Year 2 Pure · Chapter 10

Addition and Double-Angle Formulae

Specification: Edexcel Pure 5.6–5.9 — Trigonometry · AQA section E6, E7, E8, E9

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In this chapter

1. The addition formulae
2. Double-angle formulae
3. The form R cos(θ − α)
4. Modelling with the R form
Practice

Key facts — this chapter

$\sin(A \pm B) \equiv \sin A\cos B \pm \cos A\sin B$ and $\cos(A \pm B) \equiv \cos A\cos B \mp \sin A\sin B$ .
$\tan(A \pm B) \equiv \dfrac{\tan A \pm \tan B}{1 \mp \tan A\tan B}$ .
$\sin 2A \equiv 2\sin A\cos A$ , and $\cos 2A \equiv \cos^2 A - \sin^2 A \equiv 2\cos^2 A - 1 \equiv 1 - 2\sin^2 A$ .
$\tan 2A \equiv \dfrac{2\tan A}{1 - \tan^2 A}$ .
$a\cos\theta + b\sin\theta \equiv R\cos(\theta - \alpha)$ , where $R = \sqrt{a^2 + b^2}$ and $\tan\alpha = \frac{b}{a}$ . Its largest value is $R$ and its smallest is $-R$ .
To solve $a\cos\theta + b\sin\theta = c$ , write the left side in this form first.

$\sin(A + B)$ is not $\sin A + \sin B$ : try $A = B = \frac{\pi}{2}$ , which gives $0$ on the left and $2$ on the right. This chapter finds what it really is. The formulae that come out of it turn sums of sines and cosines into single waves, which is how a wave’s height and timing are found in models.

1. The addition formulae

Key result

\begin{aligned} \sin(A + B) &\equiv \sin A\cos B + \cos A\sin B \\ \sin(A - B) &\equiv \sin A\cos B - \cos A\sin B \\ \cos(A + B) &\equiv \cos A\cos B - \sin A\sin B \\ \cos(A - B) &\equiv \cos A\cos B + \sin A\sin B \end{aligned}

\tan(A \pm B) \equiv \frac{\tan A \pm \tan B}{1 \mp \tan A\tan B}

Watch the signs in the cosine formulae: a $+$ inside the bracket gives a $-$ between the terms.

A line OP of length 1 at angle A plus B above the horizontal. A right-angled triangle OQP has its right angle at Q, with OQ at angle A, so OQ = cos B and QP = sin B. Perpendiculars from P and Q to the horizontal, and a horizontal from Q, split the height of P into sin B cos A plus cos B sin A — The height of P, which is sin(A + B), split into two pieces.

A geometric proof

In the figure, $OP = 1$ makes angle $A + B$ with the horizontal, so $P$ is at height $\sin(A + B)$ . The right-angled triangle $OQP$ has angle $B$ at $O$ , so $OQ = \cos B$ and $QP = \sin B$ .

$Q$ is at height $OQ\sin A = \cos B\sin A$ .
$QP$ makes angle $A$ with the vertical, so $P$ is a further $QP\cos A = \sin B\cos A$ above $Q$ .

Adding the two pieces, $\sin(A + B) = \sin A\cos B + \cos A\sin B$ . The horizontal distances give $\cos(A + B) = \cos A\cos B - \sin A\sin B$ in the same way. Replacing $B$ with $-B$ gives the two minus formulae, and dividing $\sin(A + B)$ by $\cos(A + B)$ gives the one for $\tan$ .

Worked example 10.1 · An exact value

Find the exact value of $\cos\tfrac{\pi}{12}$ .

Write $\tfrac{\pi}{12}$ as a difference of angles you know: $\tfrac{\pi}{12} = \tfrac{\pi}{3} - \tfrac{\pi}{4}$ .
Use $\cos(A - B)$ :
$\begin{aligned} &\cos\tfrac{\pi}{12} \\ &= \cos\tfrac{\pi}{3}\cos\tfrac{\pi}{4} + \sin\tfrac{\pi}{3}\sin\tfrac{\pi}{4} \\ &= \tfrac{1}{2} \times \tfrac{\sqrt2}{2} + \tfrac{\sqrt3}{2} \times \tfrac{\sqrt2}{2} \\ &= \tfrac{\sqrt2 + \sqrt6}{4} \end{aligned}$

Worked example 10.2 · Two angles in different quadrants

$A$ is acute with $\sin A = \tfrac{3}{5}$ , and $B$ is obtuse with $\cos B = -\tfrac{5}{13}$ . Find the exact value of $\sin(A + B)$ .

Find the missing ratios, with the right signs. $A$ is acute, so $\cos A = \tfrac{4}{5}$ . $B$ is obtuse, where sine is positive, so $\sin B = \tfrac{12}{13}$ .
Substitute:
$\begin{aligned} &\sin(A + B) \\ &= \tfrac{3}{5} \times \left(-\tfrac{5}{13}\right) + \tfrac{4}{5} \times \tfrac{12}{13} \\ &= \tfrac{-15 + 48}{65} = \tfrac{33}{65} \end{aligned}$

Worked example 10.3 · Undoing the formula

Given that $\tan\left(x + \tfrac{\pi}{4}\right) = 3$ , find the exact value of $\tan x$ .

$\tan\tfrac{\pi}{4} = 1$ , so the formula gives

\begin{aligned} \frac{\tan x + 1}{1 - \tan x} &= 3 \\ \tan x + 1 &= 3 - 3\tan x \\ \tan x &= \tfrac{1}{2} \end{aligned}

2. Double-angle formulae

Put $B = A$ in the addition formulae and they become the double-angle formulae. Using $\sin^2 A + \cos^2 A \equiv 1$ turns the cosine one into three versions: choose the one that leaves only the function you want.

Key result

\begin{aligned} \sin 2A &\equiv 2\sin A\cos A \\ \cos 2A &\equiv \cos^2 A - \sin^2 A \\ &\equiv 2\cos^2 A - 1 \\ &\equiv 1 - 2\sin^2 A \\ \tan 2A &\equiv \frac{2\tan A}{1 - \tan^2 A} \end{aligned}

Rearranging the cosine versions gives $\cos^2 A \equiv \frac{1}{2}(1 + \cos 2A)$ and $\sin^2 A \equiv \frac{1}{2}(1 - \cos 2A)$ . With $A = \frac{x}{2}$ these are the half-angle formulae, and they are what you’ll use to integrate $\sin^2 x$ and $\cos^2 x$ in Chapter 15.

Worked example 10.4 · Double angles from a single ratio

Given that $\cos x = \tfrac{1}{3}$ and $x$ is acute, find the exact values of $\sin 2x$ and $\cos 2x$ .

$\sin x = \sqrt{1 - \tfrac{1}{9}} = \tfrac{2\sqrt2}{3}$ , positive because $x$ is acute.
Then
$\begin{aligned} \sin 2x &= 2 \times \tfrac{2\sqrt2}{3} \times \tfrac{1}{3} = \tfrac{4\sqrt2}{9} \\ \cos 2x &= 2 \times \tfrac{1}{9} - 1 = -\tfrac{7}{9} \end{aligned}$

Worked example 10.5 · Solving with a double angle

Solve $\sin 2x = \sin x$ for $0 \le x \le 2\pi$ .

Expand and factorise. Don’t divide by $\sin x$ , which would lose solutions:
$\begin{aligned} 2\sin x\cos x - \sin x &= 0 \\ \sin x\,(2\cos x - 1) &= 0 \end{aligned}$
$\sin x = 0$ gives $x = 0, \pi, 2\pi$ . $\cos x = \tfrac{1}{2}$ gives $x = \tfrac{\pi}{3}, \tfrac{5\pi}{3}$ .

So $x = 0,\ \tfrac{\pi}{3},\ \pi,\ \tfrac{5\pi}{3},\ 2\pi$ .

Worked example 10.6 · Choosing the right version

Solve $\cos 2x = 5\cos x - 3$ for $0 \le x < 2\pi$ .

The right side has $\cos x$ , so use $\cos 2x = 2\cos^2 x - 1$ :
$\begin{aligned} &2\cos^2 x - 1 = 5\cos x - 3 \\ &2\cos^2 x - 5\cos x + 2 = 0 \\ &(2\cos x - 1)(\cos x - 2) = 0 \end{aligned}$
$\cos x = 2$ is impossible, so $\cos x = \tfrac{1}{2}$ and $x = \tfrac{\pi}{3}$ or $\tfrac{5\pi}{3}$ .

Worked example 10.7 · Proving an identity

Prove that $\dfrac{1 - \cos 2\theta}{\sin 2\theta} \equiv \tan\theta$ .

Use $\cos 2\theta \equiv 1 - 2\sin^2\theta$ , which makes the $1$ s cancel:

\begin{aligned} \frac{1 - \cos 2\theta}{\sin 2\theta} &\equiv \frac{2\sin^2\theta}{2\sin\theta\cos\theta} \\ &\equiv \frac{\sin\theta}{\cos\theta} \equiv \tan\theta \end{aligned}

3. The form R cos(θ − α)

Adding a cosine wave and a sine wave with the same period gives another wave with that period: it’s just taller and shifted along. Writing it as a single cosine tells you its height and where its peaks are.

Key result

For positive $a$ and $b$ ,

a\cos\theta + b\sin\theta \equiv R\cos(\theta - \alpha)

where $R = \sqrt{a^2 + b^2}$ and $\tan\alpha = \frac{b}{a}$ , with $0 < \alpha < \frac{\pi}{2}$ .

The same method gives the other forms: $R\cos(\theta + \alpha)$ for $a\cos\theta - b\sin\theta$ , and $R\sin(\theta \pm \alpha)$ for $a\sin\theta \pm b\cos\theta$ .

Left: a right-angled triangle with sides 3 and 4 and hypotenuse R = 5, with the angle alpha between the side 3 and the hypotenuse. Right: the graph of y = 3 cos theta + 4 sin theta, a single wave of height 5 whose first peak is at theta = alpha, with the two separate waves 3 cos theta (dashed) and 4 sin theta (dotted) drawn faintly — 3 cos θ (dashed) plus 4 sin θ (dotted) is one wave of height R = 5, peaking at θ = α. The triangle gives R and α.

To see why, expand the right side:

\begin{aligned} R\cos(\theta - \alpha) &= R\cos\alpha\cos\theta \\ &\quad + R\sin\alpha\sin\theta \end{aligned}

Matching the coefficients with $a\cos\theta + b\sin\theta$ needs

R\cos\alpha = a \quad\text{and}\quad R\sin\alpha = b

Squaring and adding gives $R^2 = a^2 + b^2$ . Dividing gives $\tan\alpha = \frac{b}{a}$ .

Choose a and b. The faint waves are a cos θ and b sin θ; the solid one is their sum. However you choose them, the sum is a single cosine wave, and the right-angled triangle on the left gives its height R and its shift α.

Worked example 10.8 · Writing it as one wave

Express $3\cos\theta + 4\sin\theta$ in the form $R\cos(\theta - \alpha)$ , where $R > 0$ and $0 < \alpha < \tfrac{\pi}{2}$ . State its maximum value and the smallest positive value of $\theta$ where it occurs.

$R\cos\alpha = 3$ and $R\sin\alpha = 4$ , so
$\begin{aligned} R &= \sqrt{3^2 + 4^2} = 5 \\ \alpha &= \tan^{-1}\tfrac{4}{3} = 0.9272\ldots \end{aligned}$
So $3\cos\theta + 4\sin\theta = 5\cos(\theta - 0.927)$ , to 3 s.f.
The largest value of a cosine is $1$ , so the maximum is $5$ , when $\theta - \alpha = 0$ , that is $\theta = 0.927$ (3 s.f.).

Worked example 10.9 · Solving with the R form

Solve $3\cos\theta + 4\sin\theta = 2$ for $0 \le \theta < 2\pi$ , giving your answers to 3 significant figures.

From Example 10.8, $5\cos(\theta - \alpha) = 2$ , so $\cos(\theta - \alpha) = 0.4$ , where $\alpha = 0.9272\ldots$
As $\theta$ runs from $0$ to $2\pi$ , $\theta - \alpha$ runs from $-0.927$ to $5.356$ . In that interval,
$\theta - \alpha = 1.1592\ldots \text{ or } 5.1239\ldots$
( $-1.159$ is too small to be in the interval.)
Add $\alpha$ : $\theta = 2.09$ or $\theta = 6.05$ (3 s.f.).

Common mistake

Find $\alpha$ from $\tan\alpha = \frac{b}{a}$ , not $\frac{a}{b}$ : the $\sin\alpha$ coefficient goes on top. Then check by expanding: $5\cos(\theta - 0.927)$ should give back $3\cos\theta + 4\sin\theta$ .

4. Modelling with the R form

Worked example 10.10 · Temperature in a greenhouse

The temperature in a greenhouse, $T$ °C, $t$ hours after midnight is modelled by $T = 20 - 3\cos\left(\tfrac{\pi t}{12}\right) - 4\sin\left(\tfrac{\pi t}{12}\right)$ . Find the highest temperature and the time of day at which it happens.

Write $\theta = \tfrac{\pi t}{12}$ . By Example 10.8, $3\cos\theta + 4\sin\theta = 5\cos(\theta - \alpha)$ with $\alpha = 0.9272\ldots$ , so $T = 20 - 5\cos(\theta - \alpha)$ .
$T$ is largest when the cosine is $-1$ , so the highest temperature is $20 + 5 = 25$ °C.
That happens when $\theta - \alpha = \pi$ :
$\begin{aligned} \tfrac{\pi t}{12} &= \pi + 0.9272\ldots \\ t &= 15.542 \text{ (3 d.p.)} \end{aligned}$
$0.542$ hours is $33$ minutes to the nearest minute, so the greenhouse is warmest at about 3:33 pm.

Practice

Twelve questions in three tiers. Try each one properly before opening the solution — the solutions show full method, not just answers.

Warm-up

1 [3 marks]

Find the exact value of $\sin\tfrac{\pi}{12}$ .

Show solution

$\tfrac{\pi}{12} = \tfrac{\pi}{3} - \tfrac{\pi}{4}$ , so use $\sin(A - B)$ :

\begin{aligned} \sin\tfrac{\pi}{12} &= \sin\tfrac{\pi}{3}\cos\tfrac{\pi}{4} - \cos\tfrac{\pi}{3}\sin\tfrac{\pi}{4} \\ &= \tfrac{\sqrt3}{2} \times \tfrac{\sqrt2}{2} - \tfrac{1}{2} \times \tfrac{\sqrt2}{2} \\ &= \tfrac{\sqrt6 - \sqrt2}{4} \end{aligned}

2 [3 marks]

Write each of these as a single trigonometric function or an exact value.

$\sin 3x\cos x + \cos 3x\sin x$ [1]
$\cos^2 2x - \sin^2 2x$ [1]
$2\sin\tfrac{\pi}{8}\cos\tfrac{\pi}{8}$ [1]

Show solution

(a) $\sin(3x + x) = \sin 4x$ .

(b) $\cos(2 \times 2x) = \cos 4x$ .

(c) $\sin\left(2 \times \tfrac{\pi}{8}\right) = \sin\tfrac{\pi}{4} = \tfrac{\sqrt2}{2}$ .

3 [3 marks]

$A$ and $B$ are acute angles with $\tan A = \tfrac{1}{2}$ and $\tan B = \tfrac{1}{3}$ . Find $\tan(A + B)$ , and hence show that $A + B = \tfrac{\pi}{4}$ .

Show solution

\tan(A + B) = \frac{\frac12 + \frac13}{1 - \frac12 \times \frac13} = \frac{\frac56}{\frac56} = 1

$A$ and $B$ are acute, so $0 < A + B < \pi$ , and the only angle there with tangent $1$ is $\tfrac{\pi}{4}$ .

4 [2 marks]

Given that $\sin x = \tfrac{1}{4}$ , find the exact value of $\cos 2x$ .

Show solution

Use the version with $\sin$ only: $\cos 2x = 1 - 2\sin^2 x = 1 - \tfrac{2}{16} = \tfrac{7}{8}$ .

Standard

5 [4 marks]

Solve $\cos 2x = \cos x$ for $0 \le x \le 2\pi$ .

Show solution

Use $\cos 2x = 2\cos^2 x - 1$ :

\begin{aligned} 2\cos^2 x - \cos x - 1 &= 0 \\ (2\cos x + 1)(\cos x - 1) &= 0 \end{aligned}

$\cos x = 1$ gives $x = 0$ or $2\pi$ , and $\cos x = -\tfrac{1}{2}$ gives $x = \tfrac{2\pi}{3}$ or $\tfrac{4\pi}{3}$ . So $x = 0,\ \tfrac{2\pi}{3},\ \tfrac{4\pi}{3},\ 2\pi$ .

6 [4 marks]

Prove that $\tan\theta + \cot\theta \equiv 2\operatorname{cosec} 2\theta$ .

Show solution

\begin{aligned} \tan\theta + \cot\theta &\equiv \frac{\sin\theta}{\cos\theta} + \frac{\cos\theta}{\sin\theta} \\ &\equiv \frac{\sin^2\theta + \cos^2\theta}{\sin\theta\cos\theta} \\ &\equiv \frac{1}{\frac{1}{2}\sin 2\theta} \equiv 2\operatorname{cosec} 2\theta \end{aligned}

7 [6 marks]

Express $\sqrt3\cos\theta + \sin\theta$ in the form $R\cos(\theta - \alpha)$ , where $R > 0$ and $0 < \alpha < \tfrac{\pi}{2}$ , giving exact values of $R$ and $\alpha$ . [3]
Hence solve $\sqrt3\cos\theta + \sin\theta = \sqrt2$ for $0 \le \theta < 2\pi$ . [3]

Show solution

(a) $R\cos\alpha = \sqrt3$ and $R\sin\alpha = 1$ , so $R = \sqrt{3 + 1} = 2$ and $\tan\alpha = \tfrac{1}{\sqrt3}$ , giving $\alpha = \tfrac{\pi}{6}$ . So the expression is $2\cos\left(\theta - \tfrac{\pi}{6}\right)$ .

(b) $\cos\left(\theta - \tfrac{\pi}{6}\right) = \tfrac{\sqrt2}{2}$ . As $\theta$ runs from $0$ to $2\pi$ , $\theta - \tfrac{\pi}{6}$ runs from $-\tfrac{\pi}{6}$ to $\tfrac{11\pi}{6}$ , and in that interval

\theta - \tfrac{\pi}{6} = \tfrac{\pi}{4} \text{ or } \tfrac{7\pi}{4}

So $\theta = \tfrac{5\pi}{12}$ or $\tfrac{23\pi}{12}$ .

8 [4 marks]

$A$ is obtuse with $\cos A = -\tfrac{4}{5}$ , and $B$ is acute with $\sin B = \tfrac{12}{13}$ . Find the exact value of $\cos(A - B)$ .

Show solution

$A$ is obtuse, so $\sin A = \tfrac{3}{5}$ (positive). $B$ is acute, so $\cos B = \tfrac{5}{13}$ . Then

\begin{aligned} &\cos(A - B) \\ &= -\tfrac{4}{5} \times \tfrac{5}{13} + \tfrac{3}{5} \times \tfrac{12}{13} \\ &= \tfrac{-20 + 36}{65} = \tfrac{16}{65} \end{aligned}

9 [5 marks]

Show that $\cos 3x \equiv 4\cos^3 x - 3\cos x$ . [4]
Hence find the exact value of $\cos 3x$ when $\cos x = \tfrac{1}{3}$ . [1]

Show solution

(a) Write $3x$ as $2x + x$ :

\begin{aligned} &\cos 3x \\ &\equiv \cos 2x\cos x - \sin 2x\sin x \\ &\equiv (2\cos^2 x - 1)\cos x - 2\sin^2 x\cos x \\ &\equiv 2\cos^3 x - \cos x \\ &\quad - 2(1 - \cos^2 x)\cos x \\ &\equiv 4\cos^3 x - 3\cos x \end{aligned}

(b) $4 \times \tfrac{1}{27} - 3 \times \tfrac{1}{3} = \tfrac{4}{27} - 1 = -\tfrac{23}{27}$ .

Exam style

10 [8 marks]

Express $5\sin\theta - 12\cos\theta$ in the form $R\sin(\theta - \alpha)$ , where $R > 0$ and $0 < \alpha < \tfrac{\pi}{2}$ . Give $\alpha$ to 4 decimal places. [3]
State the maximum value of $5\sin\theta - 12\cos\theta$ , and the smallest positive value of $\theta$ at which it occurs. [2]
Solve $5\sin\theta - 12\cos\theta = 6.5$ for $0 \le \theta < 2\pi$ , to 3 significant figures. [3]

Show solution

(a) $R\sin(\theta - \alpha) = R\cos\alpha\sin\theta - R\sin\alpha\cos\theta$ , so $R\cos\alpha = 5$ and $R\sin\alpha = 12$ . Then $R = \sqrt{25 + 144} = 13$ and $\alpha = \tan^{-1}\tfrac{12}{5} = 1.1760$ (4 d.p.).

(b) The maximum is $13$ , when $\theta - \alpha = \tfrac{\pi}{2}$ , that is $\theta = \tfrac{\pi}{2} + 1.1760 = 2.75$ (3 s.f.).

(c) $\sin(\theta - \alpha) = \tfrac{6.5}{13} = \tfrac{1}{2}$ . As $\theta$ runs from $0$ to $2\pi$ , $\theta - \alpha$ runs from $-1.176$ to $5.107$ , and in that interval $\theta - \alpha = \tfrac{\pi}{6}$ or $\tfrac{5\pi}{6}$ . So

\begin{aligned} \theta &= 1.1760 + 0.5236 = 1.70 \\ \theta &= 1.1760 + 2.6180 = 3.79 \end{aligned}

to 3 significant figures.

11 [8 marks]

The depth of water in a harbour, $D$ metres, $t$ hours after midnight is modelled by

D = 5 + 1.5\cos\left(\tfrac{\pi t}{6}\right) + 2\sin\left(\tfrac{\pi t}{6}\right)

Write $D$ in the form $5 + R\cos\left(\tfrac{\pi t}{6} - \alpha\right)$ , giving $R$ exactly and $\alpha$ to 4 decimal places. [3]
Find the greatest depth, and the first time after midnight when it occurs, to the nearest minute. [3]
State the least depth. [1]
Give one reason why the model may not be realistic. [1]

Show solution

(a) $R\cos\alpha = 1.5$ and $R\sin\alpha = 2$ , so $R = \sqrt{2.25 + 4} = 2.5$ and $\alpha = \tan^{-1}\tfrac{4}{3} = 0.9273$ (4 d.p.). So $D = 5 + 2.5\cos\left(\tfrac{\pi t}{6} - 0.9273\right)$ .

(b) The greatest depth is $5 + 2.5 = 7.5$ m, when the cosine is $1$ , so $\tfrac{\pi t}{6} = 0.9272\ldots$ and $t = 1.7710\ldots$ Then $0.7710$ hours is $46$ minutes, so this is at about 1:46 am.

(c) $5 - 2.5 = 2.5$ m.

(d) For example, real tides don’t repeat exactly every $12$ hours, and the depth also depends on the weather.

12 [6 marks]

Prove that $\dfrac{\sin 2\theta}{1 + \cos 2\theta} \equiv \tan\theta$ . [3]
Hence solve $\dfrac{\sin 4x}{1 + \cos 4x} = \sqrt3$ for $0 \le x \le \pi$ . [3]

Show solution

(a) Use $\cos 2\theta \equiv 2\cos^2\theta - 1$ , which makes the $1$ s cancel:

\frac{\sin 2\theta}{1 + \cos 2\theta} \equiv \frac{2\sin\theta\cos\theta}{2\cos^2\theta} \equiv \tan\theta

(b) With $\theta = 2x$ , the equation is $\tan 2x = \sqrt3$ . As $x$ runs from $0$ to $\pi$ , $2x$ runs from $0$ to $2\pi$ , so $2x = \tfrac{\pi}{3}$ or $\tfrac{4\pi}{3}$ , and $x = \tfrac{\pi}{6}$ or $\tfrac{2\pi}{3}$ . At both, $1 + \cos 4x = \tfrac{1}{2} \neq 0$ , so the original fraction is defined.