Trigonometry: Grade 10 Optional Math

In 1849 a surveyor named James Nicolson carried a theodolite weighing about 500 kg (it took 12 men to carry it) across the plains of northern India and measured the angles up to a far-off summit the survey called Peak XV. He was never closer than 174 km to it, and the mountain was in Nepal, which no surveyor could enter. In 1852, working in the survey’s office, the mathematician Radhanath Sikdar worked through the angles and found that Peak XV was higher than any mountain measured before. We call it Sagarmatha, or Everest.

How can a few angles, measured from far away, give a height? And what happens to sine and cosine when angles are added, doubled, halved or combined? This unit answers both; the height comes back in 7.7.

For the SEE

Trigonometry is a content area of its own in the SEE: 17 of the 75 marks, the most of any unit in this book. In CDC’s model paper it has three 1-mark questions, a question on compound and multiple angles (write sin 2A from sin (A + B), then show $\sin 2A = 2\tan A/(1 + \tan^2 A)$ ), a heights-and-distances question with a diagram (2 + 1 marks), a conditional identity to verify (4 marks) and part of a question joining trigonometry with a straight line’s slope.

Compound angles: sin (A ± B) = sin A cos B ± cos A sin B, $\cos(A \pm B) = \cos A \cos B \mp \sin A \sin B$ , $\tan(A \pm B) = (\tan A \pm \tan B)/(1 \mp \tan A \tan B)$ .
Double angles: sin 2A = 2 sin A cos A; $\cos 2A = \cos^2 A - \sin^2 A = 1 - 2\sin^2 A = 2\cos^2 A - 1$ ; $\tan 2A = 2\tan A/(1 - \tan^2 A)$ .
Triple angles: $\sin 3A = 3\sin A - 4\sin^3 A$ , $\cos 3A = 4\cos^3 A - 3\cos A$ , $\tan 3A = (3\tan A - \tan^3 A)/(1 - 3\tan^2 A)$ .
Half angles: the same formulas with A for 2A and A/2 for A.
Product ↔ sum: 2 sin A cos B = sin (A + B) + sin (A − B) and its three partners, and sin C + sin D = 2 sin {(C + D)/2} cos {(C − D)/2} and its three partners.
In a triangle, $A + B + C = 180°$ : sin (B + C) = sin A, cos (B + C) = − cos A, tan (B + C) = − tan A.
Heights and distances: draw the figure first; tan of the angle of elevation is height over distance.

How the SEE asks it, with two questions worked mark by mark · Exam practice

7.1 Introduction

Trigonometry is the branch of mathematics that studies how the sides and angles of a triangle are related (from the Greek trigonon, “triangle”, and metron, “measure”). In Grade 9 you met its six ratios, sin θ, cos θ, tan θ and their reciprocals $\operatorname{cosec} \theta$ , $\sec \theta$ , $\cot \theta$ , and the identities between them, such as $\sin^2 \theta + \cos^2 \theta = 1$ . In Grade 10 Mathematics you used tan to find heights: Heights and distances. This unit goes further: what the ratios of A + B, 2A or A/2 are, in terms of the ratios of A and B.

Trigonometry grew out of astronomy and building. The Egyptians and Babylonians used the relation between angles and sides to build pyramids and to follow the stars. The Greek astronomer Hipparchus (second century BC) is the first person known to have made a trigonometric table: a table of chords of a circle, the straight line across the circle that an angle at its center cuts off. For this he is called the father of trigonometry. In the second century AD Ptolemy extended his table, for every half degree, in his book the Almagest.

A story from maths

From a bowstring to a pocket: the word sine

Around 499, in India, Aryabhata wrote the Aryabhatiya. It works not with the whole chord but with half of it, the ardha-jya, “half-chord”: the line from a point on the circle straight down to the diameter. That half-chord is what we call the sine. Aryabhata gave a table of sines in a single verse, for every 3¾° from 0° to 90°, and he also used the cosine (kojya). People shortened ardha-jya to jya, “bowstring”: an arc of a circle and its chord look like a bow and its string.

Arabic scholars translated the Indian books and wrote the word as jiba. Arabic is written without its short vowels, so jb was later read as jaib, “pocket” or “fold”. In the twelfth century, translators into Latin wrote jaib as sinus, Latin for a fold, and sinus became the English sine. So the word for a bowstring turned, by a misreading, into the word for a pocket.

In Baghdad, around 830, Habash al-Hasib made the first tables of the tangent and cotangent. Much later, in 1748, Leonhard Euler wrote sine and cosine as functions of a number, the form we use today.

A white statue of a standing man with long hair and a cloth over one shoulder, pointing ahead with his right arm, on a lawn in front of a building — **Figure 7.1** A statue of Aryabhata at the Inter-University Centre for Astronomy and Astrophysics in Pune, India. No picture of him from his own time survives; the sculptor imagined his face. Photo: Wikimedia Commons, Public domain

7.2 Compound angles

An angle made by adding or subtracting two or more angles is a compound angle. If A and B are angles, A + B and A − B are compound angles; so are 2A + B and A + B − C.

Why sin(A + B) is not sin A + sin B

Try $A = B = 30°$ . Then $\sin(A + B) = \sin 60° = \sqrt{3}/2 \approx 0.866$ , but sin A + sin B = 1/2 + 1/2 = 1. Try $A = B = 60°$ : $\sin 120° \approx 0.866$ , but $\sin 60° + \sin 60° \approx 1.732$ , more than any sine can ever be. A sine is a ratio of lengths in a turning picture, not a quantity you can add. So we need the true rule for sin (A + B) in terms of sin A, cos A, sin B and cos B. Finding it, once, gives every other formula in this unit.

7.2.1 Trigonometric ratios of compound angles A + B and A − B

cos(A + B) and cos(A − B)

Draw a circle of radius 1 (a unit circle) with its center at the origin O. A point on it whose radius is turned through an angle θ from the positive X-axis has coordinates ( cos θ, sin θ): that is what cosine and sine mean on a circle of radius 1.

Figure 7.3 P(1, 0), Q( cos A, sin A), R( cos (A + B), sin (A + B)) and S( cos B, − sin B) on a circle of radius 1. The chords PR and QS (blue) cut off the same angle, A + B, at the center, so they are equal.

Let ∠POQ = A and ∠QOR = B, so ∠POR = A + B. Turn OS through B the other way, below the X-axis, so ∠POS = B and ∠SOQ = A + B. The four points are:

$P(1, 0), \quad Q(\cos A, \sin A), \quad R(\cos(A + B), \sin(A + B)), \quad S(\cos B, -\sin B)$

(S is at angle −B, and cos (−B) = cos B, sin (−B) = − sin B.)

By the distance formula,

$\begin{aligned} PR^2 &= \{\cos(A + B) - 1\}^2 + \sin^2(A + B) \\ &= \cos^2(A + B) - 2\cos(A + B) + 1 + \sin^2(A + B) \\ &= 2 - 2\cos(A + B) \end{aligned}$

since $\cos^2(A + B) + \sin^2(A + B) = 1$ . And

$\begin{aligned} QS^2 &= (\cos A - \cos B)^2 + (\sin A + \sin B)^2 \\ &= \cos^2 A - 2\cos A \cos B + \cos^2 B + \sin^2 A + 2\sin A \sin B + \sin^2 B \\ &= 2 - 2\cos A \cos B + 2\sin A \sin B \end{aligned}$

The chords PR and QS both cut off the angle A + B at the center, so they are equal: $PR^2 = QS^2$ . So

$2 - 2\cos(A + B) = 2 - 2\cos A \cos B + 2\sin A \sin B$

$\boxed{\cos(A + B) = \cos A \cos B - \sin A \sin B}$

Put −B for B, with cos (−B) = cos B and sin (−B) = − sin B:

$\boxed{\cos(A - B) = \cos A \cos B + \sin A \sin B}$

Angle A: 50°

Angle B: 30°

Chords PR and QS: 1.286 and 1.286
cos(A + B), R’s x: 0.174
cos A cos B − sin A sin B: 0.174
cos A + cos B: 1.509

The two chords stay equal, so the formula always gives R’s x-coordinate, 0.174. Adding the cosines gives 1.509, outside −1 to 1, where no cosine can be.

The two chords stay equal, so the formula always gives R’s x-coordinate, 0.174. Adding the cosines gives 1.509, the same here only by chance.

The two chords stay equal, so the formula always gives R’s x-coordinate, 0.174. Adding the cosines gives 1.509, a different number.

The two chords stay equal, so the formula always gives R’s x-coordinate, −0.000. Adding the cosines gives −0.000, outside −1 to 1, where no cosine can be.

The two chords stay equal, so the formula always gives R’s x-coordinate, −0.000. Adding the cosines gives −0.000, the same here only by chance.

The two chords stay equal, so the formula always gives R’s x-coordinate, −0.000. Adding the cosines gives −0.000, a different number.

The widget shows the proof’s one idea: however you turn A and B, the two chords stay equal. Watch the last readout: cos A + cos B is almost never cos (A + B), and it can even be more than 1 or less than −1, which no cosine can be.

sin(A + B) and sin(A − B)

Use $\sin\theta = \cos(90° - \theta)$ and the cosine formula just proved:

$\begin{aligned} \sin(A + B) &= \cos\{90° - (A + B)\} = \cos\{(90° - A) - B\} \\ &= \cos(90° - A)\cos B + \sin(90° - A)\sin B \\ &= \sin A \cos B + \cos A \sin B \end{aligned}$

$\boxed{\sin(A + B) = \sin A \cos B + \cos A \sin B}$

Put −B for B:

$\boxed{\sin(A - B) = \sin A \cos B - \cos A \sin B}$

tan(A + B) and tan(A − B)

Divide the sine by the cosine, then divide the top and the bottom by cos A cos B:

$\begin{aligned} \tan(A + B) &= \frac{\sin A \cos B + \cos A \sin B}{\cos A \cos B - \sin A \sin B} \\ &= \frac{\dfrac{\sin A \cos B}{\cos A \cos B} + \dfrac{\cos A \sin B}{\cos A \cos B}}{\dfrac{\cos A \cos B}{\cos A \cos B} - \dfrac{\sin A \sin B}{\cos A \cos B}} \\ &= \frac{\tan A + \tan B}{1 - \tan A \tan B} \end{aligned}$

and, putting −B for B (tan (−B) = − tan B),

$\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$

In the same way, dividing the top and bottom of cos (A ± B)/ sin (A ± B) by sin A sin B:

$\cot(A + B) = \frac{\cot A \cot B - 1}{\cot B + \cot A}, \qquad \cot(A - B) = \frac{\cot A \cot B + 1}{\cot B - \cot A}$

In short

The compound angle formulas

$\begin{aligned} \sin(A + B) &= \sin A \cos B + \cos A \sin B \\ \sin(A - B) &= \sin A \cos B - \cos A \sin B \\ \cos(A + B) &= \cos A \cos B - \sin A \sin B \\ \cos(A - B) &= \cos A \cos B + \sin A \sin B \end{aligned}$
$\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}, \qquad \tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$
To remember them: sine keeps the sign and mixes (sin cos, cos sin); cosine keeps the letters together (cos cos, sin sin) and changes the sign.

Did you know?

Where you meet compound angles

Turning a picture. When a phone turns a photo through an angle B, a point at angle A from the center, at distance r, goes to angle A + B. Its new coordinates are r cos (A + B) and r sin (A + B), which the compound-angle formulas turn into the rotation rule x′ = x cos B − y sin B, y′ = x sin B + y cos B. Two turns one after the other add their angles, as in Unit 9; these formulas are why.
Ptolemy’s table. Ptolemy built his table of chords, every half degree, with the equivalent of these very formulas, worked out from a theorem about four points on a circle (Ptolemy’s theorem). With the chords of 72° and 60°, which geometry gives exactly, he could find the chord of 12°, then halve it again and again.

Which of these equals cos (A − B)?

Pick an answer to see why.

Worked examples

Exercise 7.1

Question 1

A. Which one is equal to sin (A + B)?

Pick an answer to see why.

B. Which is equal to cos (A − B)?

Pick an answer to see why.

C. If $A = 30°$ and $B = 60°$ , what is the value of sin (A + B)?

Pick an answer to see why.

D. If sin (A + B) = 1/2 and $B = 30°$ , what is the value of A?

Pick an answer to see why.

E. Which of the following relations is correct?

Pick an answer to see why.

Question 2

Prove, without a calculator or table:

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

Answer

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

Answer

$\tan 75° = 2 + \sqrt{3}$

Answer

$\cos 105° = \dfrac{1 - \sqrt{3}}{2\sqrt{2}}$

Answer

$\tan 105° = -(2 + \sqrt{3})$

Answer

$\cot 15° = 2 + \sqrt{3}$

Answer

Question 3

If sin A = 3/5 and cos B = 12/13 (A and B acute), find the value of each. First, from the 3, 4, 5 and 5, 12, 13 triangles, cos A = 4/5 and sin B = 5/13.

sin (A + B)

Answer

cos (A + B)

Answer

sin (A − B)

Answer

cos (A − B)

Answer

tan (A + B)

Answer

$\cot(A - B)$

Answer

Question 4

Prove that:

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

Answer

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

Answer

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

Answer

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

Answer

Question 5

Prove, without a calculator or table:

$\sin 75° - \sin 15° = \dfrac{1}{\sqrt{2}}$

Answer

$\cos 15° - \sin 15° = \dfrac{1}{\sqrt{2}}$

Answer

$\sin 105° - \cos 75° = \dfrac{1}{\sqrt{2}}$

Answer

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

Answer

$\tan 15° + \cot 15° = 4$

Answer

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

Answer

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

Answer

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

Answer

Question 6

Prove that:

$\sin(25° + \alpha)\cos(65° - \alpha) + \cos(25° + \alpha)\sin(65° - \alpha) = 1$

Answer

$\cos(25° + \alpha)\cos(65° + \alpha) - \sin(25° + \alpha)\sin(65° + \alpha) = -\sin 2\alpha$

Answer

$\cos(60° + \beta) - \cos\beta + \cos(60° - \beta) = 0$

Answer

$\dfrac{\tan(45° + \phi) - \tan\phi}{1 + \tan(45° + \phi)\tan\phi} = 1$

Answer

Question 7

If $A + B = (\pi/4)^c$ , prove that:

tan A + tan B + tan A tan B = 1

Answer

(1 + tan A)(1 + tan B) = 2

Answer

Question 8

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

Answer

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

Answer

Question 9

Prove that:

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

Answer

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

Answer

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

Answer

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

Answer

Question 10

Prove that:

$\dfrac{\tan 5A - \tan 4A}{1 + \tan 5A \tan 4A} = \tan A$

Answer

$\dfrac{\tan 7A - \tan 4A}{1 + \tan 7A \tan 4A} = \tan 3A$

Answer

$\dfrac{\tan(A + B) - \tan B}{1 + \tan(A + B)\tan B} = \tan A$

Answer

$\tan(A + B)\tan(A - B) = \dfrac{\tan^2 A - \tan^2 B}{1 - \tan^2 A \tan^2 B}$

Answer

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

Answer

$\dfrac{\cos(A + B)}{\sin A \sin B} = \cot A \cot B - 1$

Answer

Question 11

Prove that:

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

Answer

$1 - \tan 27° - \tan 18° - \tan 27° \tan 18° = 0$

Answer

$\tan 65° - \tan 20° = 1 + \tan 65° \tan 20°$

Answer

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

Answer

Question 12

Prove that:

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

Answer

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

Answer

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

Answer

$\cot 5A \cot 4A - \cot 9A \cot 5A - \cot 9A \cot 4A = 1$

Answer

Question 13

Prove that:

$\tan 50° - \tan 40° = 2\tan 10°$

Answer

$\tan 70° = 2\tan 50° + \tan 20°$

Answer

Question 14

Prove that:

$\dfrac{\sin(\beta + \gamma)}{\sin(\beta - \gamma)} = \dfrac{\tan\beta + \tan\gamma}{\tan\beta - \tan\gamma}$

Answer

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

Answer

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

Answer

Question 15

Prove that:

$\dfrac{\sin(A - B)}{\cos A \cos B} + \dfrac{\sin(B - C)}{\cos B \cos C} + \dfrac{\sin(C - A)}{\cos C \cos A} = 0$

Answer

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

Answer

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

Answer

$\tan(A + B + C) = \dfrac{\tan A + \tan B + \tan C - \tan A \tan B \tan C}{1 - \tan A \tan B - \tan B \tan C - \tan C \tan A}$

Answer

7.3 Multiple angles

If you know $\sin 15°$ , can you find $\sin 30°$ from it? It is $\sin(2 \times 15°)$ , but it is not $2\sin 15°$ (that would be about 0.518, not 0.5). And $\sin 45°$ is $\sin(3 \times 15°)$ , not $3\sin 15°$ .

An angle made by multiplying an angle by 2, 3, 4, … is a multiple angle of it: 2A, 3A, 4A are multiple angles of A. A multiple angle is just a compound angle whose two parts are equal, 2A = A + A, so every formula below comes from 7.2 by putting B = A.

7.3.1 Trigonometric ratios of multiple angles

1. The ratios of 2A

$\sin 2A = \sin(A + A) = \sin A \cos A + \cos A \sin A = 2\sin A \cos A$

$\cos 2A = \cos(A + A) = \cos A \cos A - \sin A \sin A = \cos^2 A - \sin^2 A$

Replace $\cos^2 A$ by $1 - \sin^2 A$ , or $\sin^2 A$ by $1 - \cos^2 A$ , for two more forms:

$\cos 2A = 1 - 2\sin^2 A, \qquad \cos 2A = 2\cos^2 A - 1$

And with B = A in tan (A + B):

$\tan 2A = \frac{\tan A + \tan A}{1 - \tan A \tan A} = \frac{2\tan A}{1 - \tan^2 A}$

The two last forms of cos 2A are worth turning round, because they turn a square into a plain cosine:

$\sin^2 A = \frac{1 - \cos 2A}{2}, \qquad \cos^2 A = \frac{1 + \cos 2A}{2}$

2. sin 2A and cos 2A in terms of tan A

Write 2 sin A cos A over $1 = \cos^2 A + \sin^2 A$ , then divide the top and the bottom by $\cos^2 A$ :

$\sin 2A = \frac{2\sin A \cos A}{\cos^2 A + \sin^2 A} = \frac{2\tan A}{1 + \tan^2 A}$

Another way, with $\cos^2 A = 1/\sec^2 A$ :

$2\sin A \cos A = 2\frac{\sin A}{\cos A}\cos^2 A = \frac{2\tan A}{\sec^2 A} = \frac{2\tan A}{1 + \tan^2 A}$

In the same way,

$\cos 2A = \frac{\cos^2 A - \sin^2 A}{\cos^2 A + \sin^2 A} = \frac{1 - \tan^2 A}{1 + \tan^2 A}$

So if you know only tan A, you know all three ratios of 2A. With tan A = t, draw a right-angled triangle with sides 2t and $1 - t^2$ : its hypotenuse is $1 + t^2$ , since $(2t)^2 + (1 - t^2)^2 = (1 + t^2)^2$ .

3. The ratios of 3A

$\begin{aligned} \sin 3A &= \sin(A + 2A) = \sin A \cos 2A + \cos A \sin 2A \\ &= \sin A(1 - 2\sin^2 A) + \cos A \cdot 2\sin A \cos A \\ &= \sin A - 2\sin^3 A + 2\sin A(1 - \sin^2 A) \\ &= 3\sin A - 4\sin^3 A \end{aligned}$

$\begin{aligned} \cos 3A &= \cos(A + 2A) = \cos A \cos 2A - \sin A \sin 2A \\ &= \cos A(2\cos^2 A - 1) - \sin A \cdot 2\sin A \cos A \\ &= 2\cos^3 A - \cos A - 2\cos A(1 - \cos^2 A) \\ &= 4\cos^3 A - 3\cos A \end{aligned}$

$\begin{aligned} \tan 3A &= \tan(A + 2A) = \frac{\tan A + \dfrac{2\tan A}{1 - \tan^2 A}}{1 - \tan A \cdot \dfrac{2\tan A}{1 - \tan^2 A}} \\ &= \frac{\tan A(1 - \tan^2 A) + 2\tan A}{1 - \tan^2 A - 2\tan^2 A} = \frac{3\tan A - \tan^3 A}{1 - 3\tan^2 A} \end{aligned}$

In short

The multiple angle formulas

$\begin{aligned} &\sin 2A = 2\sin A \cos A = \frac{2\tan A}{1 + \tan^2 A} \\ &\cos 2A = \cos^2 A - \sin^2 A = 1 - 2\sin^2 A = 2\cos^2 A - 1 = \frac{1 - \tan^2 A}{1 + \tan^2 A} \\ &\tan 2A = \frac{2\tan A}{1 - \tan^2 A} \\ &\sin 3A = 3\sin A - 4\sin^3 A, \qquad \cos 3A = 4\cos^3 A - 3\cos A \\ &\tan 3A = \frac{3\tan A - \tan^3 A}{1 - 3\tan^2 A} \end{aligned}$
sin 3A starts with 3 sin; cos 3A starts with 4 cos cubed.

If tan A = 1/2, what is sin 2A?

Pick an answer to see why.

Worked examples

Exercise 7.2

Question 1

A. Which of the following is equal to cos 2A?

Pick an answer to see why.

B. Which of the following is equal to sin 3θ?

Pick an answer to see why.

C. If $A = 30°$ , what is the value of sin 2A?

Pick an answer to see why.

D. What should the value of A be when cos 3A = −1?

Pick an answer to see why.

E. Which of the following is equal to tan 2A?

Pick an answer to see why.

Question 2

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

Answer

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

Answer

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

Answer

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

Answer

Question 3

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

Answer

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

Answer

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

Answer

Question 4

Prove that:

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

Answer

$\dfrac{1 - \cos 2A}{\sin 2A} = \tan A$

Answer

$\dfrac{1 - \cos 2A}{1 + \cos 2A} = \tan^2 A$

Answer

$\dfrac{1 - \tan\theta}{1 + \tan\theta} = \dfrac{1 - \sin 2\theta}{\cos 2\theta}$

Answer

$\dfrac{\cos 2\alpha}{1 + \sin 2\alpha} = \dfrac{1 - \tan\alpha}{1 + \tan\alpha}$

Answer

$\dfrac{1 + \sin 2\theta - \cos 2\theta}{1 + \sin 2\theta + \cos 2\theta} = \tan\theta$

Answer

$\dfrac{\sin\theta + \sin 2\theta}{1 + \cos\theta + \cos 2\theta} = \tan\theta$

Answer

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

Answer

$\cot A - \tan A = 2\cot 2A$

Answer

Question 5

Prove that:

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

Answer

$\dfrac{1 - \sin 2\theta}{\cos 2\theta} = \dfrac{\cos\theta - \sin\theta}{\cos\theta + \sin\theta}$

Answer

$2\sin^2(45° - A) = 1 - \sin 2A$

Answer

$\tan(45° + \theta) = \sec 2\theta + \tan 2\theta$

Answer

$2\cos^2(45° - A) = 1 + \sin 2A$

Answer

$\cos^2(45° - \theta) - \sin^2(45° - \theta) = \sin 2\theta$

Answer

$\dfrac{\sin 5A}{\sin A} - \dfrac{\cos 5A}{\cos A} = 4\cos 2A$

Answer

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

Answer

Question 6

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

Answer

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

Answer

If $\sin\beta = \frac{1}{2}\left(a + \frac{1}{a}\right)$ , prove that $\sin 3\beta = -\frac{1}{2}\left(a^3 + \frac{1}{a^3}\right)$ .

Answer

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

Answer

Question 7

Prove that:

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

Answer

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

Answer

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

Answer

Question 8

Prove that:

$\sin^2 A - \cos^2 A \cos 2B = \sin^2 B - \cos^2 B \cos 2A$

Answer

$\cos^2\alpha + \sin^2\alpha\cos 2\beta = \cos^2\beta + \sin^2\beta\cos 2\alpha$

Answer

Question 9

Prove that:

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

Answer

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

Answer

Question 10

Prove that:

$\dfrac{\cos^3 A + \sin^3 A}{\cos A + \sin A} = 1 - \dfrac{1}{2}\sin 2A$

Answer

$\cos^6\theta - \sin^6\theta = \cos 2\theta\left(1 - \frac{1}{4}\sin^2 2\theta\right)$

Answer

$4(\sin^6\theta + \cos^6\theta) = 4 - 3\sin^2 2\theta$

Answer

$\cos^6\theta - \sin^6\theta = \frac{1}{4}(\cos^3 2\theta + 3\cos 2\theta)$

Answer

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

Answer

$\sin^8\theta + \cos^8\theta = \frac{1}{8}(8 - 8\sin^2 2\theta + \sin^4 2\theta)$

Answer

Question 11

Prove that:

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

Answer

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

Answer

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

Answer

Question 12

Prove that:

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

Answer

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

Answer

$\dfrac{\sec 4A - 1}{\sec 2A - 1} = \tan 4A \cot A$

Answer

$\dfrac{\sec 8A - 1}{\sec 4A - 1} = \tan 8A \cot 2A$

Answer

Question 13

Prove that:

$\dfrac{1}{\tan 3\theta + \tan\theta} - \dfrac{1}{\cot 3\theta + \cot\theta} = \cot 4\theta$

Answer

$\dfrac{\cot\theta}{\cot\theta - \cot 3\theta} - \dfrac{\tan\theta}{\tan 3\theta - \tan\theta} = 1$

Answer

Question 14

Prove that:

$4(\cos^3 10° + \sin^3 20°) = 3(\cos 10° + \sin 20°)$

Answer

$4(\cos^3 20° + \sin^3 50°) = 3(\cos 20° + \sin 50°)$

Answer

$\cos^3 A \cos 3A + \sin^3 A \sin 3A = \cos^3 2A$

Answer

Question 15

Prove that:

$\tan\theta + \tan(60° + \theta) - \tan(60° - \theta) = 3\tan 3\theta$

Answer

$\cot\theta + \cot(60° + \theta) - \cot(60° - \theta) = 3\cot 3\theta$

Answer

$\tan\theta + 2\tan 2\theta + 4\tan 4\theta + 8\cot 8\theta = \cot\theta$

Answer

Question 16

Express cos 4θ in terms of sin θ.

Answer

Express cos 5θ in terms of cos θ.

Answer

Question 17

Prove that:

$\dfrac{\cos A}{\cos A - \sin A} - \dfrac{\cos A}{\cos A + \sin A} = \tan 2A$

Answer

$\tan 2\theta + \sec 2\theta = \dfrac{\cos\theta + \sin\theta}{\cos\theta - \sin\theta}$

Answer

$\dfrac{\cot 2\theta + \tan 2\theta}{\cot 2\theta - \tan 2\theta} = \sec 4\theta$

Answer

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

Answer

Question 18

If 2 tan α = 3 tan β, prove that:

$\tan(\alpha + \beta) = \dfrac{5\sin 2\beta}{5\cos 2\beta - 1}$

Answer

$\tan(\alpha - \beta) = \dfrac{\sin 2\beta}{5 - \cos 2\beta}$

Answer

Question 19

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

Answer

Question 20

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

Answer

Question 21

Prove that $\sin^4 A = \dfrac{3}{8} - \dfrac{1}{2}\cos 2A + \dfrac{1}{8}\cos 4A$ .

Answer

Question 22

Prove that $\sin^4\dfrac{\pi}{8} + \sin^4\dfrac{3\pi}{8} + \sin^4\dfrac{5\pi}{8} + \sin^4\dfrac{7\pi}{8} = \dfrac{3}{2}$ .

Answer

7.4 Submultiple angles

If you know $\sin 60°$ , how do you find $\sin 30°$ , half of 60°? Or $\sin 15°$ , half of 30°, or $\sin 22\tfrac{1}{2}°$ , half of 45°? 7.3 doubled angles; now we halve them.

An angle made by dividing an angle into 2, 3, 4, … equal parts is a submultiple angle of it: A/2, A/3, A/4 are submultiple angles of A. 15° and 20° are submultiple angles of 60° (a quarter and a third of it).

7.4.1 sin A, cos A and tan A in terms of A/2

Every formula of 7.3 holds for any angle, so it holds with A/2 in place of A: then 2A becomes A. For example, $\sin A = \sin\left(\frac{A}{2} + \frac{A}{2}\right) = 2\sin\frac{A}{2}\cos\frac{A}{2}$ . In the same way:

$\begin{aligned} &\text{a. } \sin A = 2\sin\frac{A}{2}\cos\frac{A}{2} &&\text{b. } \cos A = \cos^2\frac{A}{2} - \sin^2\frac{A}{2} \\ &\text{c. } \cos A = 1 - 2\sin^2\frac{A}{2} &&\text{d. } \cos A = 2\cos^2\frac{A}{2} - 1 \end{aligned}$

$\text{e. } \tan A = \frac{2\tan\frac{A}{2}}{1 - \tan^2\frac{A}{2}}, \qquad \text{f. } \sin A = \frac{2\tan\frac{A}{2}}{1 + \tan^2\frac{A}{2}}, \qquad \text{g. } \cos A = \frac{1 - \tan^2\frac{A}{2}}{1 + \tan^2\frac{A}{2}}$

Turning c and d round gives the half-angle formulas, the way to halve an angle you know:

$\sin\frac{A}{2} = \pm\sqrt{\frac{1 - \cos A}{2}}, \qquad \cos\frac{A}{2} = \pm\sqrt{\frac{1 + \cos A}{2}}, \qquad \tan\frac{A}{2} = \frac{\sin A}{1 + \cos A}$

The sign is the sign of the ratio in the quadrant where A/2 lies: for an acute half angle, both are positive.

7.4.2 sin A, cos A and tan A in terms of A/3

With A/3 for A in the triple-angle formulas:

$\sin A = 3\sin\frac{A}{3} - 4\sin^3\frac{A}{3}, \qquad \cos A = 4\cos^3\frac{A}{3} - 3\cos\frac{A}{3}, \qquad \tan A = \frac{3\tan\frac{A}{3} - \tan^3\frac{A}{3}}{1 - 3\tan^2\frac{A}{3}}$

In short

Multiple and submultiple angles, side by side

Multiple angles	Submultiple angles
sin 2A = 2 sin A cos A	$\sin A = 2\sin\frac{A}{2}\cos\frac{A}{2}$
$\cos 2A = \cos^2 A - \sin^2 A$	$\cos A = \cos^2\frac{A}{2} - \sin^2\frac{A}{2}$
$\cos 2A = 1 - 2\sin^2 A$	$\cos A = 1 - 2\sin^2\frac{A}{2}$
$\cos 2A = 2\cos^2 A - 1$	$\cos A = 2\cos^2\frac{A}{2} - 1$
$\sin 2A = \frac{2\tan A}{1 + \tan^2 A}$	$\sin A = \frac{2\tan(A/2)}{1 + \tan^2(A/2)}$
$\cos 2A = \frac{1 - \tan^2 A}{1 + \tan^2 A}$	$\cos A = \frac{1 - \tan^2(A/2)}{1 + \tan^2(A/2)}$
$\tan 2A = \frac{2\tan A}{1 - \tan^2 A}$	$\tan A = \frac{2\tan(A/2)}{1 - \tan^2(A/2)}$
$\sin 3A = 3\sin A - 4\sin^3 A$	$\sin A = 3\sin\frac{A}{3} - 4\sin^3\frac{A}{3}$
$\cos 3A = 4\cos^3 A - 3\cos A$	$\cos A = 4\cos^3\frac{A}{3} - 3\cos\frac{A}{3}$
$\tan 3A = \frac{3\tan A - \tan^3 A}{1 - 3\tan^2 A}$	$\tan A = \frac{3\tan(A/3) - \tan^3(A/3)}{1 - 3\tan^2(A/3)}$

Which formula finds $\cos 15°$ from $\cos 30°$ ?

Pick an answer to see why.

Worked examples

Exercise 7.3

Question 1

A. Which of the following is equal to sin A?

Pick an answer to see why.

B. Which of the following is equal to cos θ?

Pick an answer to see why.

C. If $A = 60°$ , what is the value of sin (A/2)?

Pick an answer to see why.

D. Which is a submultiple angle of 60°?

Pick an answer to see why.

E. Which is equal to tan A?

Pick an answer to see why.

Question 2

Define a submultiple angle, with an example.

Answer

Express sin θ in terms of tan (θ/2).

Answer

Express cos θ in terms of cos (θ/2).

Answer

Express cos θ in terms of tan (θ/2).

Answer

Express sin θ in terms of sin (θ/3).

Answer

Question 3

Take the given angle (A/2, θ/3) as acute.

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

Answer

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

Answer

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

Answer

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

Answer

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

Answer

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

Answer

Question 4

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

Answer

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

Answer

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

Answer

Question 5

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

Answer

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

Answer

Question 6

Prove that:

If $\cos\frac{\theta}{2} = \frac{1}{2}\left(a + \frac{1}{a}\right)$ , then $\cos\theta = \frac{1}{2}\left(a^2 + \frac{1}{a^2}\right)$ .

Answer

If $\sin\frac{A}{2} = \frac{1}{2}\left(a + \frac{1}{a}\right)$ , then $\cos A = -\frac{1}{2}\left(a^2 + \frac{1}{a^2}\right)$ .

Answer

If $\sin\frac{\beta}{3} = \frac{1}{2}\left(a + \frac{1}{a}\right)$ , then $\sin\beta = -\frac{1}{2}\left(a^3 + \frac{1}{a^3}\right)$ .

Answer

If $\cos\frac{\beta}{3} = \frac{1}{2}\left(a + \frac{1}{a}\right)$ , then $\cos\beta = \frac{1}{2}\left(a^3 + \frac{1}{a^3}\right)$ .

Answer

Question 7

Prove that (with s = sin (A/2), c = cos (A/2) where it helps):

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

Answer

$\dfrac{\sin A}{1 + \cos A} = \tan\dfrac{A}{2}$

Answer

$\dfrac{1 - \cos A}{\sin A} = \tan\dfrac{A}{2}$

Answer

$\dfrac{\sin A}{1 - \cos A} = \cot\dfrac{A}{2}$

Answer

$\dfrac{1 + \sin A}{\cos A} = \dfrac{\cos\frac{A}{2} + \sin\frac{A}{2}}{\cos\frac{A}{2} - \sin\frac{A}{2}}$

Answer

$\cos A = \dfrac{1 - \tan^2\frac{A}{2}}{1 + \tan^2\frac{A}{2}}$

Answer

$\dfrac{1 - \tan\frac{\theta}{2}}{1 + \tan\frac{\theta}{2}} = \dfrac{1 - \sin\theta}{\cos\theta}$

Answer

$\dfrac{\cos\alpha}{1 + \sin\alpha} = \dfrac{1 - \tan\frac{\alpha}{2}}{1 + \tan\frac{\alpha}{2}}$

Answer

$\dfrac{1 + \sin\theta - \cos\theta}{1 + \sin\theta + \cos\theta} = \tan\dfrac{\theta}{2}$

Answer

$\dfrac{\sin\theta - \cos\frac{\theta}{2}}{1 - \sin\frac{\theta}{2} - \cos\theta} = \cot\dfrac{\theta}{2}$

Answer

$\sin A + \tan A = \dfrac{4\tan\frac{A}{2}}{1 - \tan^4\frac{A}{2}}$

Answer

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

Answer

Question 8

Prove that:

$\tan\left(45° + \dfrac{A}{2}\right) = \dfrac{\cos A}{1 - \sin A}$

Answer

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

Answer

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

Answer

$\cos^2\left(45° - \dfrac{\theta}{2}\right) - \sin^2\left(45° - \dfrac{\theta}{2}\right) = \sin\theta$

Answer

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

Answer

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

Answer

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

Answer

$\dfrac{\sin 2A}{1 + \cos 2A} \times \dfrac{\cos A}{1 + \cos A} = \tan\dfrac{A}{2}$

Answer

Question 9

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

Answer

7.5 Transformation of trigonometric identities

Sometimes a product such as $2\sin 50° \cos 30°$ is easier to work with as a sum, and sometimes a sum such as $\sin 50° + \sin 30°$ is easier as a product. This section turns one into the other. Both come straight from the four compound-angle formulas, by adding and subtracting them.

7.5.1 Transformation of a product into a sum or difference

Write the two sine formulas one under the other:

$\begin{aligned} \sin(A + B) &= \sin A \cos B + \cos A \sin B \quad \ldots \text{(i)} \\ \sin(A - B) &= \sin A \cos B - \cos A \sin B \quad \ldots \text{(ii)} \end{aligned}$

Adding (i) and (ii), the cos A sin B terms cancel; subtracting (ii) from (i), the sin A cos B terms cancel:

$2\sin A \cos B = \sin(A + B) + \sin(A - B) \quad \ldots \text{(iii)}$
$2\cos A \sin B = \sin(A + B) - \sin(A - B) \quad \ldots \text{(iv)}$

Do the same with the two cosine formulas:

$\begin{aligned} \cos(A + B) &= \cos A \cos B - \sin A \sin B \quad \ldots \text{(v)} \\ \cos(A - B) &= \cos A \cos B + \sin A \sin B \quad \ldots \text{(vi)} \end{aligned}$

$2\cos A \cos B = \cos(A + B) + \cos(A - B) \quad \ldots \text{(vii)}$
$2\sin A \sin B = \cos(A - B) - \cos(A + B) \quad \ldots \text{(viii)}$

Mind the order in (viii): the difference angle comes first. (Written the other way, cos (A + B) − cos (A − B) = −2 sin A sin B.)

7.5.2 Transformation of a sum or difference into a product

Read the four formulas from right to left. Let

$A + B = C \quad \text{and} \quad A - B = D$

Adding and subtracting these, A = (C + D)/2 and B = (C − D)/2. Put them into (iii), (iv), (vii) and (viii):

In short

Sum to product

$\begin{aligned} \sin C + \sin D &= 2\sin\frac{C + D}{2}\cos\frac{C - D}{2} \\ \sin C - \sin D &= 2\cos\frac{C + D}{2}\sin\frac{C - D}{2} \\ \cos C + \cos D &= 2\cos\frac{C + D}{2}\cos\frac{C - D}{2} \\ \cos C - \cos D &= -2\sin\frac{C + D}{2}\sin\frac{C - D}{2} \end{aligned}$
In words: half the sum, half the difference. Only the last has a minus sign; it can also be written $\cos C - \cos D = 2\sin\frac{C + D}{2}\sin\frac{D - C}{2}$ .

A story from maths

Multiplying by adding: how astronomers used these formulas

Around 1580, before logarithms, astronomers had to multiply long numbers by hand thousands of times to work out where the planets would be. A trick called prosthaphaeresis (Greek for “adding and taking away”) turned each multiplication into an addition, using exactly the product-to-sum formula:

$\cos A \cos B = \tfrac{1}{2}\{\cos(A + B) + \cos(A - B)\}$

To multiply two numbers, look up the angles whose cosines they are in a table, add and subtract the angles, look up two cosines, and halve their sum. The Danish astronomer Tycho Brahe used it for his famous measurements of the planets. For about 25 years it was the only general shortcut for multiplying, until John Napier’s logarithms (1614) replaced it.

A hand-colored old print: a huge quarter-circle scale fixed to a wall, an assistant at its right edge looking along it, a bearded man in the middle pointing upward, and rooms of an observatory painted behind them — **Figure 7.4** Tycho Brahe’s great mural quadrant, a quarter circle fixed to a wall of his observatory: the assistant at the right looks along it, through an opening high in the wall, to read the angle of a star. Tycho himself is painted on the wall inside the arc. Every angle measured with it went into long calculations. Photo: Unknown author, Public domain

What is $\sin 70° + \sin 50°$ as a product?

Pick an answer to see why.

Worked examples

Exercise 7.4

Question 1

A. Which expression is equal to 2 sin A cos B?

Pick an answer to see why.

B. Which expression is equal to sin A − sin B?

Pick an answer to see why.

C. Which expression is equal to cos C − cos D?

Pick an answer to see why.

D. What is cos 6A + cos 4A as a product?

Pick an answer to see why.

E. $\sin 60° + \sin 40°$ is equal to which of these?

Pick an answer to see why.

Question 2

Write each product as a sum or difference.

$\sin 50° \cos 30°$

Answer

$\cos 25° \cos 15°$

Answer

cos 10A sin 6A

Answer

$\sin 27° \sin 32°$

Answer

Question 3

Write each sum or difference as a product.

$\sin 50° + \sin 30°$

Answer

$\cos 25° + \cos 15°$

Answer

cos 7A − cos 5A

Answer

$\sin 67° - \sin 43°$

Answer

Question 4

Prove that:

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

Answer

$\sin 50° - \sin 70° = -\sin 10°$

Answer

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

Answer

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

Answer

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

Answer

Question 5

Prove that:

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

Answer

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

Answer

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

Answer

$\dfrac{\sin A - \sin B}{\sin A + \sin B} = \cot\dfrac{A + B}{2}\tan\dfrac{A - B}{2}$

Answer

$\dfrac{\cos B - \cos A}{\sin A + \sin B} = \tan\dfrac{A - B}{2}$

Answer

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

Answer

$2\sin(45° + \phi)\cos(45° - \phi) = 1 + \sin 2\phi$

Answer

$2\cos(45° + \theta)\cos(45° - \theta) = \cos 2\theta$

Answer

Question 6

Prove that:

$\dfrac{\sin\theta + \sin 3\theta - \sin 2\theta}{\cos\theta - \cos 2\theta + \cos 3\theta} = \tan 2\theta$

Answer

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

Answer

$\dfrac{\sin A - \sin 3A + \sin 5A - \sin 7A}{\cos A - \cos 3A - \cos 5A + \cos 7A} = \cot 2A$

Answer

$\dfrac{\cos 2A \cos 3A - \cos 2A \cos 7A + \cos A \cos 10A}{\sin 4A \sin 3A - \sin 2A \sin 5A + \sin 4A \sin 7A} = \cot 6A \cot 5A$

Answer

Question 7

Prove that:

$4\cos\theta\cos(60° + \theta)\cos(60° - \theta) = \cos 3\theta$

Answer

$\cot\theta\cot(60° + \theta)\cot(60° - \theta) = \cot 3\theta$

Answer

$\sec\left(45° + \dfrac{\theta}{2}\right)\sec\left(45° - \dfrac{\theta}{2}\right) = 2\sec\theta$

Answer

$\tan\left(\dfrac{\pi}{4} + \theta\right) - \tan\left(\dfrac{\pi}{4} - \theta\right) = 2\tan 2\theta$

Answer

Question 8

Prove that:

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

Answer

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

Answer

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

Answer

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

Answer

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

Answer

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

Answer

7.6 Conditional identities

A relation that is true for every angle, such as sin 2θ = 2 sin θ cos θ, is a trigonometric identity. A relation that is true only when the angles satisfy some condition, such as $\tan A = \cot B$ , true when $A + B = 90°$ , is a conditional identity.

The condition you will meet most is that A, B and C are the angles of a triangle, so $A + B + C = 180°$ . Then $B + C = 180° - A$ , and the ratios of an angle and its supplement give:

$\sin(B + C) = \sin A, \qquad \cos(B + C) = -\cos A, \qquad \tan(B + C) = -\tan A$

and, since $\frac{B + C}{2} = 90° - \frac{A}{2}$ ,

$\sin\frac{B + C}{2} = \cos\frac{A}{2}, \qquad \cos\frac{B + C}{2} = \sin\frac{A}{2}$

Proving one for every triangle

Checking with numbers shows that a relation holds for those numbers; a proof shows it holds for every triangle. For example, if $A + B + C = 180°$ , then $A + B = 180° - C$ , and

$\tan(A + B) = -\tan C \;\Rightarrow\; \frac{\tan A + \tan B}{1 - \tan A \tan B} = -\tan C$

$\tan A + \tan B = -\tan C + \tan A \tan B \tan C \;\Rightarrow\; \tan A + \tan B + \tan C = \tan A \tan B \tan C$

(as long as none of the angles is 90°, where tan is not defined). So in every triangle without a right angle, the sum of the three tangents equals their product: a surprising fact you can test with any triangle you draw.

Which of these is a conditional identity, not an identity?

Pick an answer to see why.

Worked examples

Exercise 7.5

Question 1

A. Which of the following statements is correct?

Pick an answer to see why.

B. If tan A = 1, what is the value of A (A acute)?

Pick an answer to see why.

C. Under what condition is sin A = cos B true?

Pick an answer to see why.

D. Which of the following is a conditional identity?

Pick an answer to see why.

E. If $A + B + C = 180°$ , which one is correct?

Pick an answer to see why.

Question 2

The angles of a triangle are $\alpha = 30°$ , $\beta = 90°$ and $\gamma = 60°$ . Verify that:

$\sin(\alpha + \beta) = \sin\gamma$

Answer

$\cos(\beta + \gamma) = -\cos\alpha$

Answer

$\tan(\alpha + \beta) = -\tan\gamma$

Answer

$\cot(\alpha + \gamma) = -\cot\beta$

Answer

$\sin\left(\dfrac{\alpha}{2} + \dfrac{\beta}{2}\right) = \cos\dfrac{\gamma}{2}$

Answer

$\cos\left(\dfrac{\alpha}{2} + \dfrac{\beta}{2}\right) = \sin\dfrac{\gamma}{2}$

Answer

Question 3

If $A = 30°$ , $B = 120°$ and $C = 30°$ , verify that:

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

Answer

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

Answer

$\cot A \cot B + \cot B \cot C + \cot C \cot A = 1$

Answer

$\cot 2A \cot 2B + \cot 2B \cot 2C + \cot 2C \cot 2A = 1$

Answer

$\tan\dfrac{A}{2}\tan\dfrac{B}{2} + \tan\dfrac{B}{2}\tan\dfrac{C}{2} + \tan\dfrac{C}{2}\tan\dfrac{A}{2} = 1$

Answer

Question 4

If the angles of △ABC are $A = 30°$ , $B = 90°$ and $C = 60°$ , verify that:

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

Answer

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

Answer

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

Answer

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

Answer

Question 5

The angles of a triangle are $\alpha = 45°$ , $\beta = 45°$ and $\gamma = 90°$ . Verify that:

$\sin^2\alpha + \sin^2\beta + \sin^2\gamma = 4\sin\alpha\sin\beta\sin\gamma$

Answer

$\sin^2\alpha + \sin^2\beta - \sin^2\gamma = 1 - 2\cos\alpha\cos\beta\sin\gamma$

Answer

$\cos^2\alpha - \cos^2\beta + \cos^2\gamma = 1 - 2\sin\alpha\cos\beta\sin\gamma$

Answer

$\cos^2\alpha - \cos^2\beta - \cos^2\gamma = 2\cos\alpha\sin\beta\sin\gamma - 1$

Answer

Question 6

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

Answer

7.7 Height and distance

How tall is your school building? How far apart are two electric poles across a river? How high is a mountain you can’t climb? Measuring these directly is hard or impossible. But an angle is easy to measure from where you stand, and with the trigonometric ratios of a right-angled triangle an angle and one distance give the rest. In Grade 10 Mathematics you met the method: Heights and distances. Here it is used again, often with two angles at once.

7.7.1 Angle of elevation and angle of depression

When you look at something above your eye level, the angle between your line of sight and the horizontal is the angle of elevation. Looking at the top of a house, a pole or a tree from the ground, you measure an angle of elevation.

Figure 7.5 The angle of elevation θ: from the horizontal at the eye E up to the line of sight EA.

When you look at something below your eye level, the angle between the line of sight and the horizontal is the angle of depression: looking down at a bird in the yard from a roof, or at a car from a tower.

Figure 7.6 The angle of depression from A down to E equals the angle of elevation from E up to A: the two horizontals are parallel, and the line of sight cuts them at alternate angles.

That equality is the key to most questions: an angle of depression measured from the top can be drawn at the bottom, inside the triangle, as an angle of elevation.

Worked examples

The same idea works for any two angles α (from the far point) and β (from the near point), d apart. The near point is h/ tan β from the foot and the far one h/ tan α, so

$d = \frac{h}{\tan\alpha} - \frac{h}{\tan\beta} \;\Rightarrow\; h = \frac{d\tan\alpha\tan\beta}{\tan\beta - \tan\alpha}$

This is how surveyors measure a mountain whose foot they can’t reach: the point straight below the summit is inside the rock.

Back to Peak XV

Now the opening question. From a station on the plains, a surveyor measured the angle of elevation of Peak XV. If the Earth were flat and the light came straight, the height would be the distance times the tangent of that angle, as in this unit. Two things make it harder over 174 km:

The Earth curves. Over a distance d the ground falls away below a level line by about $d^2/(2R)$ , where $R \approx 6371$ km is the Earth’s radius. For 174 km that is $174^2/12742 \approx 2.4$ km: from Nicolson’s nearest station, the bottom 2.4 km of the mountain lay below his level line. The angle he saw was only about 2°.
The air bends light. Light coming down through denser and denser air curves, so the summit looks a little higher than it is. Nicolson’s first average, about 9,200 m, did not allow for this.

The survey’s mathematicians corrected for both, and for the stations’ own heights, and checked the figures for years. In 1852 Sikdar’s calculations showed Peak XV higher than Kangchenjunga; in March 1856 the survey announced 29,002 ft (8,840 m). The calculation had given exactly 29,000 ft; the story goes that Waugh added 2 ft so that it would not look like a rounded guess.

Later measurements, by better instruments and then by satellite, came very close: the Survey of India’s triangulation of 1952 to 1954 gave 8,847.73 m. After the 2015 earthquake, Nepal’s own survey department planned a new measurement, and in 2020 Nepal and China announced the result of a joint survey: 8,848.86 m. The method of a few angles and a measured base line, used on Peak XV in the 1840s, is the method of this unit.

The dark rock pyramid of Mount Everest rising behind snow-covered ridges and a jumbled icefall, under a clear blue sky — **Figure 7.12** Sagarmatha (Mount Everest) from Kala Patthar in Khumbu, Nepal: 8,848.86 m by the joint survey of 2020. Photo: Photo by Pavel Novak, CC BY-SA 2.5

Exercise 7.6

Question 1

A. What is the angle between the line of sight and the horizontal called, when you look at an object below eye level?

Pick an answer to see why.

B. What is the angle between the line of sight and the horizontal called, when you look at an object above eye level?

Pick an answer to see why.

C. If the angle of elevation of an object seen from a place is 0°, where is the object?

Pick an answer to see why.

D. The angle of elevation of the top of a pole from a place is 45°. How do the pole’s height and its distance from the place compare?

Pick an answer to see why.

E. As the angle of elevation of the top of a tower seen from a place increases, what happens to the distance from the tower?

Pick an answer to see why.

F. What is the angle of elevation of the top of a tree $10\sqrt{3}$ m high, seen from a point 10 m from its foot?

Pick an answer to see why.

Question 2

Find the values of x, y and θ in each figure.

Figure 7.13

Answer

Figure 7.14

Answer

Figure 7.15

Answer

Figure 7.16

Answer

Figure 7.17

Answer

Figure 7.18

Answer

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

How the SEE asks it

In the SEE’s Optional Mathematics paper (75 marks, 3 hours), Trigonometry is a content area on its own, with more marks than any other unit. CDC’s specification grid gives it 17 marks, in three kinds of question:

Group	Kind of question	Trigonometry’s marks	Levels
A	multiple choice, 1 mark each	3	knowledge 1, application 1, higher ability 1
B	within the content area: questions on trigonometry alone, each of several parts	9	knowledge 2, understanding 2, application 4, higher ability 1
C	across content areas: one question joining trigonometry with another area	5	understanding 2, application 2, higher ability 1

In CDC’s model paper:

Group A: the value of $\cos 20° + \cos 40°$ in another form (sum to product); sin (A + B) written out; and what happens to the angle of elevation of a tree’s top as you walk towards it.
Group B: write sin 2A from sin (A + B) (1 mark), then show $\sin 2A = 2\tan A/(1 + \tan^2 A)$ (1 mark); a cliff with two angles of depression: draw the diagram (1 mark) and find a distance (2 marks); a conditional identity in a triangle: write a product as a sum (1 mark), verify the identity for given angles (2 marks), and say with a reason why it fails for angles that are not a triangle’s (1 mark).
Group C: a straight line’s slope written as tan of an angle, joined with the double-angle formula.

What each kind asks, and where its marks are:

Proving an identity: start from one side (usually the harder one) and reach the other; each correct formula used is a step mark.
Values without a calculator: split the angle into 30°, 45°, 60° and 90° pieces, or halve a known one.
Verifying a conditional identity: both sides worked out separately with the given angles; the reason it fails elsewhere is the condition.
Heights and distances: the figure (a mark of its own), the right ratio in the right triangle, the answer with its unit.

Here is a question of the longer kind, written for this book, with how it is marked.

Miscellaneous exercise

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

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

Given $\cos 45° = 1/\sqrt{2}$ :

a.Express cos A in terms of sin (A/2).
b.Write any two submultiple angles of 15°.
c.Prove that $\sin 22\tfrac{1}{2}° = \tfrac{1}{2}\sqrt{2 - \sqrt{2}}$ .

Answer

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

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 $\tan A = \cot B$ true for this triangle? Give a reason.

Answer

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

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

Worksheet to print (PDF, 4 pages) Answer key for teachers

Harder worksheet (PDF, 3 pages) Answer key for teachers

Quiz yourself on this unit →

Practice every question in the book, and keep track of the ones you miss →

Where these facts come from

Peak XV: the Great Trigonometrical Survey begun in 1802, its theodolites of about 500 kg carried by 12 men; James Nicolson’s observations of 1849 from Jirol (190 km) and five stations, the nearest 174 km away; his raw average of about 9,200 m without the correction for refraction; Radhanath Sikdar identifying it as the highest peak in 1852; Waugh’s announcement of 29,002 ft (8,840 m) in 1856; the Survey of India’s 8,847.73 m (1952 to 1954); Nepal’s planned survey after the 2015 earthquake; the joint Nepal–China height of 8,848.86 m (2020) (the opening, Back to Peak XV): Wikipedia, “Mount Everest”, read October 2026.
Radhanath Sikdar (1813 to 1870), his calculation of exactly 29,000 ft and the 2 ft Waugh added: Wikipedia, “Radhanath Sikdar”, read October 2026.
Triangulation by measured baselines and angles; the large theodolites of the survey: Wikipedia, “Great Trigonometrical Survey”, read October 2026.
Hipparchus’s table of chords, Ptolemy’s table for every half degree in the Almagest and his use of what we call the sum and difference formulas, his half-angle result; Habash al-Hasib’s first tables of the tangent and cotangent (about 830); Madhava’s series; Euler’s modern form (1748) (7.1, 7.2, 7.4): Wikipedia, “History of trigonometry”, read October 2026.
Hipparchus as the first known maker of a trigonometric table: Wikipedia, “Hipparchus”, read October 2026.
Aryabhata (476 to 550), the Aryabhatiya (499), his table of sines in a single verse at intervals of 3¾°, ardha-jya and jya, kojya: Wikipedia, “Aryabhata”, read October 2026.
The word sine from jya through Arabic jiba, read as jaib, and Latin sinus; cosinus from complementi sinus: Wikipedia, “Sine and cosine” (Etymology), read October 2026.
Prosthaphaeresis: multiplication by the product-to-sum formulas from the 1580s, used by Tycho Brahe, the only general shortcut for about 25 years before logarithms (1614) (7.5): Wikipedia, “Prosthaphaeresis”, read October 2026.
Beats: two close frequencies heard as one sound whose loudness varies at the difference of the frequencies; tuning by beats (7.5): Wikipedia, “Beat (acoustics)”, read October 2026.
A projectile’s range $R = v^2\sin 2\theta/g$ on level ground without air resistance, greatest at 45° (7.3): Wikipedia, “Range of a projectile”, read October 2026.
Rotation by an angle as x′ = x cos θ − y sin θ, y′ = x sin θ + y cos θ, and two rotations composing (7.2): Wikipedia, “Rotation matrix”, read October 2026.
The curvature drop $d^2/(2R)$ over 174 km and the angle of about 2° (Back to Peak XV): worked out for this book from the distances above and the Earth’s mean radius, 6,371 km.
The statue of Aryabhata at IUCAA, Tycho Brahe’s mural quadrant and Everest from Kala Patthar: their Wikimedia Commons file pages.
CDC’s specification grid and model paper for Optional Mathematics, 2082 (How the SEE asks it): the textbook folder, Specification Grids/grid-Opt. Math-G10.

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