Compound and Double Angles – A Level Maths Year 2

1 The Addition Formulae

The tempting mistake. $\sin(A + B)$ is not $\sin A + \sin B$. Try $A = B = 90^\circ$: $\sin 180^\circ = 0$, but $\sin 90^\circ + \sin 90^\circ = 2$.

Compound angle formulae

$\sin(A \pm B) \equiv \sin A\cos B \pm \cos A\sin B$
$\cos(A \pm B) \equiv \cos A\cos B \mp \sin A\sin B$
$\tan(A \pm B) \equiv \dfrac{\tan A \pm \tan B}{1 \mp \tan A\tan B}$

Watch the signs in the cosine formula. $\cos(A + B)$ has a minus in the middle and $\cos(A - B)$ has a plus. The sine formula keeps the same sign as the bracket.

Proof — $\tan(A + B)$ from the sine and cosine formulae

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

②Divide every term, top and bottom, by $\cos A\cos B$:

③$= \dfrac{\frac{\sin A}{\cos A} + \frac{\sin B}{\cos B}}{1 - \frac{\sin A}{\cos A}\cdot\frac{\sin B}{\cos B}} = \dfrac{\tan A + \tan B}{1 - \tan A\tan B}$ $\blacksquare$

Replacing $B$ by $-B$, and using $\sin(-B) = -\sin B$, $\cos(-B) = \cos B$, $\tan(-B) = -\tan B$, gives every "minus" version from the "plus" version.

Worked Example 1 — Exact values

Find the exact values of $\sin 75^\circ$ and $\tan 15^\circ$.

①$\sin 75^\circ = \sin(45^\circ + 30^\circ) = \dfrac{\sqrt2}{2}\cdot\dfrac{\sqrt3}{2} + \dfrac{\sqrt2}{2}\cdot\dfrac12 = \dfrac{\sqrt6 + \sqrt2}{4}$.

②$\tan 15^\circ = \tan(45^\circ - 30^\circ) = \dfrac{1 - \frac{1}{\sqrt3}}{1 + \frac{1}{\sqrt3}} = \dfrac{\sqrt3 - 1}{\sqrt3 + 1}$.

③Rationalise: $\dfrac{(\sqrt3 - 1)^2}{3 - 1} = \dfrac{4 - 2\sqrt3}{2} = 2 - \sqrt3$.

Worked Example 2 — Using given ratios

$A$ and $B$ are acute, with $\sin A = \tfrac35$ and $\cos B = \tfrac{5}{13}$. Find the exact values of $\sin(A + B)$ and $\cos(A - B)$.

①Right-angled triangles (3-4-5 and 5-12-13) give $\cos A = \tfrac45$ and $\sin B = \tfrac{12}{13}$ — both positive, as the angles are acute.

②$\sin(A + B) = \tfrac35\cdot\tfrac{5}{13} + \tfrac45\cdot\tfrac{12}{13} = \tfrac{15 + 48}{65} = \tfrac{63}{65}$.

③$\cos(A - B) = \tfrac45\cdot\tfrac{5}{13} + \tfrac35\cdot\tfrac{12}{13} = \tfrac{20 + 36}{65} = \tfrac{56}{65}$.

2 Double Angle Formulae

Put $B = A$ in the addition formulae:

Double angle formulae

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

Choosing the form of $\cos 2A$. Pick the version that leaves you with only one trig function. If the rest of the equation has $\sin x$, use $1 - 2\sin^2 x$; if it has $\cos x$, use $2\cos^2 x - 1$.

Rearranged — essential for integration later

$\cos^2 A \equiv \tfrac12(1 + \cos 2A) \qquad \sin^2 A \equiv \tfrac12(1 - \cos 2A)$

Worked Example 3 — Double angles from a given ratio

Given $\cos x = -\tfrac13$ and $\tfrac{\pi}{2} < x < \pi$, find the exact values of $\sin 2x$ and $\cos 2x$.

①$\sin^2 x = 1 - \tfrac19 = \tfrac89$. In the second quadrant $\sin x > 0$, so $\sin x = \tfrac{2\sqrt2}{3}$.

②$\sin 2x = 2\cdot\tfrac{2\sqrt2}{3}\cdot\left(-\tfrac13\right) = -\tfrac{4\sqrt2}{9}$.

③$\cos 2x = 2\cos^2 x - 1 = \tfrac29 - 1 = -\tfrac79$.

Worked Example 4 — A triple angle

Show that $\cos 3x \equiv 4\cos^3 x - 3\cos x$.

①$\cos 3x = \cos(2x + x) = \cos 2x\cos x - \sin 2x\sin x$

②$= (2\cos^2 x - 1)\cos x - 2\sin^2 x\cos x$

③$= 2\cos^3 x - \cos x - 2(1 - \cos^2 x)\cos x$

④$= 2\cos^3 x - \cos x - 2\cos x + 2\cos^3 x = 4\cos^3 x - 3\cos x$ $\blacksquare$

Worked Example 5 — Proving an identity

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

①Use $1 - \cos 2x = 2\sin^2 x$ (choosing the form that cancels the $1$).

②LHS $= \dfrac{2\sin^2 x}{2\sin x\cos x} = \dfrac{\sin x}{\cos x} = \tan x = $ RHS $\blacksquare$

3 Solving Equations

Use a double angle formula to get every term in terms of $x$, not $2x$.
Rearrange to $= 0$ and factorise. Never divide by $\sin x$ or $\cos x$ — you would lose the solutions where it is zero.
Solve each factor in the given interval.

Worked Example 6 — Factorise, don't divide

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

①$2\sin x\cos x - \sin x = 0 \Rightarrow \sin x(2\cos x - 1) = 0$.

②$\sin x = 0$: $x = 0, \pi, 2\pi$.

③$\cos x = \tfrac12$: $x = \tfrac{\pi}{3}, \tfrac{5\pi}{3}$.

④$x = 0, \tfrac{\pi}{3}, \pi, \tfrac{5\pi}{3}, 2\pi$.

Dividing both sides by $\sin x$ gives only $\cos x = \tfrac12$ and loses three of the five answers.

Worked Example 7 — A quadratic in $\cos x$

Solve $\cos 2x + 3\cos x + 2 = 0$ for $0 \leqslant x \leqslant 2\pi$.

①The other term is in $\cos x$, so use $2\cos^2 x - 1$: $\;2\cos^2 x + 3\cos x + 1 = 0$.

②$(2\cos x + 1)(\cos x + 1) = 0$.

③$\cos x = -\tfrac12$: $x = \tfrac{2\pi}{3}, \tfrac{4\pi}{3}$. $\cos x = -1$: $x = \pi$.

Worked Example 8 — Expanding a compound angle

Solve $\sin\left(x + \tfrac{\pi}{6}\right) = 2\cos x$ for $0 \leqslant x \leqslant 2\pi$.

①$\sin x\cdot\tfrac{\sqrt3}{2} + \cos x\cdot\tfrac12 = 2\cos x$.

②$\tfrac{\sqrt3}{2}\sin x = \tfrac32\cos x$. Here $\cos x \neq 0$ (otherwise $\sin x = 0$ too, which is impossible), so divide by $\cos x$:

③$\tan x = \dfrac{3}{\sqrt3} = \sqrt3 \Rightarrow x = \tfrac{\pi}{3}, \tfrac{4\pi}{3}$.

Dividing to get $\tan x$ is fine here because you have justified that $\cos x$ cannot be zero. That justification is worth writing down.

4 Quick Reference

sin(A ± B)

$\sin A\cos B \pm \cos A\sin B$

cos(A ± B)

$\cos A\cos B \mp \sin A\sin B$ — sign flips.

tan(A ± B)

$\dfrac{\tan A \pm \tan B}{1 \mp \tan A\tan B}$

sin 2A

$2\sin A\cos A$

cos 2A

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

tan 2A

$\dfrac{2\tan A}{1 - \tan^2 A}$

Squares

$\cos^2 A = \tfrac12(1 + \cos 2A)$, $\sin^2 A = \tfrac12(1 - \cos 2A)$

Equations

Factorise; never divide by a function that could be zero.

Exact values

$75^\circ = 45^\circ + 30^\circ$, $15^\circ = 45^\circ - 30^\circ$

5 Practice Questions

Question 1

Find the exact value of $\cos 105^\circ$.

▶ Show solution

$\cos(60^\circ + 45^\circ) = \tfrac12\cdot\tfrac{\sqrt2}{2} - \tfrac{\sqrt3}{2}\cdot\tfrac{\sqrt2}{2} = \dfrac{\sqrt2 - \sqrt6}{4}$.

Question 2

Simplify $\sin 3x\cos x - \cos 3x\sin x$.

▶ Show solution

This is $\sin(A - B)$ with $A = 3x$, $B = x$: $\;\sin(3x - x) = \sin 2x$.

Question 3

Show that $\tan\left(x + \tfrac{\pi}{4}\right) \equiv \dfrac{1 + \tan x}{1 - \tan x}$, and find its value when $\tan x = \tfrac13$.

▶ Show solution

$\tan\tfrac{\pi}{4} = 1$, so $\dfrac{\tan x + 1}{1 - \tan x \times 1} = \dfrac{1 + \tan x}{1 - \tan x}$ ✓

With $\tan x = \tfrac13$: $\dfrac{4/3}{2/3} = 2$.

Question 4

$A$ is obtuse with $\sin A = \tfrac45$, and $B$ is acute with $\cos B = \tfrac{12}{13}$. Find the exact value of $\cos(A + B)$.

▶ Show solution

$A$ obtuse: $\cos A = -\tfrac35$. $B$ acute: $\sin B = \tfrac{5}{13}$.

$\cos(A + B) = \left(-\tfrac35\right)\tfrac{12}{13} - \tfrac45\cdot\tfrac{5}{13} = -\tfrac{36}{65} - \tfrac{20}{65} = -\tfrac{56}{65}$.

Question 5

Given that $\tan x = \tfrac34$ and $\pi < x < \tfrac{3\pi}{2}$, find the exact values of $\sin 2x$ and $\tan 2x$.

▶ Show solution

Third quadrant: $\sin x = -\tfrac35$, $\cos x = -\tfrac45$.

$\sin 2x = 2\left(-\tfrac35\right)\left(-\tfrac45\right) = \tfrac{24}{25}$.

$\tan 2x = \dfrac{2\cdot\frac34}{1 - \frac{9}{16}} = \dfrac{3/2}{7/16} = \dfrac{24}{7}$.

Question 6

Solve $\cos 2x = \sin x$ for $0^\circ \leqslant x \leqslant 360^\circ$.

▶ Show solution

$1 - 2\sin^2 x = \sin x \Rightarrow 2\sin^2 x + \sin x - 1 = 0 \Rightarrow (2\sin x - 1)(\sin x + 1) = 0$.

$\sin x = \tfrac12$: $x = 30^\circ, 150^\circ$. $\sin x = -1$: $x = 270^\circ$.

Question 7

Prove that $\dfrac{\sin 2x}{1 + \cos 2x} \equiv \tan x$.

▶ Show solution

$1 + \cos 2x = 2\cos^2 x$, so LHS $= \dfrac{2\sin x\cos x}{2\cos^2 x} = \dfrac{\sin x}{\cos x} = \tan x$ $\blacksquare$

Question 8

Prove that $\sin 3x \equiv 3\sin x - 4\sin^3 x$.

▶ Show solution

$\sin 3x = \sin 2x\cos x + \cos 2x\sin x = 2\sin x\cos^2 x + (1 - 2\sin^2 x)\sin x$

$= 2\sin x(1 - \sin^2 x) + \sin x - 2\sin^3 x = 3\sin x - 4\sin^3 x$ $\blacksquare$

Question 9

Solve $3\cos 2x - \cos x + 2 = 0$ for $0^\circ \leqslant x \leqslant 360^\circ$, giving answers to $1$ decimal place.

▶ Show solution

$3(2\cos^2 x - 1) - \cos x + 2 = 0 \Rightarrow 6\cos^2 x - \cos x - 1 = 0 \Rightarrow (2\cos x - 1)(3\cos x + 1) = 0$.

$\cos x = \tfrac12$: $x = 60^\circ, 300^\circ$.

$\cos x = -\tfrac13$: $x = 109.5^\circ$, $360^\circ - 109.5^\circ = 250.5^\circ$.

Question 10

(a) Prove that $\cos^4 x - \sin^4 x \equiv \cos 2x$. (b) Hence solve $\cos^4 x - \sin^4 x = \tfrac12$ for $0 \leqslant x \leqslant \pi$. (c) Use a double angle formula to show that $\cos\dfrac{\pi}{8} = \dfrac{\sqrt{2 + \sqrt2}}{2}$.

▶ Show solution

(a) Difference of two squares: $(\cos^2 x - \sin^2 x)(\cos^2 x + \sin^2 x) = \cos 2x \times 1 = \cos 2x$ $\blacksquare$

(b) $\cos 2x = \tfrac12$ with $0 \leqslant 2x \leqslant 2\pi$: $2x = \tfrac{\pi}{3}, \tfrac{5\pi}{3}$, so $x = \tfrac{\pi}{6}, \tfrac{5\pi}{6}$.

(c) $\cos\tfrac{\pi}{4} = 2\cos^2\tfrac{\pi}{8} - 1$, so $\cos^2\tfrac{\pi}{8} = \dfrac{1 + \frac{\sqrt2}{2}}{2} = \dfrac{2 + \sqrt2}{4}$.

$\tfrac{\pi}{8}$ is acute, so take the positive root: $\cos\tfrac{\pi}{8} = \dfrac{\sqrt{2 + \sqrt2}}{2}$ $\blacksquare$

Compound and Double Angles · A Level Maths Year 2 · Created with MathJax