Unit 10 · Trigonometric Identities
Use $\sin^2\theta + \cos^2\theta = 1$, the compound-angle formulae, and the double-angle and half-angle formulae to find exact values and prove identities.
An identity is an equation true for every value of the variable for which both sides are defined, unlike an equation, which is true only for some values. The base identity comes straight from the unit circle, where $(\cos\theta, \sin\theta)$ lies on $x^2 + y^2 = 1$: $$\sin^2\theta + \cos^2\theta = 1.$$ Dividing by $\cos^2\theta$ gives $\tan^2\theta + 1 = \sec^2\theta$; dividing by $\sin^2\theta$ gives $1 + \cot^2\theta = \operatorname{cosec}^2\theta$.
These express functions of $A \pm B$ through functions of $A$ and $B$ separately: $$\sin(A+B) = \sin A\cos B + \cos A\sin B, \qquad \sin(A-B) = \sin A\cos B - \cos A\sin B,$$ $$\cos(A+B) = \cos A\cos B - \sin A\sin B, \qquad \cos(A-B) = \cos A\cos B + \sin A\sin B,$$ $$\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A\tan B}, \qquad \tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A\tan B}.$$ Notice that $\cos$ changes the sign between the terms, while $\sin$ keeps it. The formula for $\cos(A-B)$ is the one proved first (from the distance between two points on the unit circle); the others follow by replacing $B$ by $-B$ or by $\dfrac{\pi}{2} - B$. The tangent formula is the quotient of the sine and cosine formulae after dividing top and bottom by $\cos A\cos B$.
Example: Find the exact value of $\sin 75^\circ$. Write $75^\circ = 45^\circ + 30^\circ$: $\sin 75^\circ = \sin 45^\circ\cos 30^\circ + \cos 45^\circ\sin 30^\circ = \dfrac{\sqrt2}{2}\cdot\dfrac{\sqrt3}{2} + \dfrac{\sqrt2}{2}\cdot\dfrac12 = \dfrac{\sqrt6 + \sqrt2}{4}$.
Example: If $\tan A = \dfrac12$ and $\tan B = \dfrac13$, with $A$ and $B$ acute, find $A + B$. $\tan(A+B) = \dfrac{\frac12 + \frac13}{1 - \frac16} = 1$. Both angles are less than $\dfrac{\pi}{4}$ (their tangents are less than $1$), so $0 < A + B < \dfrac{\pi}{2}$, which gives $A + B = \dfrac{\pi}{4}$.
Put $B = A$ in the compound formulae: $$\sin 2A = 2\sin A\cos A, \qquad \tan 2A = \frac{2\tan A}{1 - \tan^2 A},$$ $$\cos 2A = \cos^2 A - \sin^2 A = 2\cos^2 A - 1 = 1 - 2\sin^2 A.$$ The three forms of $\cos 2A$ are chosen to suit the problem: use $2\cos^2 A - 1$ when only $\cos A$ is known, and $1 - 2\sin^2 A$ when only $\sin A$ is known. Putting $A$ for $2A$ and $\dfrac{A}{2}$ for $A$ gives the half-angle forms $$\cos^2 A = \frac{1 + \cos 2A}{2}, \qquad \sin^2 A = \frac{1 - \cos 2A}{2}, \qquad \tan^2 A = \frac{1 - \cos 2A}{1 + \cos 2A}.$$ These are the key to reducing powers of $\sin$ and $\cos$ before integrating.
The $t$-formulae. With $t = \tan\dfrac{A}{2}$: $$\sin A = \frac{2t}{1+t^2}, \qquad \cos A = \frac{1-t^2}{1+t^2}, \qquad \tan A = \frac{2t}{1-t^2}.$$
Example: If $\cos\theta = \dfrac{7}{25}$ with $\theta$ acute, find $\sin\dfrac{\theta}{2}$ and $\cos\dfrac{\theta}{2}$. Use $\cos\theta = 1 - 2\sin^2\dfrac{\theta}{2}$: $\sin^2\dfrac{\theta}{2} = \dfrac{1 - 7/25}{2} = \dfrac{9}{25}$, and $\dfrac{\theta}{2}$ is acute, so $\sin\dfrac{\theta}{2} = \dfrac35$. Then $\cos^2\dfrac{\theta}{2} = \dfrac{1 + 7/25}{2} = \dfrac{16}{25}$, so $\cos\dfrac{\theta}{2} = \dfrac45$.
Example: Find $\sin 2\theta$ and $\cos 2\theta$ if $\tan\theta = \dfrac12$. Use the $t$-formulae with $A = 2\theta$ and $t = \tan\theta$: $\sin 2\theta = \dfrac{2t}{1+t^2} = \dfrac{1}{5/4} = \dfrac45$ and $\cos 2\theta = \dfrac{1-t^2}{1+t^2} = \dfrac{3/4}{5/4} = \dfrac35$.
Strategy:
Example: Prove $\dfrac{\sin\theta}{1 + \cos\theta} + \dfrac{1 + \cos\theta}{\sin\theta} = \dfrac{2}{\sin\theta}$. Combine over $\sin\theta(1 + \cos\theta)$: numerator $\sin^2\theta + (1+\cos\theta)^2 = \sin^2\theta + 1 + 2\cos\theta + \cos^2\theta = 2 + 2\cos\theta = 2(1+\cos\theta)$. So LHS $= \dfrac{2(1+\cos\theta)}{\sin\theta(1+\cos\theta)} = \dfrac{2}{\sin\theta}$.
Example: Prove $\dfrac{\sin 2\theta}{1 + \cos 2\theta} = \tan\theta$. Use $\sin 2\theta = 2\sin\theta\cos\theta$ and $1 + \cos 2\theta = 2\cos^2\theta$: LHS $= \dfrac{2\sin\theta\cos\theta}{2\cos^2\theta} = \tan\theta$.
Exam tip: Identity proofs are worth many marks in Paper I and are marked line by line, so write each step. A frequent route is "express $\sin 3x$ or $\cos 3x$ in terms of $\sin x$ or $\cos x$": write $3x = 2x + x$, expand with the compound formula, then use the double-angle formulae. Never work on both sides at once and equate them.