theta)

Answer

From b, with sin⁡22θ=1−cos⁡22θ \sin^2 2\theta = 1 - \cos^2 2\theta : cos⁡2θ(34+14cos⁡22θ)=14(cos⁡32θ+3cos⁡2θ) \cos 2\theta\left(\frac{3}{4} + \frac{1}{4}\cos^2 2\theta\right) = \frac{1}{4}(\cos^3 2\theta + 3\cos 2\theta) .

e.

cos⁡6θ+sin⁡6θ=18(5+3cos⁡4θ) \cos^6\theta + \sin^6\theta = \frac{1}{8}(5 + 3\cos 4\theta)

Answer

This is Example 7, with θ for ϕ \phi .

f.

sin⁡8θ+cos⁡8θ=18(8−8sin⁡22θ+sin⁡42θ) \sin^8\theta + \cos^8\theta = \frac{1}{8}(8 - 8\sin^2 2\theta + \sin^4 2\theta)

Answer

sin⁡4θ+cos⁡4θ=(sin⁡2θ+cos⁡2θ)2−2sin⁡2θcos⁡2θ=1−12sin⁡22θ \sin^4\theta + \cos^4\theta = (\sin^2\theta + \cos^2\theta)^2 - 2\sin^2\theta\cos^2\theta = 1 - \frac{1}{2}\sin^2 2\theta . Then
sin⁡8θ+cos⁡8θ=(sin⁡4θ+cos⁡4θ)2−2sin⁡4θcos⁡4θ=(1−12sin⁡22θ)2−18sin⁡42θ \sin^8\theta + \cos^8\theta = (\sin^4\theta + \cos^4\theta)^2 - 2\sin^4\theta\cos^4\theta = \left(1 - \tfrac{1}{2}\sin^2 2\theta\right)^2 - \tfrac{1}{8}\sin^4 2\theta
=1−sin⁡22θ+14sin⁡42θ−18sin⁡42θ=18(8−8sin⁡22θ+sin⁡42θ) = 1 - \sin^2 2\theta + \tfrac{1}{4}\sin^4 2\theta - \tfrac{1}{8}\sin^4 2\theta = \tfrac{1}{8}(8 - 8\sin^2 2\theta + \sin^4 2\theta)

Question 11

a.

cosec⁡10°−3sec⁡10°=4 \operatorname{cosec} 10° - \sqrt{3}\sec 10° = 4

Answer

cos⁡10°−3sin⁡10°sin⁡10°cos⁡10°=2(sin⁡30°cos⁡10°−cos⁡30°sin⁡10°)12sin⁡20°=4sin⁡20°sin⁡20°=4 \frac{\cos 10° - \sqrt{3}\sin 10°}{\sin 10° \cos 10°} = \frac{2(\sin 30° \cos 10° - \cos 30° \sin 10°)}{\frac{1}{2}\sin 20°} = \frac{4\sin 20°}{\sin 20°} = 4

b.

3cosec⁡20°−sec⁡20°=4 \sqrt{3}\operatorname{cosec} 20° - \sec 20° = 4

Answer

3cos⁡20°−sin⁡20°sin⁡20°cos⁡20°=2(sin⁡60°cos⁡20°−cos⁡60°sin⁡20°)12sin⁡40°=4sin⁡40°sin⁡40°=4 \frac{\sqrt{3}\cos 20° - \sin 20°}{\sin 20° \cos 20°} = \frac{2(\sin 60° \cos 20° - \cos 60° \sin 20°)}{\frac{1}{2}\sin 40°} = \frac{4\sin 40°}{\sin 40°} = 4

c.

sec⁡40°+3cosec⁡40°=4 \sec 40° + \sqrt{3}\operatorname{cosec} 40° = 4

Answer

sin⁡40°+3cos⁡40°sin⁡40°cos⁡40°=2(cos⁡60°sin⁡40°+sin⁡60°cos⁡40°)12sin⁡80°=4sin⁡100°sin⁡80°=4 \frac{\sin 40° + \sqrt{3}\cos 40°}{\sin 40° \cos 40°} = \frac{2(\cos 60° \sin 40° + \sin 60° \cos 40°)}{\frac{1}{2}\sin 80°} = \frac{4\sin 100°}{\sin 80°} = 4
since sin⁡100°=sin⁡80° \sin 100° = \sin 80° .

Question 12

a.

cosec⁡2A+cot⁡4A=cot⁡A−cosec⁡4A \operatorname{cosec} 2A + \cot 4A = \cot A - \operatorname{cosec} 4A

Answer

First, cosec⁡x+cot⁡x=1+cos⁡xsin⁡x=2cos⁡2(x/2)2sin⁡(x/2)cos⁡(x/2)=cot⁡x2 \operatorname{cosec} x + \cot x = \dfrac{1 + \cos x}{\sin x} = \dfrac{2\cos^2(x/2)}{2\sin(x/2)\cos(x/2)} = \cot\dfrac{x}{2} . So
cosec⁡4A+cot⁡4A=cot⁡2A,cosec⁡2A+cot⁡2A=cot⁡A \operatorname{cosec} 4A + \cot 4A = \cot 2A, \qquad \operatorname{cosec} 2A + \cot 2A = \cot A
Add cosec⁡4A \operatorname{cosec} 4A to the left side of the question: cosec⁡2A+(cot⁡4A+cosec⁡4A)=cosec⁡2A+cot⁡2A=cot⁡A \operatorname{cosec} 2A + (\cot 4A + \operatorname{cosec} 4A) = \operatorname{cosec} 2A + \cot 2A = \cot A . Take cosec⁡4A \operatorname{cosec} 4A away again.

b.

cot⁡8A+cosec⁡4A=cot⁡2A−cosec⁡8A \cot 8A + \operatorname{cosec} 4A = \cot 2A - \operatorname{cosec} 8A

Answer

In the same way: (cot⁡8A+cosec⁡8A)+cosec⁡4A=cot⁡4A+cosec⁡4A=cot⁡2A (\cot 8A + \operatorname{cosec} 8A) + \operatorname{cosec} 4A = \cot 4A + \operatorname{cosec} 4A = \cot 2A .

c.

sec⁡4A−1sec⁡2A−1=tan⁡4Acot⁡A \dfrac{\sec 4A - 1}{\sec 2A - 1} = \tan 4A \cot A

Answer

sec⁡2x−1=1−cos⁡2xcos⁡2x=2sin⁡2xcos⁡2x=tan⁡2xtan⁡x \sec 2x - 1 = \dfrac{1 - \cos 2x}{\cos 2x} = \dfrac{2\sin^2 x}{\cos 2x} = \tan 2x \tan x . So the left side is tan⁡4Atan⁡2Atan⁡2Atan⁡A=tan⁡4Acot⁡A \dfrac{\tan 4A \tan 2A}{\tan 2A \tan A} = \tan 4A \cot A .

d.

sec⁡8A−1sec⁡4A−1=tan⁡8Acot⁡2A \dfrac{\sec 8A - 1}{\sec 4A - 1} = \tan 8A \cot 2A

Answer

The same: tan⁡8Atan⁡4Atan⁡4Atan⁡2A=tan⁡8Acot⁡2A \dfrac{\tan 8A \tan 4A}{\tan 4A \tan 2A} = \tan 8A \cot 2A .

Question 13

a.

1tan⁡3θ+tan⁡θ−1cot⁡3θ+cot⁡θ=cot⁡4θ \dfrac{1}{\tan 3\theta + \tan\theta} - \dfrac{1}{\cot 3\theta + \cot\theta} = \cot 4\theta

Answer

tan⁡3θ+tan⁡θ=sin⁡4θcos⁡3θcos⁡θ \tan 3\theta + \tan\theta = \dfrac{\sin 4\theta}{\cos 3\theta\cos\theta} and cot⁡3θ+cot⁡θ=sin⁡4θsin⁡3θsin⁡θ \cot 3\theta + \cot\theta = \dfrac{\sin 4\theta}{\sin 3\theta\sin\theta} (add the fractions; the top is a compound sine). So
LHS=cos⁡3θcos⁡θ−sin⁡3θsin⁡θsin⁡4θ=cos⁡4θsin⁡4θ=cot⁡4θ \text{LHS} = \frac{\cos 3\theta\cos\theta - \sin 3\theta\sin\theta}{\sin 4\theta} = \frac{\cos 4\theta}{\sin 4\theta} = \cot 4\theta

b.

cot⁡θcot⁡θ−cot⁡3θ−tan⁡θtan⁡3θ−tan⁡θ=1 \dfrac{\cot\theta}{\cot\theta - \cot 3\theta} - \dfrac{\tan\theta}{\tan 3\theta - \tan\theta} = 1

Answer

cot⁡θ−cot⁡3θ=sin⁡2θsin⁡θsin⁡3θ \cot\theta - \cot 3\theta = \dfrac{\sin 2\theta}{\sin\theta\sin 3\theta} and tan⁡3θ−tan⁡θ=sin⁡2θcos⁡3θcos⁡θ \tan 3\theta - \tan\theta = \dfrac{\sin 2\theta}{\cos 3\theta\cos\theta} . So
LHS=cos⁡θsin⁡3θsin⁡2θ−sin⁡θcos⁡3θsin⁡2θ=sin⁡(3θ−θ)sin⁡2θ=1 \text{LHS} = \frac{\cos\theta\sin 3\theta}{\sin 2\theta} - \frac{\sin\theta\cos 3\theta}{\sin 2\theta} = \frac{\sin(3\theta - \theta)}{\sin 2\theta} = 1

Question 14

a.

4(cos⁡310°+sin⁡320°)=3(cos⁡10°+sin⁡20°) 4(\cos^3 10° + \sin^3 20°) = 3(\cos 10° + \sin 20°)

Answer

From the triple angles, 4cos⁡3x=cos⁡3x+3cos⁡x 4\cos^3 x = \cos 3x + 3\cos x and 4sin⁡3x=3sin⁡x−sin⁡3x 4\sin^3 x = 3\sin x - \sin 3x . So
LHS=cos⁡30°+3cos⁡10°+3sin⁡20°−sin⁡60°=3(cos⁡10°+sin⁡20°) \text{LHS} = \cos 30° + 3\cos 10° + 3\sin 20° - \sin 60° = 3(\cos 10° + \sin 20°)
since cos⁡30°=sin⁡60° \cos 30° = \sin 60° .

b.

4(cos⁡320°+sin⁡350°)=3(cos⁡20°+sin⁡50°) 4(\cos^3 20° + \sin^3 50°) = 3(\cos 20° + \sin 50°)

Answer

LHS=cos⁡60°+3cos⁡20°+3sin⁡50°−sin⁡150°=3(cos⁡20°+sin⁡50°) \text{LHS} = \cos 60° + 3\cos 20° + 3\sin 50° - \sin 150° = 3(\cos 20° + \sin 50°) , since cos⁡60°=sin⁡150°=12 \cos 60° = \sin 150° = \frac{1}{2} .

c.

cos⁡3Acos⁡3A+sin⁡3Asin⁡3A=cos⁡32A \cos^3 A \cos 3A + \sin^3 A \sin 3A = \cos^3 2A

Answer

Put cos⁡3A=14(cos⁡3A+3cos⁡A) \cos^3 A = \frac{1}{4}(\cos 3A + 3\cos A) and sin⁡3A=14(3sin⁡A−sin⁡3A) \sin^3 A = \frac{1}{4}(3\sin A - \sin 3A) :
LHS=14{cos⁡23A+3cos⁡3Acos⁡A+3sin⁡3Asin⁡A−sin⁡23A}=14{cos⁡6A+3cos⁡2A}=14{4cos⁡32A−3cos⁡2A+3cos⁡2A}=cos⁡32A \begin{aligned} \text{LHS} &= \tfrac{1}{4}\{\cos^2 3A + 3\cos 3A\cos A + 3\sin 3A\sin A - \sin^2 3A\} \\ &= \tfrac{1}{4}\{\cos 6A + 3\cos 2A\} = \tfrac{1}{4}\{4\cos^3 2A - 3\cos 2A + 3\cos 2A\} = \cos^3 2A \end{aligned}

Question 15

a.

tan⁡θ+tan⁡(60°+θ)−tan⁡(60°−θ)=3tan⁡3θ \tan\theta + \tan(60° + \theta) - \tan(60° - \theta) = 3\tan 3\theta

Answer

With t =  tan θ and tan⁡60°=3 \tan 60° = \sqrt{3} :
tan⁡(60°+θ)−tan⁡(60°−θ)=3+t1−3t−3−t1+3t=8t1−3t2 \tan(60° + \theta) - \tan(60° - \theta) = \frac{\sqrt{3} + t}{1 - \sqrt{3}t} - \frac{\sqrt{3} - t}{1 + \sqrt{3}t} = \frac{8t}{1 - 3t^2}
t+8t1−3t2=9t−3t31−3t2=3⋅3t−t31−3t2=3tan⁡3θ t + \frac{8t}{1 - 3t^2} = \frac{9t - 3t^3}{1 - 3t^2} = 3 \cdot \frac{3t - t^3}{1 - 3t^2} = 3\tan 3\theta

b.

cot⁡θ+cot⁡(60°+θ)−cot⁡(60°−θ)=3cot⁡3θ \cot\theta + \cot(60° + \theta) - \cot(60° - \theta) = 3\cot 3\theta

Answer

With c=cot⁡θ c = \cot\theta and cot⁡60°=1/3 \cot 60° = 1/\sqrt{3} , cot⁡(60°±θ)=c∓33c±1 \cot(60° \pm \theta) = \dfrac{c \mp \sqrt{3}}{\sqrt{3}c \pm 1} (from cot⁡(A±B) \cot(A \pm B) , multiplied through by 3 \sqrt{3} ). Then
cot⁡(60°+θ)−cot⁡(60°−θ)=c−33c+1−c+33c−1=−8c3c2−1 \cot(60° + \theta) - \cot(60° - \theta) = \frac{c - \sqrt{3}}{\sqrt{3}c + 1} - \frac{c + \sqrt{3}}{\sqrt{3}c - 1} = \frac{-8c}{3c^2 - 1}
c−8c3c2−1=3c3−9c3c2−1=3⋅c3−3c3c2−1=3cot⁡3θ c - \frac{8c}{3c^2 - 1} = \frac{3c^3 - 9c}{3c^2 - 1} = 3 \cdot \frac{c^3 - 3c}{3c^2 - 1} = 3\cot 3\theta
by Example 3.

c.

tan⁡θ+2tan⁡2θ+4tan⁡4θ+8cot⁡8θ=cot⁡θ \tan\theta + 2\tan 2\theta + 4\tan 4\theta + 8\cot 8\theta = \cot\theta

Answer

From Question 4 i, tan⁡x=cot⁡x−2cot⁡2x \tan x = \cot x - 2\cot 2x . So
tan⁡θ+2tan⁡2θ+4tan⁡4θ=(cot⁡θ−2cot⁡2θ)+(2cot⁡2θ−4cot⁡4θ)+(4cot⁡4θ−8cot⁡8θ)=cot⁡θ−8cot⁡8θ \tan\theta + 2\tan 2\theta + 4\tan 4\theta = (\cot\theta - 2\cot 2\theta) + (2\cot 2\theta - 4\cot 4\theta) + (4\cot 4\theta - 8\cot 8\theta) = \cot\theta - 8\cot 8\theta
Add 8cot⁡8θ 8\cot 8\theta : cot⁡θ \cot\theta . (Each pair cancels the next, like a folding telescope.)

Question 16

a.

Express cos 4θ in terms of sin θ.

Answer

cos⁡4θ=1−2sin⁡22θ=1−8sin⁡2θcos⁡2θ=1−8sin⁡2θ(1−sin⁡2θ) \cos 4\theta = 1 - 2\sin^2 2\theta = 1 - 8\sin^2\theta\cos^2\theta = 1 - 8\sin^2\theta(1 - \sin^2\theta)
cos⁡4θ=1−8sin⁡2θ+8sin⁡4θ \cos 4\theta = 1 - 8\sin^2\theta + 8\sin^4\theta . Check with θ=30° \theta = 30° : 1−2+12=−12=cos⁡120° 1 - 2 + \frac{1}{2} = -\frac{1}{2} = \cos 120° ✓.

b.

Express cos 5θ in terms of cos θ.

Answer

With c =  cos θ: cos 5θ =  cos (3θ + 2θ) =  cos 3θ cos 2θ −  sin 3θ sin 2θ.
cos⁡3θcos⁡2θ=(4c3−3c)(2c2−1)=8c5−10c3+3c \cos 3\theta\cos 2\theta = (4c^3 - 3c)(2c^2 - 1) = 8c^5 - 10c^3 + 3c
sin⁡3θsin⁡2θ=(3−4sin⁡2θ)sin⁡θ⋅2sin⁡θcos⁡θ=2c(1−c2)(4c2−1)=−8c5+10c3−2c \sin 3\theta\sin 2\theta = (3 - 4\sin^2\theta)\sin\theta \cdot 2\sin\theta\cos\theta = 2c(1 - c^2)(4c^2 - 1) = -8c^5 + 10c^3 - 2c
cos⁡5θ=16c5−20c3+5c \cos 5\theta = 16c^5 - 20c^3 + 5c . Check with θ = 0: 16 − 20 + 5 = 1 ✓.

Question 17

a.

cos⁡Acos⁡A−sin⁡A−cos⁡Acos⁡A+sin⁡A=tan⁡2A \dfrac{\cos A}{\cos A - \sin A} - \dfrac{\cos A}{\cos A + \sin A} = \tan 2A

Answer

cos⁡A(cos⁡A+sin⁡A)−cos⁡A(cos⁡A−sin⁡A)cos⁡2A−sin⁡2A=2sin⁡Acos⁡Acos⁡2A=tan⁡2A \begin{aligned} &\frac{\cos A(\cos A + \sin A) - \cos A(\cos A - \sin A)}{\cos^2 A - \sin^2 A} \\ &= \frac{2\sin A \cos A}{\cos 2A} = \tan 2A \end{aligned}

b.

tan⁡2θ+sec⁡2θ=cos⁡θ+sin⁡θcos⁡θ−sin⁡θ \tan 2\theta + \sec 2\theta = \dfrac{\cos\theta + \sin\theta}{\cos\theta - \sin\theta}

Answer

This is Question 5 d, read from the other end.

c.

cot⁡2θ+tan⁡2θcot⁡2θ−tan⁡2θ=sec⁡4θ \dfrac{\cot 2\theta + \tan 2\theta}{\cot 2\theta - \tan 2\theta} = \sec 4\theta

Answer

Multiply the top and bottom by sin 2θ cos 2θ: cos⁡22θ+sin⁡22θcos⁡22θ−sin⁡22θ=1cos⁡4θ=sec⁡4θ \dfrac{\cos^2 2\theta + \sin^2 2\theta}{\cos^2 2\theta - \sin^2 2\theta} = \dfrac{1}{\cos 4\theta} = \sec 4\theta .

d.

cos⁡2A1+sin⁡2A=tan⁡(45°−A)=1−tan⁡A1+tan⁡A \dfrac{\cos 2A}{1 + \sin 2A} = \tan(45° - A) = \dfrac{1 - \tan A}{1 + \tan A}

Answer

The first equals the last by Question 4 e (with A for α), and (1−tan⁡A)/(1+tan⁡A)=tan⁡(45°−A) (1 - \tan A)/(1 + \tan A) = \tan(45° - A) by tan (A − B) with tan⁡45°=1 \tan 45° = 1 .

Question 18

a.

tan⁡(α+β)=5sin⁡2β5cos⁡2β−1 \tan(\alpha + \beta) = \dfrac{5\sin 2\beta}{5\cos 2\beta - 1}

Answer

Let t =  tan β; then tan⁡α=32t \tan\alpha = \frac{3}{2}t , and tan⁡(α+β)=52t1−32t2=5t2−3t2 \tan(\alpha + \beta) = \dfrac{\frac{5}{2}t}{1 - \frac{3}{2}t^2} = \dfrac{5t}{2 - 3t^2} . The right side, with sin⁡2β=2t/(1+t2) \sin 2\beta = 2t/(1 + t^2) and cos⁡2β=(1−t2)/(1+t2) \cos 2\beta = (1 - t^2)/(1 + t^2) :
10t5(1−t2)−(1+t2)=10t4−6t2=5t2−3t2 \frac{10t}{5(1 - t^2) - (1 + t^2)} = \frac{10t}{4 - 6t^2} = \frac{5t}{2 - 3t^2}
Equal.

b.

tan⁡(α−β)=sin⁡2β5−cos⁡2β \tan(\alpha - \beta) = \dfrac{\sin 2\beta}{5 - \cos 2\beta}

Answer

tan⁡(α−β)=12t1+32t2=t2+3t2 \tan(\alpha - \beta) = \dfrac{\frac{1}{2}t}{1 + \frac{3}{2}t^2} = \dfrac{t}{2 + 3t^2} , and
2t5(1+t2)−(1−t2)=2t4+6t2=t2+3t2 \frac{2t}{5(1 + t^2) - (1 - t^2)} = \frac{2t}{4 + 6t^2} = \frac{t}{2 + 3t^2}

Question 19

a.

If tan A = b/a, prove that a cos 2A + b sin 2A = a.

Answer

a⋅1−b2/a21+b2/a2+b⋅2b/a1+b2/a2=a(a2−b2)+2ab2a2+b2=a(a2+b2)a2+b2=a a \cdot \frac{1 - b^2/a^2}{1 + b^2/a^2} + b \cdot \frac{2b/a}{1 + b^2/a^2} = \frac{a(a^2 - b^2) + 2ab^2}{a^2 + b^2} = \frac{a(a^2 + b^2)}{a^2 + b^2} = a

Question 20

a.

Prove that sin⁡A−1+sin⁡2Acos⁡A−1+sin⁡2A=cot⁡A \dfrac{\sin A - \sqrt{1 + \sin 2A}}{\cos A - \sqrt{1 + \sin 2A}} = \cot A , where sin A +  cos A > 0.

Answer

1+sin⁡2A=(sin⁡A+cos⁡A)2 1 + \sin 2A = (\sin A + \cos A)^2 , so 1+sin⁡2A=sin⁡A+cos⁡A \sqrt{1 + \sin 2A} = \sin A + \cos A . Then
sin⁡A−sin⁡A−cos⁡Acos⁡A−sin⁡A−cos⁡A=−cos⁡A−sin⁡A=cot⁡A \frac{\sin A - \sin A - \cos A}{\cos A - \sin A - \cos A} = \frac{-\cos A}{-\sin A} = \cot A

Question 21

a.

Prove that sin⁡4A=38−12cos⁡2A+18cos⁡4A \sin^4 A = \dfrac{3}{8} - \dfrac{1}{2}\cos 2A + \dfrac{1}{8}\cos 4A .

Answer

sin⁡4A=(1−cos⁡2A2)2=1−2cos⁡2A+cos⁡22A4=1−2cos⁡2A+12(1+cos⁡4A)4=38−12cos⁡2A+18cos⁡4A \sin^4 A = \left(\frac{1 - \cos 2A}{2}\right)^2 = \frac{1 - 2\cos 2A + \cos^2 2A}{4} = \frac{1 - 2\cos 2A + \frac{1}{2}(1 + \cos 4A)}{4} = \frac{3}{8} - \frac{1}{2}\cos 2A + \frac{1}{8}\cos 4A

Question 22

a.

Prove that sin⁡4π8+sin⁡43π8+sin⁡45π8+sin⁡47π8=32 \sin^4\dfrac{\pi}{8} + \sin^4\dfrac{3\pi}{8} + \sin^4\dfrac{5\pi}{8} + \sin^4\dfrac{7\pi}{8} = \dfrac{3}{2} .

Answer

In degrees the angles are 22½°, 67½°, 112½° and 157½°. sin⁡15712°=sin⁡2212° \sin 157\tfrac{1}{2}° = \sin 22\tfrac{1}{2}° , sin⁡11212°=sin⁡6712°=cos⁡2212° \sin 112\tfrac{1}{2}° = \sin 67\tfrac{1}{2}° = \cos 22\tfrac{1}{2}° . So
LHS=2(sin⁡42212°+cos⁡42212°)=2(1−12sin⁡245°)=2(1−14)=32 \text{LHS} = 2\left(\sin^4 22\tfrac{1}{2}° + \cos^4 22\tfrac{1}{2}°\right) = 2\left(1 - \tfrac{1}{2}\sin^2 45°\right) = 2\left(1 - \tfrac{1}{4}\right) = \tfrac{3}{2}
using sin⁡4x+cos⁡4x=1−12sin⁡22x \sin^4 x + \cos^4 x = 1 - \frac{1}{2}\sin^2 2x (Question 10 f).

Exercise 7.3

Question 2

a.

Define a submultiple angle, with an example.

Answer

An angle got by dividing an angle into 2, 3, 4, … equal parts. For example, 30° and 20° are submultiple angles of 60°: half and a third of it.

b.

Express sin θ in terms of tan (θ/2).

Answer

sin⁡θ=2tan⁡θ21+tan⁡2θ2 \sin\theta = \frac{2\tan\frac{\theta}{2}}{1 + \tan^2\frac{\theta}{2}}

c.

Express cos θ in terms of cos (θ/2).

Answer

cos⁡θ=2cos⁡2(θ/2)−1 \cos\theta = 2\cos^2(\theta/2) - 1 .

d.

Express cos θ in terms of tan (θ/2).

Answer

cos⁡θ=1−tan⁡2θ21+tan⁡2θ2 \cos\theta = \frac{1 - \tan^2\frac{\theta}{2}}{1 + \tan^2\frac{\theta}{2}}

e.

Express sin θ in terms of sin (θ/3).

Answer

sin⁡θ=3sin⁡(θ/3)−4sin⁡3(θ/3) \sin\theta = 3\sin(\theta/3) - 4\sin^3(\theta/3) .

Question 3

a.

If sin (A/2) = 4/5, find sin A, cos A and tan A.

Answer

This is Example 1: 24/25, −7/25, −24/7.

b.

If tan (A/2) = 3/4, find sin A, cos A and tan A.

Answer

sin⁡A=2⋅341+916=3/225/16=2425,cos⁡A=1−9161+916=725,tan⁡A=247 \sin A = \frac{2 \cdot \frac{3}{4}}{1 + \frac{9}{16}} = \frac{3/2}{25/16} = \frac{24}{25}, \qquad \cos A = \frac{1 - \frac{9}{16}}{1 + \frac{9}{16}} = \frac{7}{25}, \qquad \tan A = \frac{24}{7}

c.

If cos (A/2) = 5/13, find sin A, cos A and tan A.

Answer

sin (A/2) = 12/13. sin⁡A=2⋅1213⋅513=120169 \sin A = 2 \cdot \frac{12}{13} \cdot \frac{5}{13} = \frac{120}{169} ; cos⁡A=25−144169=−119169 \cos A = \frac{25 - 144}{169} = -\frac{119}{169} ; tan⁡A=−120119 \tan A = -\frac{120}{119} .

d.

If sin (θ/3) = 1/2, find sin θ, cos θ and tan θ.

Answer

sin⁡θ=3⋅12−4⋅18=1 \sin\theta = 3 \cdot \frac{1}{2} - 4 \cdot \frac{1}{8} = 1 ; cos⁡θ=4(32)3−3⋅32=0 \cos\theta = 4\left(\frac{\sqrt{3}}{2}\right)^3 - 3\cdot\frac{\sqrt{3}}{2} = 0 . So θ=90° \theta = 90° and tan θ is not defined (it would be 1 divided by 0).

e.

If cos (θ/3) = 4/5, find sin θ, cos θ and tan θ.

Answer

sin (θ/3) = 3/5. As in Exercise 7.2 Question 2 c: sin θ = 117/125, cos θ = −44/125, tan θ = −117/44.

f.

If tan (θ/3) = 1/2, find the value of tan θ.

Answer

tan⁡θ=3⋅12−181−34=112 \tan\theta = \frac{3 \cdot \frac{1}{2} - \frac{1}{8}}{1 - \frac{3}{4}} = \frac{11}{2}

Question 4

a.

If cos A = 7/25, find the value of sin (A/2).

Answer

sin⁡(A/2)=(1−7/25)/2=9/25=3/5 \sin(A/2) = \sqrt{(1 - 7/25)/2} = \sqrt{9/25} = 3/5 .

b.

If cos A = 13/36, find the value of cos (A/2).

Answer

cos⁡(A/2)=(1+13/36)/2=49/72=7/(62)=72/12 \cos(A/2) = \sqrt{(1 + 13/36)/2} = \sqrt{49/72} = 7/(6\sqrt{2}) = 7\sqrt{2}/12 .

c.

If tan A = 4/3, find the value of tan (A/2).

Answer

With t =  tan (A/2): 2t/(1−t2)=4/3 2t/(1 - t^2) = 4/3 , so 2t2+3t−2=0 2t^2 + 3t - 2 = 0 , (2t − 1)(t + 2) = 0. t = 1/2 or −2. If A is acute, A/2 is acute too and tan (A/2) = 1/2. (Check: sin A = 4/5, cos A = 3/5, and sin A/(1 +  cos A) = (4/5)/(8/5) = 1/2.)

Question 5

a.

If cos⁡30°=3/2 \cos 30° = \sqrt{3}/2 , prove that: i. sin⁡15°=3−122 \sin 15° = \dfrac{\sqrt{3} - 1}{2\sqrt{2}} ii. cos⁡15°=3+122 \cos 15° = \dfrac{\sqrt{3} + 1}{2\sqrt{2}} iii. tan⁡15°=2−3 \tan 15° = 2 - \sqrt{3} iv. cot⁡15°=2+3 \cot 15° = 2 + \sqrt{3}

Answer
  • i.This is Example 3.
  • ii.cos⁡215°=1+3/22=2+34=(3+1)28 \cos^2 15° = \dfrac{1 + \sqrt{3}/2}{2} = \dfrac{2 + \sqrt{3}}{4} = \dfrac{(\sqrt{3} + 1)^2}{8} , so cos⁡15°=3+122 \cos 15° = \dfrac{\sqrt{3} + 1}{2\sqrt{2}} .
  • iii.tan⁡15°=sin⁡30°1+cos⁡30°=1/21+3/2=12+3=2−3 \tan 15° = \dfrac{\sin 30°}{1 + \cos 30°} = \dfrac{1/2}{1 + \sqrt{3}/2} = \dfrac{1}{2 + \sqrt{3}} = 2 - \sqrt{3} .
  • iv.cot⁡15°=12−3=2+3 \cot 15° = \dfrac{1}{2 - \sqrt{3}} = 2 + \sqrt{3} .
b.

If cos⁡45°=1/2 \cos 45° = 1/\sqrt{2} , prove that: i. cos⁡2212°=122+2 \cos 22\tfrac{1}{2}° = \tfrac{1}{2}\sqrt{2 + \sqrt{2}} ii. sin⁡2212°=122−2 \sin 22\tfrac{1}{2}° = \tfrac{1}{2}\sqrt{2 - \sqrt{2}} iii. tan⁡2212°=3−22 \tan 22\tfrac{1}{2}° = \sqrt{3 - 2\sqrt{2}} iv. cot⁡2212°=2+1 \cot 22\tfrac{1}{2}° = \sqrt{2} + 1

Answer
  • i.cos⁡22212°=1+1/22=2+24 \cos^2 22\tfrac{1}{2}° = \dfrac{1 + 1/\sqrt{2}}{2} = \dfrac{2 + \sqrt{2}}{4} : take the root.
  • ii.sin⁡22212°=2−24 \sin^2 22\tfrac{1}{2}° = \dfrac{2 - \sqrt{2}}{4} : take the root.
  • iii.tan⁡22212°=2−22+2=(2−2)22=6−422=3−22 \tan^2 22\tfrac{1}{2}° = \dfrac{2 - \sqrt{2}}{2 + \sqrt{2}} = \dfrac{(2 - \sqrt{2})^2}{2} = \dfrac{6 - 4\sqrt{2}}{2} = 3 - 2\sqrt{2} . Also 3−22=(2−1)2 3 - 2\sqrt{2} = (\sqrt{2} - 1)^2 , so tan⁡2212°=2−1≈0.4142 \tan 22\tfrac{1}{2}° = \sqrt{2} - 1 \approx 0.4142 .
  • iv.cot⁡2212°=12−1=2+1 \cot 22\tfrac{1}{2}° = \dfrac{1}{\sqrt{2} - 1} = \sqrt{2} + 1 .

Question 6

a.

If cos⁡θ2=12(a+1a) \cos\frac{\theta}{2} = \frac{1}{2}\left(a + \frac{1}{a}\right) , then cos⁡θ=12(a2+1a2) \cos\theta = \frac{1}{2}\left(a^2 + \frac{1}{a^2}\right) .

Answer

Exercise 7.2 Question 6 a with θ/2 for θ: cos⁡θ=2cos⁡2θ2−1=12(a2+2+1a2)−1 \cos\theta = 2\cos^2\frac{\theta}{2} - 1 = \frac{1}{2}\left(a^2 + 2 + \frac{1}{a^2}\right) - 1 .

b.

If sin⁡A2=12(a+1a) \sin\frac{A}{2} = \frac{1}{2}\left(a + \frac{1}{a}\right) , then cos⁡A=−12(a2+1a2) \cos A = -\frac{1}{2}\left(a^2 + \frac{1}{a^2}\right) .

Answer

cos⁡A=1−2sin⁡2A2=1−12(a2+2+1a2)=−12(a2+1a2) \cos A = 1 - 2\sin^2\frac{A}{2} = 1 - \frac{1}{2}\left(a^2 + 2 + \frac{1}{a^2}\right) = -\frac{1}{2}\left(a^2 + \frac{1}{a^2}\right) .

c.

If sin⁡β3=12(a+1a) \sin\frac{\beta}{3} = \frac{1}{2}\left(a + \frac{1}{a}\right) , then sin⁡β=−12(a3+1a3) \sin\beta = -\frac{1}{2}\left(a^3 + \frac{1}{a^3}\right) .

Answer

sin⁡β=3sin⁡β3−4sin⁡3β3 \sin\beta = 3\sin\frac{\beta}{3} - 4\sin^3\frac{\beta}{3} : the working of Exercise 7.2 Question 6 c.

d.

If cos⁡β3=12(a+1a) \cos\frac{\beta}{3} = \frac{1}{2}\left(a + \frac{1}{a}\right) , then cos⁡β=12(a3+1a3) \cos\beta = \frac{1}{2}\left(a^3 + \frac{1}{a^3}\right) .

Answer

cos⁡β=4cos⁡3β3−3cos⁡β3 \cos\beta = 4\cos^3\frac{\beta}{3} - 3\cos\frac{\beta}{3} : the working of Exercise 7.2 Question 6 d.

Question 7

a.

1+cos⁡Asin⁡A=cot⁡A2 \dfrac{1 + \cos A}{\sin A} = \cot\dfrac{A}{2}

Answer

This is Example 4: 2c22sc=cs \dfrac{2c^2}{2sc} = \dfrac{c}{s} .

b.

sin⁡A1+cos⁡A=tan⁡A2 \dfrac{\sin A}{1 + \cos A} = \tan\dfrac{A}{2}

Answer

2sc2c2=sc=tan⁡A2 \dfrac{2sc}{2c^2} = \dfrac{s}{c} = \tan\dfrac{A}{2} .

c.

1−cos⁡Asin⁡A=tan⁡A2 \dfrac{1 - \cos A}{\sin A} = \tan\dfrac{A}{2}

Answer

2s22sc=tan⁡A2 \dfrac{2s^2}{2sc} = \tan\dfrac{A}{2} .

d.

sin⁡A1−cos⁡A=cot⁡A2 \dfrac{\sin A}{1 - \cos A} = \cot\dfrac{A}{2}

Answer

2sc2s2=cot⁡A2 \dfrac{2sc}{2s^2} = \cot\dfrac{A}{2} .

e.

1+sin⁡Acos⁡A=cos⁡A2+sin⁡A2cos⁡A2−sin⁡A2 \dfrac{1 + \sin A}{\cos A} = \dfrac{\cos\frac{A}{2} + \sin\frac{A}{2}}{\cos\frac{A}{2} - \sin\frac{A}{2}}

Answer

c2+s2+2scc2−s2=(c+s)2(c−s)(c+s)=c+sc−s \dfrac{c^2 + s^2 + 2sc}{c^2 - s^2} = \dfrac{(c + s)^2}{(c - s)(c + s)} = \dfrac{c + s}{c - s} .

f.

cos⁡A=1−tan⁡2A21+tan⁡2A2 \cos A = \dfrac{1 - \tan^2\frac{A}{2}}{1 + \tan^2\frac{A}{2}}

Answer

cos⁡A=c2−s2c2+s2 \cos A = \dfrac{c^2 - s^2}{c^2 + s^2} ; divide the top and the bottom by c2 c^2 .

g.

1−tan⁡θ21+tan⁡θ2=1−sin⁡θcos⁡θ \dfrac{1 - \tan\frac{\theta}{2}}{1 + \tan\frac{\theta}{2}} = \dfrac{1 - \sin\theta}{\cos\theta}

Answer

Exercise 7.2 Question 4 d with θ/2 for θ.

h.

cos⁡α1+sin⁡α=1−tan⁡α21+tan⁡α2 \dfrac{\cos\alpha}{1 + \sin\alpha} = \dfrac{1 - \tan\frac{\alpha}{2}}{1 + \tan\frac{\alpha}{2}}

Answer

Exercise 7.2 Question 4 e with α/2 for α: c2−s2(c+s)2=c−sc+s \dfrac{c^2 - s^2}{(c + s)^2} = \dfrac{c - s}{c + s} , then divide by c.

i.

1+sin⁡θ−cos⁡θ1+sin⁡θ+cos⁡θ=tan⁡θ2 \dfrac{1 + \sin\theta - \cos\theta}{1 + \sin\theta + \cos\theta} = \tan\dfrac{\theta}{2}

Answer

This is Example 5.

j.

sin⁡θ−cos⁡θ21−sin⁡θ2−cos⁡θ=cot⁡θ2 \dfrac{\sin\theta - \cos\frac{\theta}{2}}{1 - \sin\frac{\theta}{2} - \cos\theta} = \cot\dfrac{\theta}{2}

Answer

With s =  sin (θ/2), c =  cos (θ/2): the top is 2sc − c = c(2s − 1), and the bottom is (1−cos⁡θ)−s=2s2−s=s(2s−1) (1 - \cos\theta) - s = 2s^2 - s = s(2s - 1) . The quotient is c/s=cot⁡(θ/2) c/s = \cot(\theta/2) .

k.

sin⁡A+tan⁡A=4tan⁡A21−tan⁡4A2 \sin A + \tan A = \dfrac{4\tan\frac{A}{2}}{1 - \tan^4\frac{A}{2}}

Answer

With t =  tan (A/2):
2t1+t2+2t1−t2=2t(1−t2)+2t(1+t2)1−t4=4t1−t4 \frac{2t}{1 + t^2} + \frac{2t}{1 - t^2} = \frac{2t(1 - t^2) + 2t(1 + t^2)}{1 - t^4} = \frac{4t}{1 - t^4}

l.

tan⁡A2+cot⁡A2=2cosec⁡A \tan\dfrac{A}{2} + \cot\dfrac{A}{2} = 2\operatorname{cosec} A

Answer

Exercise 7.2 Question 4 h with A/2 for A: s2+c2sc=22sc=2sin⁡A \dfrac{s^2 + c^2}{sc} = \dfrac{2}{2sc} = \dfrac{2}{\sin A} .

Question 8

a.

tan⁡(45°+A2)=cos⁡A1−sin⁡A \tan\left(45° + \dfrac{A}{2}\right) = \dfrac{\cos A}{1 - \sin A}

Answer

Exercise 7.2 Question 5 a with A/2 for A.

b.

2sin⁡2(45°−A2)=1−sin⁡A 2\sin^2\left(45° - \dfrac{A}{2}\right) = 1 - \sin A

Answer

2sin⁡2x=1−cos⁡2x 2\sin^2 x = 1 - \cos 2x : 1−cos⁡(90°−A)=1−sin⁡A 1 - \cos(90° - A) = 1 - \sin A .

c.

2cos⁡2(45°−A2)=1+sin⁡A 2\cos^2\left(45° - \dfrac{A}{2}\right) = 1 + \sin A

Answer

2cos⁡2x=1+cos⁡2x 2\cos^2 x = 1 + \cos 2x : 1+cos⁡(90°−A)=1+sin⁡A 1 + \cos(90° - A) = 1 + \sin A .

d.

cos⁡2(45°−θ2)−sin⁡2(45°−θ2)=sin⁡θ \cos^2\left(45° - \dfrac{\theta}{2}\right) - \sin^2\left(45° - \dfrac{\theta}{2}\right) = \sin\theta

Answer

It is cos⁡(90°−θ)=sin⁡θ \cos(90° - \theta) = \sin\theta .

e.

1−tan⁡2(45°−A2)1+tan⁡2(45°−A2)=sin⁡A \dfrac{1 - \tan^2(45° - \frac{A}{2})}{1 + \tan^2(45° - \frac{A}{2})} = \sin A

Answer

(1−tan⁡2x)/(1+tan⁡2x)=cos⁡2x (1 - \tan^2 x)/(1 + \tan^2 x) = \cos 2x : cos⁡(90°−A)=sin⁡A \cos(90° - A) = \sin A .

f.

1−tan⁡2(45°−A)1+tan⁡2(45°−A)=sin⁡2A \dfrac{1 - \tan^2(45° - A)}{1 + \tan^2(45° - A)} = \sin 2A

Answer

cos⁡2(45°−A)=cos⁡(90°−2A)=sin⁡2A \cos 2(45° - A) = \cos(90° - 2A) = \sin 2A .

g.

tan⁡(45°+A2)=sec⁡A+tan⁡A=1+sin⁡A1−sin⁡A \tan\left(45° + \dfrac{A}{2}\right) = \sec A + \tan A = \sqrt{\dfrac{1 + \sin A}{1 - \sin A}} (for cos A > 0)

Answer

sec⁡A+tan⁡A=1+sin⁡Acos⁡A \sec A + \tan A = \dfrac{1 + \sin A}{\cos A} , which is tan⁡(45°+A/2) \tan(45° + A/2) by Exercise 7.2 Question 5 d with A/2 for θ. Its square is (1+sin⁡A)2cos⁡2A=(1+sin⁡A)2(1−sin⁡A)(1+sin⁡A)=1+sin⁡A1−sin⁡A \dfrac{(1 + \sin A)^2}{\cos^2 A} = \dfrac{(1 + \sin A)^2}{(1 - \sin A)(1 + \sin A)} = \dfrac{1 + \sin A}{1 - \sin A} ; with cos A > 0, 1 +  sin A and cos A are both positive, so take the positive root.

h.

sin⁡2A1+cos⁡2A×cos⁡A1+cos⁡A=tan⁡A2 \dfrac{\sin 2A}{1 + \cos 2A} \times \dfrac{\cos A}{1 + \cos A} = \tan\dfrac{A}{2}

Answer

sin⁡2A1+cos⁡2A=tan⁡A \dfrac{\sin 2A}{1 + \cos 2A} = \tan A (Question 7 b with 2A), and tan⁡A⋅cos⁡A1+cos⁡A=sin⁡A1+cos⁡A=tan⁡A2 \tan A \cdot \dfrac{\cos A}{1 + \cos A} = \dfrac{\sin A}{1 + \cos A} = \tan\dfrac{A}{2} .

Question 9

a.

Prove that
(1+cos⁡π8)(1+cos⁡3π8)×(1+cos⁡5π8)(1+cos⁡7π8)=18 \begin{aligned} &\left(1 + \cos\frac{\pi}{8}\right)\left(1 + \cos\frac{3\pi}{8}\right) \\ &\times \left(1 + \cos\frac{5\pi}{8}\right)\left(1 + \cos\frac{7\pi}{8}\right) = \frac{1}{8} \end{aligned}

Answer

cos⁡7π8=−cos⁡π8 \cos\frac{7\pi}{8} = -\cos\frac{\pi}{8} and cos⁡5π8=−cos⁡3π8 \cos\frac{5\pi}{8} = -\cos\frac{3\pi}{8} . Pair them:
(1−cos⁡2π8)(1−cos⁡23π8)=sin⁡2π8sin⁡23π8=sin⁡2π8cos⁡2π8 \left(1 - \cos^2\tfrac{\pi}{8}\right)\left(1 - \cos^2\tfrac{3\pi}{8}\right) = \sin^2\tfrac{\pi}{8}\sin^2\tfrac{3\pi}{8} = \sin^2\tfrac{\pi}{8}\cos^2\tfrac{\pi}{8}
since 3π8=π2−π8 \frac{3\pi}{8} = \frac{\pi}{2} - \frac{\pi}{8} . That is (12sin⁡π4)2=14⋅12=18 \left(\tfrac{1}{2}\sin\tfrac{\pi}{4}\right)^2 = \tfrac{1}{4} \cdot \tfrac{1}{2} = \tfrac{1}{8} .

Exercise 7.4

Question 2

a.

sin⁡50°cos⁡30° \sin 50° \cos 30°

Answer

12(sin⁡80°+sin⁡20°) \tfrac{1}{2}(\sin 80° + \sin 20°)

b.

cos⁡25°cos⁡15° \cos 25° \cos 15°

Answer

12(cos⁡40°+cos⁡10°) \tfrac{1}{2}(\cos 40° + \cos 10°)

c.

cos 10A sin 6A

Answer

By (iv): 12(sin⁡16A−sin⁡4A) \tfrac{1}{2}(\sin 16A - \sin 4A)

d.

sin⁡27°sin⁡32° \sin 27° \sin 32°

Answer

By (viii): 12{cos⁡(32°−27°)−cos⁡(32°+27°)}=12(cos⁡5°−cos⁡59°) \tfrac{1}{2}\{\cos(32° - 27°) - \cos(32° + 27°)\} = \tfrac{1}{2}(\cos 5° - \cos 59°)

Question 3

a.

sin⁡50°+sin⁡30° \sin 50° + \sin 30°

Answer

2sin⁡40°cos⁡10° 2\sin 40° \cos 10°

b.

cos⁡25°+cos⁡15° \cos 25° + \cos 15°

Answer

2cos⁡20°cos⁡5° 2\cos 20° \cos 5°

c.

cos 7A −  cos 5A

Answer

−2 sin 6A sin A

d.

sin⁡67°−sin⁡43° \sin 67° - \sin 43°

Answer

2cos⁡55°sin⁡12° 2\cos 55° \sin 12°

Question 4

a.

cos⁡50°+cos⁡40°=2cos⁡5° \cos 50° + \cos 40° = \sqrt{2}\cos 5°

Answer

2cos⁡45°cos⁡5°=2⋅12cos⁡5°=2cos⁡5° 2\cos 45° \cos 5° = 2 \cdot \tfrac{1}{\sqrt{2}}\cos 5° = \sqrt{2}\cos 5° .

b.

sin⁡50°−sin⁡70°=−sin⁡10° \sin 50° - \sin 70° = -\sin 10°

Answer

2cos⁡60°sin⁡(−10°)=2⋅12⋅(−sin⁡10°)=−sin⁡10° 2\cos 60° \sin(-10°) = 2 \cdot \tfrac{1}{2} \cdot (-\sin 10°) = -\sin 10° .

c.

sin⁡10°+sin⁡70°=3sin⁡40° \sin 10° + \sin 70° = \sqrt{3}\sin 40°

Answer

2sin⁡40°cos⁡(−30°)=2sin⁡40°⋅32=3sin⁡40° 2\sin 40° \cos(-30°) = 2\sin 40° \cdot \tfrac{\sqrt{3}}{2} = \sqrt{3}\sin 40° .

d.

cos⁡40°−sin⁡40°=2sin⁡5° \cos 40° - \sin 40° = \sqrt{2}\sin 5°

Answer

sin⁡40°=cos⁡50° \sin 40° = \cos 50° : cos⁡40°−cos⁡50°=−2sin⁡45°sin⁡(−5°)=2sin⁡5° \cos 40° - \cos 50° = -2\sin 45° \sin(-5°) = \sqrt{2}\sin 5° .

e.

sin⁡37°−cos⁡67°=3sin⁡7° \sin 37° - \cos 67° = \sqrt{3}\sin 7°

Answer

cos⁡67°=sin⁡23° \cos 67° = \sin 23° : sin⁡37°−sin⁡23°=2cos⁡30°sin⁡7°=3sin⁡7° \sin 37° - \sin 23° = 2\cos 30° \sin 7° = \sqrt{3}\sin 7° .

Question 5

a.

cos⁡10°−sin⁡10°cos⁡10°+sin⁡10°=tan⁡35° \dfrac{\cos 10° - \sin 10°}{\cos 10° + \sin 10°} = \tan 35°

Answer

sin⁡10°=cos⁡80° \sin 10° = \cos 80° : cos⁡10°−cos⁡80°cos⁡10°+cos⁡80°=2sin⁡45°sin⁡35°2cos⁡45°cos⁡35°=tan⁡35° \dfrac{\cos 10° - \cos 80°}{\cos 10° + \cos 80°} = \dfrac{2\sin 45° \sin 35°}{2\cos 45° \cos 35°} = \tan 35° .

b.

cos⁡9°−sin⁡9°cos⁡9°+sin⁡9°=cot⁡54° \dfrac{\cos 9° - \sin 9°}{\cos 9° + \sin 9°} = \cot 54°

Answer

cos⁡9°−cos⁡81°cos⁡9°+cos⁡81°=2sin⁡45°sin⁡36°2cos⁡45°cos⁡36°=tan⁡36°=cot⁡54° \dfrac{\cos 9° - \cos 81°}{\cos 9° + \cos 81°} = \dfrac{2\sin 45° \sin 36°}{2\cos 45° \cos 36°} = \tan 36° = \cot 54° .

c.

cos⁡12°−sin⁡12°cos⁡12°+sin⁡12°=tan⁡33° \dfrac{\cos 12° - \sin 12°}{\cos 12° + \sin 12°} = \tan 33°

Answer

cos⁡12°−cos⁡78°cos⁡12°+cos⁡78°=2sin⁡45°sin⁡33°2cos⁡45°cos⁡33°=tan⁡33° \dfrac{\cos 12° - \cos 78°}{\cos 12° + \cos 78°} = \dfrac{2\sin 45° \sin 33°}{2\cos 45° \cos 33°} = \tan 33° .

d.

sin⁡A−sin⁡Bsin⁡A+sin⁡B=cot⁡A+B2tan⁡A−B2 \dfrac{\sin A - \sin B}{\sin A + \sin B} = \cot\dfrac{A + B}{2}\tan\dfrac{A - B}{2}

Answer

2cos⁡A+B2sin⁡A−B22sin⁡A+B2cos⁡A−B2=cot⁡A+B2tan⁡A−B2 \frac{2\cos\frac{A + B}{2}\sin\frac{A - B}{2}}{2\sin\frac{A + B}{2}\cos\frac{A - B}{2}} = \cot\frac{A + B}{2}\tan\frac{A - B}{2}

e.

cos⁡B−cos⁡Asin⁡A+sin⁡B=tan⁡A−B2 \dfrac{\cos B - \cos A}{\sin A + \sin B} = \tan\dfrac{A - B}{2}

Answer

cos⁡B−cos⁡A=−2sin⁡B+A2sin⁡B−A2=2sin⁡A+B2sin⁡A−B2 \cos B - \cos A = -2\sin\frac{B + A}{2}\sin\frac{B - A}{2} = 2\sin\frac{A + B}{2}\sin\frac{A - B}{2}
Divide by sin⁡A+sin⁡B=2sin⁡A+B2cos⁡A−B2 \sin A + \sin B = 2\sin\frac{A + B}{2}\cos\frac{A - B}{2} : tan⁡A−B2 \tan\frac{A - B}{2} .

f.

sin⁡(60°+A)+sin⁡(60°−A)cos⁡(60°+A)+cos⁡(60°−A)=3 \dfrac{\sin(60° + A) + \sin(60° - A)}{\cos(60° + A) + \cos(60° - A)} = \sqrt{3}

Answer

2sin⁡60°cos⁡A2cos⁡60°cos⁡A=tan⁡60°=3 \dfrac{2\sin 60° \cos A}{2\cos 60° \cos A} = \tan 60° = \sqrt{3} .

g.

2sin⁡(45°+ϕ)cos⁡(45°−ϕ)=1+sin⁡2ϕ 2\sin(45° + \phi)\cos(45° - \phi) = 1 + \sin 2\phi

Answer

sin⁡90°+sin⁡2ϕ=1+sin⁡2ϕ \sin 90° + \sin 2\phi = 1 + \sin 2\phi .

h.

2cos⁡(45°+θ)cos⁡(45°−θ)=cos⁡2θ 2\cos(45° + \theta)\cos(45° - \theta) = \cos 2\theta

Answer

cos⁡90°+cos⁡2θ=cos⁡2θ \cos 90° + \cos 2\theta = \cos 2\theta .

Question 6

a.

sin⁡θ+sin⁡3θ−sin⁡2θcos⁡θ−cos⁡2θ+cos⁡3θ=tan⁡2θ \dfrac{\sin\theta + \sin 3\theta - \sin 2\theta}{\cos\theta - \cos 2\theta + \cos 3\theta} = \tan 2\theta

Answer

As in Example 6: sin⁡2θ(2cos⁡θ−1)cos⁡2θ(2cos⁡θ−1)=tan⁡2θ \dfrac{\sin 2\theta(2\cos\theta - 1)}{\cos 2\theta(2\cos\theta - 1)} = \tan 2\theta .

b.

sin⁡2A+sin⁡3A+sin⁡5A+sin⁡6Acos⁡2A+cos⁡3A+cos⁡5A+cos⁡6A=tan⁡4A \dfrac{\sin 2A + \sin 3A + \sin 5A + \sin 6A}{\cos 2A + \cos 3A + \cos 5A + \cos 6A} = \tan 4A

Answer

Pair 2A with 6A and 3A with 5A: every pair averages 4A.
2sin⁡4Acos⁡2A+2sin⁡4Acos⁡A2cos⁡4Acos⁡2A+2cos⁡4Acos⁡A=sin⁡4A(cos⁡2A+cos⁡A)cos⁡4A(cos⁡2A+cos⁡A)=tan⁡4A \frac{2\sin 4A \cos 2A + 2\sin 4A \cos A}{2\cos 4A \cos 2A + 2\cos 4A \cos A} = \frac{\sin 4A(\cos 2A + \cos A)}{\cos 4A(\cos 2A + \cos A)} = \tan 4A

c.

sin⁡A−sin⁡3A+sin⁡5A−sin⁡7Acos⁡A−cos⁡3A−cos⁡5A+cos⁡7A=cot⁡2A \dfrac{\sin A - \sin 3A + \sin 5A - \sin 7A}{\cos A - \cos 3A - \cos 5A + \cos 7A} = \cot 2A

Answer

Pair A with 7A and 3A with 5A:
top=(sin⁡A−sin⁡7A)−(sin⁡3A−sin⁡5A)=−2cos⁡4Asin⁡3A+2cos⁡4Asin⁡A=−4cos⁡4Acos⁡2Asin⁡Abottom=(cos⁡A+cos⁡7A)−(cos⁡3A+cos⁡5A)=2cos⁡4Acos⁡3A−2cos⁡4Acos⁡A=−4cos⁡4Asin⁡2Asin⁡A \begin{aligned} \text{top} &= (\sin A - \sin 7A) - (\sin 3A - \sin 5A) = -2\cos 4A \sin 3A + 2\cos 4A \sin A = -4\cos 4A \cos 2A \sin A \\ \text{bottom} &= (\cos A + \cos 7A) - (\cos 3A + \cos 5A) = 2\cos 4A \cos 3A - 2\cos 4A \cos A = -4\cos 4A \sin 2A \sin A \end{aligned}
The quotient is cos⁡2A/sin⁡2A=cot⁡2A \cos 2A/\sin 2A = \cot 2A .

d.

cos⁡2Acos⁡3A−cos⁡2Acos⁡7A+cos⁡Acos⁡10Asin⁡4Asin⁡3A−sin⁡2Asin⁡5A+sin⁡4Asin⁡7A=cot⁡6Acot⁡5A \dfrac{\cos 2A \cos 3A - \cos 2A \cos 7A + \cos A \cos 10A}{\sin 4A \sin 3A - \sin 2A \sin 5A + \sin 4A \sin 7A} = \cot 6A \cot 5A

Answer

Double each product and turn it into a sum:
2×top=(cos⁡5A+cos⁡A)−(cos⁡9A+cos⁡5A)+(cos⁡11A+cos⁡9A)=cos⁡A+cos⁡11A=2cos⁡6Acos⁡5A2×bottom=(cos⁡A−cos⁡7A)−(cos⁡3A−cos⁡7A)+(cos⁡3A−cos⁡11A)=cos⁡A−cos⁡11A=2sin⁡6Asin⁡5A \begin{aligned} 2 \times \text{top} &= (\cos 5A + \cos A) - (\cos 9A + \cos 5A) + (\cos 11A + \cos 9A) = \cos A + \cos 11A = 2\cos 6A \cos 5A \\ 2 \times \text{bottom} &= (\cos A - \cos 7A) - (\cos 3A - \cos 7A) + (\cos 3A - \cos 11A) = \cos A - \cos 11A = 2\sin 6A \sin 5A \end{aligned}
The quotient is cot⁡6Acot⁡5A \cot 6A \cot 5A .

Question 7

a.

4cos⁡θcos⁡(60°+θ)cos⁡(60°−θ)=cos⁡3θ 4\cos\theta\cos(60° + \theta)\cos(60° - \theta) = \cos 3\theta

Answer

2cos⁡(60°+θ)cos⁡(60°−θ)=cos⁡120°+cos⁡2θ=2cos⁡2θ−32 2\cos(60° + \theta)\cos(60° - \theta) = \cos 120° + \cos 2\theta = 2\cos^2\theta - \tfrac{3}{2} . So the left side is 2cos⁡θ(2cos⁡2θ−32)=4cos⁡3θ−3cos⁡θ=cos⁡3θ 2\cos\theta(2\cos^2\theta - \tfrac{3}{2}) = 4\cos^3\theta - 3\cos\theta = \cos 3\theta .

b.

cot⁡θcot⁡(60°+θ)cot⁡(60°−θ)=cot⁡3θ \cot\theta\cot(60° + \theta)\cot(60° - \theta) = \cot 3\theta

Answer

From Example 5, sin⁡θsin⁡(60°+θ)sin⁡(60°−θ)=14sin⁡3θ \sin\theta\sin(60° + \theta)\sin(60° - \theta) = \tfrac{1}{4}\sin 3\theta , and from a, cos⁡θcos⁡(60°+θ)cos⁡(60°−θ)=14cos⁡3θ \cos\theta\cos(60° + \theta)\cos(60° - \theta) = \tfrac{1}{4}\cos 3\theta . Divide a’s product by Example 5’s: cot⁡3θ \cot 3\theta .

c.

sec⁡(45°+θ2)sec⁡(45°−θ2)=2sec⁡θ \sec\left(45° + \dfrac{\theta}{2}\right)\sec\left(45° - \dfrac{\theta}{2}\right) = 2\sec\theta

Answer

2cos⁡(45°+θ2)cos⁡(45°−θ2)=cos⁡90°+cos⁡θ=cos⁡θ 2\cos(45° + \frac{\theta}{2})\cos(45° - \frac{\theta}{2}) = \cos 90° + \cos\theta = \cos\theta . So the product of the cosines is 12cos⁡θ \frac{1}{2}\cos\theta , and the product of the secants is 2/cos⁡θ=2sec⁡θ 2/\cos\theta = 2\sec\theta .

d.

tan⁡(π4+θ)−tan⁡(π4−θ)=2tan⁡2θ \tan\left(\dfrac{\pi}{4} + \theta\right) - \tan\left(\dfrac{\pi}{4} - \theta\right) = 2\tan 2\theta

Answer

With t =  tan θ: 1+t1−t−1−t1+t=(1+t)2−(1−t)21−t2=4t1−t2=2tan⁡2θ \dfrac{1 + t}{1 - t} - \dfrac{1 - t}{1 + t} = \dfrac{(1 + t)^2 - (1 - t)^2}{1 - t^2} = \dfrac{4t}{1 - t^2} = 2\tan 2\theta .

Question 8

a.

sin⁡10°sin⁡50°sin⁡70°=18 \sin 10° \sin 50° \sin 70° = \dfrac{1}{8}

Answer

sin⁡50°sin⁡70°=12(cos⁡20°−cos⁡120°)=12cos⁡20°+14 \sin 50° \sin 70° = \tfrac{1}{2}(\cos 20° - \cos 120°) = \tfrac{1}{2}\cos 20° + \tfrac{1}{4} . Then
sin⁡10°(12cos⁡20°+14)=14(sin⁡30°−sin⁡10°)+14sin⁡10°=14sin⁡30°=18 \sin 10°\left(\tfrac{1}{2}\cos 20° + \tfrac{1}{4}\right) = \tfrac{1}{4}(\sin 30° - \sin 10°) + \tfrac{1}{4}\sin 10° = \tfrac{1}{4}\sin 30° = \tfrac{1}{8}
(Or Example 5 with θ=10° \theta = 10° .)

b.

cos⁡10°cos⁡50°cos⁡70°=38 \cos 10° \cos 50° \cos 70° = \dfrac{\sqrt{3}}{8}

Answer

Question 7 a with θ=10° \theta = 10° : cos⁡10°cos⁡70°cos⁡50°=14cos⁡30°=38 \cos 10° \cos 70° \cos 50° = \tfrac{1}{4}\cos 30° = \tfrac{\sqrt{3}}{8} .

c.

sin⁡10°sin⁡30°sin⁡50°sin⁡70°=116 \sin 10° \sin 30° \sin 50° \sin 70° = \dfrac{1}{16}

Answer

sin⁡30°=12 \sin 30° = \tfrac{1}{2} , times a: 12⋅18=116 \tfrac{1}{2} \cdot \tfrac{1}{8} = \tfrac{1}{16} .

d.

sin⁡10°sin⁡50°sin⁡60°sin⁡70°=316 \sin 10° \sin 50° \sin 60° \sin 70° = \dfrac{\sqrt{3}}{16}

Answer

sin⁡60°=32 \sin 60° = \tfrac{\sqrt{3}}{2} , times a: 316 \tfrac{\sqrt{3}}{16} .

e.

cos⁡20°cos⁡30°cos⁡40°cos⁡80°=316 \cos 20° \cos 30° \cos 40° \cos 80° = \dfrac{\sqrt{3}}{16}

Answer

cos⁡30°=32 \cos 30° = \tfrac{\sqrt{3}}{2} , times f: 32⋅18=316 \tfrac{\sqrt{3}}{2} \cdot \tfrac{1}{8} = \tfrac{\sqrt{3}}{16} .

f.

cos⁡20°cos⁡40°cos⁡80°=18 \cos 20° \cos 40° \cos 80° = \dfrac{1}{8}

Answer

Question 7 a with θ=20° \theta = 20° : cos⁡20°cos⁡80°cos⁡40°=14cos⁡60°=18 \cos 20° \cos 80° \cos 40° = \tfrac{1}{4}\cos 60° = \tfrac{1}{8} . Or multiply by sin⁡20° \sin 20° and use 2 sin x cos x =  sin 2x three times: 8sin⁡20°cos⁡20°cos⁡40°cos⁡80°=sin⁡160°=sin⁡20° 8\sin 20° \cos 20° \cos 40° \cos 80° = \sin 160° = \sin 20° .

Exercise 7.5

Question 2

a.

sin⁡(α+β)=sin⁡γ \sin(\alpha + \beta) = \sin\gamma

Answer

sin⁡120°=32=sin⁡60° \sin 120° = \tfrac{\sqrt{3}}{2} = \sin 60° .

b.

cos⁡(β+γ)=−cos⁡α \cos(\beta + \gamma) = -\cos\alpha

Answer

cos⁡150°=−32=−cos⁡30° \cos 150° = -\tfrac{\sqrt{3}}{2} = -\cos 30° .

c.

tan⁡(α+β)=−tan⁡γ \tan(\alpha + \beta) = -\tan\gamma

Answer

tan⁡120°=−3=−tan⁡60° \tan 120° = -\sqrt{3} = -\tan 60° .

d.

cot⁡(α+γ)=−cot⁡β \cot(\alpha + \gamma) = -\cot\beta

Answer

cot⁡90°=0 \cot 90° = 0 , and −cot⁡90°=0 -\cot 90° = 0 .

e.

sin⁡(α2+β2)=cos⁡γ2 \sin\left(\dfrac{\alpha}{2} + \dfrac{\beta}{2}\right) = \cos\dfrac{\gamma}{2}

Answer

sin⁡(15°+45°)=sin⁡60°=32=cos⁡30° \sin(15° + 45°) = \sin 60° = \tfrac{\sqrt{3}}{2} = \cos 30° .

f.

cos⁡(α2+β2)=sin⁡γ2 \cos\left(\dfrac{\alpha}{2} + \dfrac{\beta}{2}\right) = \sin\dfrac{\gamma}{2}

Answer

cos⁡60°=12=sin⁡30° \cos 60° = \tfrac{1}{2} = \sin 30° .

Question 3

a.

tan A +  tan B +  tan C =  tan A tan B tan C

Answer

This is Example 5: both sides are −1/3 -1/\sqrt{3} .

b.

tan 2A +  tan 2B +  tan 2C =  tan 2A tan 2B tan 2C

Answer

tan⁡60°=3 \tan 60° = \sqrt{3} and tan⁡240°=tan⁡60°=3 \tan 240° = \tan 60° = \sqrt{3} . LHS =33 = 3\sqrt{3} ; RHS =(3)3=33 = (\sqrt{3})^3 = 3\sqrt{3} .
(2A+2B+2C=360° 2A + 2B + 2C = 360° , and the relation also holds when the angles add up to 360°.)

c.

cot⁡Acot⁡B+cot⁡Bcot⁡C+cot⁡Ccot⁡A=1 \cot A \cot B + \cot B \cot C + \cot C \cot A = 1

Answer

cot⁡30°=3 \cot 30° = \sqrt{3} , cot⁡120°=−13 \cot 120° = -\tfrac{1}{\sqrt{3}} : −1 − 1 + 3 = 1.

d.

cot⁡2Acot⁡2B+cot⁡2Bcot⁡2C+cot⁡2Ccot⁡2A=1 \cot 2A \cot 2B + \cot 2B \cot 2C + \cot 2C \cot 2A = 1

Answer

cot⁡60°=cot⁡240°=13 \cot 60° = \cot 240° = \tfrac{1}{\sqrt{3}} : each product is 13 \tfrac{1}{3} , and three of them make 1.

e.

tan⁡A2tan⁡B2+tan⁡B2tan⁡C2+tan⁡C2tan⁡A2=1 \tan\dfrac{A}{2}\tan\dfrac{B}{2} + \tan\dfrac{B}{2}\tan\dfrac{C}{2} + \tan\dfrac{C}{2}\tan\dfrac{A}{2} = 1

Answer

tan⁡15°=2−3 \tan 15° = 2 - \sqrt{3} , tan⁡60°=3 \tan 60° = \sqrt{3} :
23(2−3)+(2−3)2=43−6+7−43=1 2\sqrt{3}(2 - \sqrt{3}) + (2 - \sqrt{3})^2 = 4\sqrt{3} - 6 + 7 - 4\sqrt{3} = 1
(It holds in every triangle: the half angles add up to 90°, and this is Exercise 7.1 Question 12 c’s pattern.)

Question 4

a.

sin 2A −  sin 2B +  sin 2C = 4 cos A sin B cos C

Answer

LHS =32−0+32=3 = \tfrac{\sqrt{3}}{2} - 0 + \tfrac{\sqrt{3}}{2} = \sqrt{3} . RHS =4⋅32⋅1⋅12=3 = 4 \cdot \tfrac{\sqrt{3}}{2} \cdot 1 \cdot \tfrac{1}{2} = \sqrt{3} .

b.

sin 2A −  sin 2B −  sin 2C = 4 sin A cos B cos C

Answer

LHS =32−0−32=0 = \tfrac{\sqrt{3}}{2} - 0 - \tfrac{\sqrt{3}}{2} = 0 . RHS has cos⁡90°=0 \cos 90° = 0 : 0.
Both sides are 0 here because B=90° B = 90° , but this one is not true in every triangle: the identity that always holds is sin 2A −  sin 2B −  sin 2C = −4 sin A cos B cos C (with 60°, 60°, 60° the left side is −32 -\tfrac{\sqrt{3}}{2} and 4sin⁡Acos⁡Bcos⁡C=+32 4\sin A \cos B \cos C = +\tfrac{\sqrt{3}}{2} ).

c.

cos 2A +  cos 2B +  cos 2C = −1 − 4 cos A cos B cos C

Answer

LHS =12−1−12=−1 = \tfrac{1}{2} - 1 - \tfrac{1}{2} = -1 . RHS = −1 − 0 = −1.

d.

cos 2A −  cos 2B +  cos 2C = 1 − 4 sin A cos B sin C

Answer

LHS =12+1−12=1 = \tfrac{1}{2} + 1 - \tfrac{1}{2} = 1 . RHS = 1 − 0 = 1.

Question 5

a.

sin⁡2α+sin⁡2β+sin⁡2γ=4sin⁡αsin⁡βsin⁡γ \sin^2\alpha + \sin^2\beta + \sin^2\gamma = 4\sin\alpha\sin\beta\sin\gamma

Answer

LHS =12+12+1=2 = \tfrac{1}{2} + \tfrac{1}{2} + 1 = 2 . RHS =4⋅12⋅12⋅1=2 = 4 \cdot \tfrac{1}{\sqrt{2}} \cdot \tfrac{1}{\sqrt{2}} \cdot 1 = 2 .
This one holds for these angles, but not in every triangle: with 60°, 60°, 60° the left side is 94 \tfrac{9}{4} and the right 332≈2.60 \tfrac{3\sqrt{3}}{2} \approx 2.60 .

b.

sin⁡2α+sin⁡2β−sin⁡2γ=1−2cos⁡αcos⁡βsin⁡γ \sin^2\alpha + \sin^2\beta - \sin^2\gamma = 1 - 2\cos\alpha\cos\beta\sin\gamma

Answer

LHS =12+12−1=0 = \tfrac{1}{2} + \tfrac{1}{2} - 1 = 0 . RHS =1−2⋅12⋅1=0 = 1 - 2 \cdot \tfrac{1}{2} \cdot 1 = 0 .
This one too holds for these angles only: the identity for every triangle is sin⁡2α+sin⁡2β−sin⁡2γ=2sin⁡αsin⁡βcos⁡γ \sin^2\alpha + \sin^2\beta - \sin^2\gamma = 2\sin\alpha\sin\beta\cos\gamma (with 30°, 90°, 60° the left side is 12 \tfrac{1}{2} and 1−2cos⁡αcos⁡βsin⁡γ=1 1 - 2\cos\alpha\cos\beta\sin\gamma = 1 ). Parts c and d hold in every triangle.

c.

cos⁡2α−cos⁡2β+cos⁡2γ=1−2sin⁡αcos⁡βsin⁡γ \cos^2\alpha - \cos^2\beta + \cos^2\gamma = 1 - 2\sin\alpha\cos\beta\sin\gamma

Answer

LHS =12−12+0=0 = \tfrac{1}{2} - \tfrac{1}{2} + 0 = 0 . RHS =1−2⋅12⋅1=0 = 1 - 2 \cdot \tfrac{1}{2} \cdot 1 = 0 .

d.

cos⁡2α−cos⁡2β−cos⁡2γ=2cos⁡αsin⁡βsin⁡γ−1 \cos^2\alpha - \cos^2\beta - \cos^2\gamma = 2\cos\alpha\sin\beta\sin\gamma - 1

Answer

LHS = 0. RHS =2⋅12⋅1−1=0 = 2 \cdot \tfrac{1}{2} \cdot 1 - 1 = 0 .

Question 6

a.

For which of these conditions is sin 2A +  sin 2B +  sin 2C = 4 sin A sin B sin C valid? a. A+B=90° A + B = 90° and C=90° C = 90° b. A+B+C=180° A + B + C = 180° c. A+B+C=360° A + B + C = 360° d. A+B+C=90° A + B + C = 90°

Answer

b, A+B+C=180° A + B + C = 180° . Proof: sin 2A +  sin 2B = 2 sin (A + B) cos (A − B) = 2 sin C cos (A − B), and sin 2C = 2 sin C cos C = −2 sin C cos (A + B). So
LHS=2sin⁡C{cos⁡(A−B)−cos⁡(A+B)}=2sin⁡C⋅2sin⁡Asin⁡B=4sin⁡Asin⁡Bsin⁡C \text{LHS} = 2\sin C\{\cos(A - B) - \cos(A + B)\} = 2\sin C \cdot 2\sin A \sin B = 4\sin A \sin B \sin C
Condition a is only a special case of b (a right-angled triangle). For c and d it fails: try A=B=C=120° A = B = C = 120° , where the left side is −332 -\tfrac{3\sqrt{3}}{2} and the right 332 \tfrac{3\sqrt{3}}{2} .

Exercise 7.6

Question 2

a.
45°20 myABC
Figure 7.13
Answer

sin⁡45°=AB/AC \sin 45° = AB/AC : y=20/sin⁡45°=202≈28.28 y = 20/\sin 45° = 20\sqrt{2} \approx 28.28 m.

b.
30°x10 mPQR
Figure 7.14
Answer

tan⁡30°=PQ/QR \tan 30° = PQ/QR : x=10/3≈5.77 x = 10/\sqrt{3} \approx 5.77 m.

c.
30°θx20 m10 mABCD
Figure 7.15
Answer

DB = 20 + 10 = 30 m. In △ABD: x=30tan⁡30°=30/3=103≈17.32 x = 30\tan 30° = 30/\sqrt{3} = 10\sqrt{3} \approx 17.32 m.
In △ABC: tan⁡θ=103/10=3 \tan\theta = 10\sqrt{3}/10 = \sqrt{3} , so θ=60° \theta = 60° .

d.
30°45°45 myxLBPN
Figure 7.16
Answer

In △LPB, tan⁡45°=1 \tan 45° = 1 , so y = x. In △LNB: tan⁡30°=x/(x+45) \tan 30° = x/(x + 45) , so 3x=x+45 \sqrt{3}x = x + 45 , x=45/(3−1) x = 45/(\sqrt{3} - 1) . x=y≈61.47 x = y \approx 61.47 m, as in Example 2.

e.
30°45°x50 myABCDFE
Figure 7.17
Answer

FD = CB = y. In △ABC, tan⁡45°=1 \tan 45° = 1 : y = AB = 50 m. In △AFD: AF=50tan⁡30°=50/3≈28.87 AF = 50\tan 30° = 50/\sqrt{3} \approx 28.87 m. So x = FB = 50 − 28.87 = 21.13 m.

f.
30°45°40 myxWTUV
Figure 7.18
Answer

In △WTV, tan⁡45°=1 \tan 45° = 1 : x = 40 m. In △WTU: y=40/tan⁡30°=403≈69.28 y = 40/\tan 30° = 40\sqrt{3} \approx 69.28 m.

Questions 3 to 13

3 a.

From a point on the ground, the angle of elevation of a point on the roof of a house is 60°. From 50 m farther away it is 30°. Find the height of the house.

Answer

With height h: the near point is h/3 h/\sqrt{3} from the house and the far one 3h \sqrt{3}h .
3h−h3=50  ⇒  2h3=50  ⇒  h=253≈43.30 m \sqrt{3}h - \frac{h}{\sqrt{3}} = 50 \;\Rightarrow\; \frac{2h}{\sqrt{3}} = 50 \;\Rightarrow\; h = 25\sqrt{3} \approx 43.30 \text{ m}

3 b.

The angles of elevation of the top of a 30 ft pole from two points on the same side are 60° and 45°. Find the distance between the points.

Answer

30/tan⁡45°−30/tan⁡60°=30−103=30−17.32= 30/\tan 45° - 30/\tan 60° = 30 - 10\sqrt{3} = 30 - 17.32 = 12.68 ft.

4 a.

From the top of a hill 500 m high, two points on the ground on the same side are seen at angles of depression of 60° and 30°. Find the distance between the points.

Answer

5003−500/3=10003≈ 500\sqrt{3} - 500/\sqrt{3} = \dfrac{1000}{\sqrt{3}} \approx 577.35 m.

4 b.

From the roof of a house, two people sitting on the ground 10 ft apart, on the same side, are seen at angles of depression of 60° and 45°. Find the height of the house.

Answer

h−h3=10  ⇒  h=1033−1=5(3+3)≈23.66 ft h - \frac{h}{\sqrt{3}} = 10 \;\Rightarrow\; h = \frac{10\sqrt{3}}{\sqrt{3} - 1} = 5(3 + \sqrt{3}) \approx 23.66 \text{ ft}

5 a.

The angles of elevation of the top of a pole from two points 40 m and 36 m from its foot, on the same side, are complementary. Find the height of the pole.

Answer

As in Example 4: h=40×36=1440=1210≈ h = \sqrt{40 \times 36} = \sqrt{1440} = 12\sqrt{10} \approx 37.95 m.

5 b.

Two poles stand 50 m apart, one twice as tall as the other. From the point midway between them, the angles of elevation of their tops are complementary. Find the heights of the poles.

Answer

Each top is 25 m away. With heights h and 2h: tan θ = h/25 and cot⁡θ=2h/25 \cot\theta = 2h/25 . Multiply: 1=2h2/625 1 = 2h^2/625 , h2=312.5 h^2 = 312.5 , h≈17.68 h \approx 17.68 m.
The poles are about 17.68 m and 35.36 m high.

6 a.

The shadow of an upright pole on level ground is 50 m longer when the sun’s altitude is 30° than when it is 60°. Find the height of the pole.

Answer

The sun’s altitude is the angle of elevation of the sun, so shadow = height ÷ tan(altitude): 3h−h/3=50 \sqrt{3}h - h/\sqrt{3} = 50 , h=253≈ h = 25\sqrt{3} \approx 43.30 m (the same numbers as 3 a).

6 b.

When the sun’s altitude is 30°, the shadow of a pole 35 ft high is x ft longer than when the altitude is 45°. Find x.

Answer

x=353−35=35(3−1)≈ x = 35\sqrt{3} - 35 = 35(\sqrt{3} - 1) \approx 25.62 ft.

7 a.

From the roof of a building 30 m high, the angles of depression of the top and the foot of a column are 45° and 60°. Find the height of the column.

Answer

The foot at 60°: distance =30/3=103≈17.32 = 30/\sqrt{3} = 10\sqrt{3} \approx 17.32 m. The top at 45°: it is 17.32 m below the roof. Column = 30 − 17.32 = 12.68 m.

7 b.

The pinnacle of a temple is seen from the roof of a building 20 m high at an angle of depression of 60°, and from the foot of the building at an angle of elevation of 30°. Find the height of the temple.

Answer

Let the temple be h high and d away. From the foot: h=d/3 h = d/\sqrt{3} . From the roof: 20−h=3d 20 - h = \sqrt{3}d . So 20=d(3+1/3)=4d/3 20 = d(\sqrt{3} + 1/\sqrt{3}) = 4d/\sqrt{3} , d=53 d = 5\sqrt{3} , and h = 5 m.

8 a.

A flag of Nepal flies on top of a tower. From a point 100 m from the foot of the tower, the angles of elevation of the top of the tower and the top of the flag are 45° and 60°. Find the height of the flag.

Answer

Tower =100tan⁡45°=100 = 100\tan 45° = 100 m; top of the flag =100tan⁡60°=173.2 = 100\tan 60° = 173.2 m. The flag is 73.2 m tall, counted from the tower’s top to the flag’s top.

8 b.

A man whose eye is 1.8 m above the ground sees the roof of a house at an angle of elevation of 45°, and the top of a window in the house, 20 m above the ground, at 30°. Find the height of the house.

Answer

Above his eye the window’s top is 20 − 1.8 = 18.2 m, so he is 18.2/tan⁡30°=18.23≈31.52 18.2/\tan 30° = 18.2\sqrt{3} \approx 31.52 m from the house. The roof at 45° is 31.52 m above his eye. House = 31.52 + 1.8 = 33.32 m.

9 a.

From the foot and the roof of a house 15 m high, the angles of elevation of the top of a column are 60° and 45°. Find the height of the column.

Answer

With column h and distance d: h=3d h = \sqrt{3}d and h − 15 = d. So (3−1)d=15 (\sqrt{3} - 1)d = 15 , d≈20.49 d \approx 20.49 m, and h=3d≈35.49 h = \sqrt{3}d \approx 35.49 m.

9 b.

An airplane flying level at a height of 6,000 m is seen from a point on the ground at an angle of elevation of 30°. Ten seconds later, from the same point, it is at 45°. Find its speed in km/h.

Answer

Its distance along the ground goes from 60003≈10392.3 6000\sqrt{3} \approx 10392.3 m to 6,000 m: it flies 4392.3 m in 10 s, 439.23 m/s.
439.23×36001000≈1581.23 km/h 439.23 \times \frac{3600}{1000} \approx 1581.23 \text{ km/h}

10 a.

Two poles of equal height stand on opposite sides of a road 30 m wide. From a point on the road between them, the angles of elevation of their tops are 60° and 30°. Find the height of the poles and the point’s distances from them.

Answer

Let the point be d from the first pole: h=3d=(30−d)/3 h = \sqrt{3}d = (30 - d)/\sqrt{3} . So 3d = 30 − d, d = 7.5 m.
Height =7.53≈12.99 = 7.5\sqrt{3} \approx 12.99 m; distances 7.5 m and 22.5 m.

10 b.

A man whose eye is 1.6 m above the ground stands midway between two poles and sees their tops at angles of elevation of 60° and 45°. The shorter pole is 41.6 m high. Find the height of the taller pole and the distance between the poles.

Answer

The shorter pole’s top is 40 m above his eye, at 45°: he is 40 m from it, and so also 40 m from the taller pole. The taller one’s top is 40tan⁡60°=403≈69.28 40\tan 60° = 40\sqrt{3} \approx 69.28 m above his eye.
Taller pole ≈69.28+1.6=70.88 \approx 69.28 + 1.6 = 70.88 m; the poles are 80 m apart.

11 a.

A point divides a pole in the ratio 4 : 5 from the ground. Both parts subtend equal angles at a point 20 m from the foot. Find the height of the pole.

Answer

Let the parts be 4k and 5k, each subtending θ. Then tan θ = 4k/20 and tan 2θ = 9k/20. By the double angle:
2⋅k51−k225=9k20  ⇒  1−k225=89  ⇒  k=53 \frac{2 \cdot \frac{k}{5}}{1 - \frac{k^2}{25}} = \frac{9k}{20} \;\Rightarrow\; 1 - \frac{k^2}{25} = \frac{8}{9} \;\Rightarrow\; k = \frac{5}{3}
The pole is 9k = 15 m high. (The lower part is 623 6\tfrac{2}{3} m.)

11 b.

A flag 5 m high is fixed on top of a tower. From a point on the ground, the tower and the flag subtend angles of 45° and 15°. Find the height of the tower.

Answer

The tower’s top is at 45° and the flag’s top at 45° + 15° = 60°. With tower h: the distance is h, and h+5=3h h + 5 = \sqrt{3}h .
h=5/(3−1)=2.5(3+1)≈6.83 h = 5/(\sqrt{3} - 1) = 2.5(\sqrt{3} + 1) \approx 6.83 m.

12.

Two poles are of equal height. A person standing between them sees each top at an angle of elevation of 45°. After moving 20 m closer to one pole, the angle of elevation of its top becomes 60°.
a. Represent the context in a diagram. b. Find the height of the poles. c. Find the distance between the poles. d. As the observer moves closer to a pole, the angle of elevation increases. Explain why.

Answer

a.

60°45°45°h20 mABCDOP
Figure 7.19
  • b.Both tops at 45° means he is h from each pole, midway. After 20 m: h=3(h−20) h = \sqrt{3}(h - 20) , so h=2033−1=10(3+3)≈ h = \dfrac{20\sqrt{3}}{\sqrt{3} - 1} = 10(3 + \sqrt{3}) \approx 47.32 m.
  • c.2h≈94.64 2h \approx 94.64 m.
  • d.tan θ = h/d: with h fixed, a smaller distance d makes tan θ larger, and a larger tangent means a larger acute angle. Near the foot, the line of sight has to tilt up more steeply to reach the same top.
13.

A ladder 10 m long leans against the outside wall of a building with its top 4 m below the roof. From the foot of the ladder, the angle of elevation of the roof is 60°.
a. Draw a diagram. b. Find the height where the ladder’s top touches the wall. c. Find the distance from the foot of the ladder to the building. d. Find the angle of elevation of the ladder’s top from its foot.

Answer

a.

60°4 m10 mdyRTBF
Figure 7.20

Let the foot be d from the wall and the ladder’s top y up. The roof is 3d \sqrt{3}d high, so y=3d−4 y = \sqrt{3}d - 4 , and y2+d2=102 y^2 + d^2 = 10^2 :
(3d−4)2+d2=100  ⇒  4d2−83d−84=0  ⇒  d2−23d−21=0 (\sqrt{3}d - 4)^2 + d^2 = 100 \;\Rightarrow\; 4d^2 - 8\sqrt{3}d - 84 = 0 \;\Rightarrow\; d^2 - 2\sqrt{3}d - 21 = 0
d=3+3+21=3+26≈6.63 m d = \sqrt{3} + \sqrt{3 + 21} = \sqrt{3} + 2\sqrt{6} \approx 6.63 \text{ m}

  • b.y=3(6.631)−4≈7.49 y = \sqrt{3}(6.631) - 4 \approx 7.49 m.
  • c.About 6.63 m.
  • d.tan⁡θ=7.485/6.631≈1.129 \tan\theta = 7.485/6.631 \approx 1.129 , so θ≈48.5° \theta \approx 48.5° (the angle the ladder makes with the ground).

Miscellaneous exercise

1.

Given 2cos⁡Asin⁡B=12 \sqrt{2}\cos A \sin B = \dfrac{1}{2} and tan⁡A+cot⁡B=2 \tan A + \cot B = 2 :

  • a.Write any two compound angles of A and B.
  • b.Find the value of A − B.
  • c.Is cos (A − B) =  cos A cos B +  sin A sin B true for all values of A and B? Give a reason.
Answer
  • a.A + B and A − B (also 2A + B, and so on).
  • b.Write the second condition with sines and cosines and add the fractions:

sin⁡Acos⁡A+cos⁡Bsin⁡B=sin⁡Asin⁡B+cos⁡Acos⁡Bcos⁡Asin⁡B=cos⁡(A−B)cos⁡Asin⁡B=2 \frac{\sin A}{\cos A} + \frac{\cos B}{\sin B} = \frac{\sin A \sin B + \cos A \cos B}{\cos A \sin B} = \frac{\cos(A - B)}{\cos A \sin B} = 2
So cos⁡(A−B)=2cos⁡Asin⁡B=2⋅122=12 \cos(A - B) = 2\cos A \sin B = 2 \cdot \dfrac{1}{2\sqrt{2}} = \dfrac{1}{\sqrt{2}} , and A − B is 45° or −45°.

Which one? Use the first condition as a sum: cos⁡Asin⁡B=12[sin⁡(A+B)−sin⁡(A−B)]=122 \cos A \sin B = \tfrac{1}{2}[\sin(A + B) - \sin(A - B)] = \dfrac{1}{2\sqrt{2}} , so sin⁡(A+B)=12+sin⁡(A−B) \sin(A + B) = \dfrac{1}{\sqrt{2}} + \sin(A - B) .

  • If A−B=45° A - B = 45° , then sin⁡(A+B)=2 \sin(A + B) = \sqrt{2} , more than 1: impossible.
  • If A−B=−45° A - B = -45° , then sin (A + B) = 0: possible, for example A+B=180° A + B = 180° , so A=6712° A = 67\tfrac{1}{2}° , B=11212° B = 112\tfrac{1}{2}° .

A−B=−45° A - B = -45° (B is 45° more than A). Check with A=6712° A = 67\tfrac{1}{2}° , B=11212° B = 112\tfrac{1}{2}° : 2cos⁡Asin⁡B=0.5 \sqrt{2}\cos A \sin B = 0.5 and tan⁡A+cot⁡B=2.414−0.414=2 \tan A + \cot B = 2.414 - 0.414 = 2 .

45° is what cos⁡(A−B)=12 \cos(A - B) = \frac{1}{\sqrt{2}} alone suggests, and some answer keys give it, but the first condition rules it out.
c. Yes: it is an identity, proved in 7.2 for every A and B, with no condition.

2.

Given cos⁡45°=1/2 \cos 45° = 1/\sqrt{2} :

  • a.Express cos A in terms of sin (A/2).
  • b.Write any two submultiple angles of 15°.
  • c.Prove that sin⁡2212°=122−2 \sin 22\tfrac{1}{2}° = \tfrac{1}{2}\sqrt{2 - \sqrt{2}} .
Answer
  • a.cos⁡A=1−2sin⁡2(A/2) \cos A = 1 - 2\sin^2(A/2) .
  • b.For example 712° 7\tfrac{1}{2}° (half of 15°) and 5° (a third of it).
  • c.sin⁡22212°=1−cos⁡45°2=1−1/22=2−24 \sin^2 22\tfrac{1}{2}° = \dfrac{1 - \cos 45°}{2} = \dfrac{1 - 1/\sqrt{2}}{2} = \dfrac{2 - \sqrt{2}}{4} , and the sine of an acute angle is positive: sin⁡2212°=122−2 \sin 22\tfrac{1}{2}° = \tfrac{1}{2}\sqrt{2 - \sqrt{2}} .
3.

The angles of △ABC are A=120° A = 120° , B=30° B = 30° and C=30° C = 30° .

  • a.Write cos 2A in terms of cos A.
  • b.Verify with these angles that tan A +  tan B +  tan C =  tan A tan B tan C.
  • c.Is tan⁡A=cot⁡B \tan A = \cot B true for this triangle? Give a reason.
Answer
  • a.cos⁡2A=2cos⁡2A−1 \cos 2A = 2\cos^2 A - 1 .
  • b.LHS =−3+13+13=−13 = -\sqrt{3} + \tfrac{1}{\sqrt{3}} + \tfrac{1}{\sqrt{3}} = -\tfrac{1}{\sqrt{3}} ; RHS =−3⋅13⋅13=−13 = -\sqrt{3} \cdot \tfrac{1}{\sqrt{3}} \cdot \tfrac{1}{\sqrt{3}} = -\tfrac{1}{\sqrt{3}} . Equal.
  • c.No. tan⁡120°=−3 \tan 120° = -\sqrt{3} but cot⁡30°=3 \cot 30° = \sqrt{3} . tan⁡A=cot⁡B \tan A = \cot B needs A+B=90° A + B = 90° , and here A+B=150° A + B = 150° .
4.

A ladder 9 m long leans against the outside wall of a house, with its top 9 m below the roof, making an angle of 60° with the ground.

  • a.Show this in a diagram.
  • b.How high up the wall does the ladder reach?
  • c.How far is the foot of the ladder from the house?
  • d.The foot is moved out until the ladder makes 30° with the ground. How far below the roof is its top now?
Answer

a.

60°9 m9 mxRTBF
Figure 7.22
  • b.9sin⁡60°=4.53≈ 9\sin 60° = 4.5\sqrt{3} \approx 7.79 m.
  • c.9cos⁡60°= 9\cos 60° = 4.5 m.
  • d.The roof is 7.79 + 9 = 16.79 m high. At 30° the top is 9sin⁡30°=4.5 9\sin 30° = 4.5 m up, so it is 16.79−4.5≈12.29 16.79 - 4.5 \approx 12.29 m below the roof.
5.

Two poles are of equal height. A person standing midway between them sees the top of each at an angle of elevation of 60°. After walking 60 m towards one pole, he sees the top of the other pole at 45°.

  • a.Define angle of elevation.
  • b.Show this in a diagram.
  • c.Find the height of the poles.
  • d.How are the distance from a pole and the angle of elevation of its top related?
Answer

a. The angle between the horizontal at the eye and the line of sight to an object above eye level.
b.

60°60°45°h60 mABCDOP
Figure 7.23

c. Let each pole be h high and half the gap d: h=3d h = \sqrt{3}d . After 60 m he is d + 60 from the far pole, at 45°: h = d + 60. So 3d=d+60 \sqrt{3}d = d + 60 , d=603−1≈81.96 d = \dfrac{60}{\sqrt{3} - 1} \approx 81.96 m, and
h=d+60≈141.96 h = d + 60 \approx 141.96 m.
d. tan θ = h/d: the tangent of the angle is inversely proportional to the distance. The nearer the pole, the larger the angle (but the angle itself is not inversely proportional: halving the distance does not double it).