3.10 Compound angle identitiesIB Maths: Analysis and Approaches HL: Revision notes

The compound angle identities (in the formula booklet) are $\sin(A\pm B)=\sin A\cos B\pm\cos A\sin B$ $\cos(A\pm B)=\cos A\cos B\mp\sin A\sin B$ $\tan(A\pm B)=\frac{\tan A\pm\tan B}{1\mp\tan A\tan B}$ Notice the sign flips in the cosine identity and in the denominator of the tangent identity.

They give exact values for angles such as $\frac{5\pi}{12}=\frac{\pi}{4}+\frac{\pi}{6}$: $\sin\frac{5\pi}{12}=\frac{\sqrt{6}+\sqrt{2}}{4}$, and $\tan\frac{\pi}{12}=2-\sqrt{3}$.

Setting $A=B=\theta$ in the compound identities gives the double angle identities:

• $\sin2\theta=2\sin\theta\cos\theta$
• $\cos2\theta=\cos^{2}\theta-\sin^{2}\theta$, which becomes $2\cos^{2}\theta-1$ or $1-2\sin^{2}\theta$ using $\sin^{2}\theta+\cos^{2}\theta=1$.

You are expected to be able to derive these, not just quote them. A typical 'show that' starts from $\cos(A+B)$, substitutes $A=B=\theta$, then replaces $\cos^{2}\theta$ by $1-\sin^{2}\theta$.

Setting $A=B=\theta$ in the tan identity: $\tan2\theta=\frac{2\tan\theta}{1-\tan^{2}\theta}.$ It is undefined when $\tan^{2}\theta=1$, since then $2\theta$ is an odd multiple of $\frac{\pi}{2}$.

Example: if $\tan x=\frac{3}{4}$, then $\tan2x=\frac{3/2}{7/16}=\frac{24}{7}$, whatever the quadrant of $x$, because only $\tan x$ is used.

When you know one ratio and the quadrant, find the others first, with signs. If $\pi<x<\frac{3\pi}{2}$ and $\tan x=\frac{3}{4}$, use a 3–4–5 triangle: $\sin x=-\frac{3}{5}$, $\cos x=-\frac{4}{5}$ (both negative in the third quadrant). Then

• $\sin2x=2\left(-\frac{3}{5}\right)\left(-\frac{4}{5}\right)=\frac{24}{25}$
• $\cos2x=\frac{16}{25}-\frac{9}{25}=\frac{7}{25}$
• $\cos\left(x+\frac{\pi}{3}\right)=\frac{3\sqrt{3}-4}{10}$

Check consistency: $\frac{\sin2x}{\cos2x}=\frac{24}{7}=\tan2x$.

Compound and double angle identities combine to build new results. Writing $3\theta=2\theta+\theta$: $\sin3\theta=\sin2\theta\cos\theta+\cos2\theta\sin\theta=3\sin\theta-4\sin^{3}\theta.$

In applications, an angle is often a difference of two angles. A painting $1$ m to $3$ m above eye level, seen from $x$ m away, subtends $\theta=\alpha-\beta$ with $\tan\alpha=\frac{3}{x}$, $\tan\beta=\frac{1}{x}$, so $\tan\theta=\frac{2x}{x^{2}+3}$. Because $x^{2}+3\ge2\sqrt{3}x$, the greatest angle is $\frac{\pi}{6}$, at $x=\sqrt{3}$.