What are the Compound Angle Identities?

In addition to the basic trigonometric identities and the reciprocal identities there are the compound angle identities including the double angle identities. The compound angle identities (sometimes called the addition angle identities) are as follows:

$\begin{array}{lll}\sin(A+B)\equiv \sin(A)\cos(B)+\cos(A)\sin(B)\\\cos(A+B)\equiv \cos(A)\cos(B)-\sin(A)\sin(B)\\\tan(A+B)\equiv\frac{\tan(A)+\tan(B)}{1-\tan(A)\tan(B)}\end{array}$

Proof of the first two identities follows from considering two compound triangles and proof of the third comes from using the first two identities. By replacing $B$ with $-B$ and noting that $\sin(-B)=-\sin(B)$ , $\cos(-B)=\cos(B)$ and $\tan(-B)=-\tan(B)$ we also have:

$\begin{array}{lll}\sin(A-B)\equiv \sin(A)\cos(B)-\cos(A)\sin(B)\\\cos(A-B)\equiv \cos(A)\cos(B)+\sin(A)\sin(B)\\\tan(A-B)\equiv\frac{\tan(A)-\tan(B)}{1+\tan(A)\tan(B)}\end{array}$

See Example 1 and videos for examples using the compound angle formulae.

Double Angle Formulae

The double angle formulae (or identities) follow from the compound angle identities. By setting $B=A$ in the first set of identities above we obtain:

$\begin{array}{lll}\sin(2A)\equiv 2\sin(A)\cos(A)\\\cos(2A)\equiv \cos^2(A)-\sin^2(A)\equiv 2\cos^2(A)-1\equiv 1-2\sin^2(A)\\\tan(2A)\equiv\frac{2\tan(A)}{1-\tan^2(A)}\end{array}$

Note that there are two additional identities for $\cos(2A)$ here. These come from using $\sin^2(A)+\cos^2(A)=1$ (see more trigonometric identities). It is worth memorising the double angle identities as it can save time when answering exam questions. See Example 2.

Expressions of the form asin(x)+bcos(x)

We can use the compound angle identities above to solve equations that are given in the form $a\sin(x)+b\cos(x)=c$ , for example. We do this by converting the trigonometric expression into the form $R\sin(x+\alpha)$ , for example. Using the identity $\sin(A+B)\equiv\sin(A)\cos(B)+\cos(A)\sin(B)$ , we can write:

$R\sin(x+\alpha)\equiv R\sin(x)\cos(\alpha)+R\cos(x)\sin(\alpha)$

This expression now has separate $\sin(x)$ and $\cos(x)$ terms, the coefficients of which are $R\cos(\alpha)$ and $R\sin(\alpha)$ . Comparing this with the expression $a\sin(x)+b\cos(x)$ we must equate the coefficients and so:

$\begin{array}{lll}a&=&R\cos(\alpha)\\b&=&R\sin(\alpha)\end{array}$

You might recognise these expressions from the right-angled triangle with hypoteneuse $R$ and angle $\alpha$ . The sides can be found using SOHCAHTOA (what is SOHCAHTOA?). The opposite is $R\sin(\alpha)$ (which is $a$ ) and the adjacent is $R\cos(\alpha)$ (which is $b$ ). It follows that expressions of the form $a\sin(x)+b\cos(x)$ can be converted to expressions of the form $R\sin(x+\alpha)$ by setting $R=\sqrt{a^2+b^2}$ and $\alpha=\tan^{-1}\left(\frac{b}{a}\right)$ . See Example 3. Note that the above process can be applied to any of the compound angle identities.

Examples

Given that $\sin(x)=\frac{5}{13}$ for obtuse $x$ and $\cos(y)=\frac{4}{5}$ for acute $y$ , find the value of $\sin(x-y)$ .
Show that $\cos(75^\circ)=\frac{\sqrt{6}-\sqrt{2}}{4}$ .

Solution:

1. Using the identity $\sin(A-B)=\sin(A)\cos(B)-\cos(A)\sin(B)$ we can write:

$\begin{array}{lll}\sin(x-y)&=&\sin(x)\cos(y)-\cos(x)\sin(y)\\&=&\frac{5}{13}\times\frac{4}{5}-\cos(x)\sin(y)\end{array}$

To get expressions for $\cos(x)$ and $\sin(y)$ we use the identity $\sin^2(x)+\cos^2(x)=1$ . More on trigonometric identities. Hence, $\cos(x)=\pm\sqrt{1-\sin^2(x)}$ and $\sin(y)=\pm\sqrt{1-\cos^2(y)}$ . Since $x$ is obtuse $\cos(x)$ is negative and so $\cos(x)=-\sqrt{1-\left(\frac{5}{13}\right)^2}=-\frac{12}{13}$ . Also, since $y$ is acute $\sin(y)$ is positive and so $\sin(y)=\sqrt{1-\left(\frac{4}{5}\right)^2}=\frac{3}{5}$ . It follows that $\sin(x-y)=\frac{5}{13}\times\frac{4}{5}+\frac{12}{13}\times\frac{3}{5}=\frac{56}{65}$ .

2. By splitting $75^\circ$ into two familiar angles such as $45^\circ$ and $30^\circ$ , we can use the identity to write $\cos(45^\circ+30^\circ)=\cos(45^\circ)\cos(30^\circ)-\sin(45^\circ)\sin(30^\circ)$ We already know these trigonometric ratios and so

$\begin{array}{lll}\cos(45^\circ+30^\circ)&=&\frac{\sqrt{2}}{2}\times\frac{\sqrt{3}}{2}-\frac{\sqrt{2}}{2}\times\frac{1}{2}\\&=&\frac{\sqrt{6}}{4}-\frac{\sqrt{2}}{4}=\frac{\sqrt{6}-\sqrt{2}}{4}\end{array}$

as required.

Solve $4\sin(2x)=\sin(x)$ for $0\leq x<2\pi$
Given the parametric equations $x=2\cos(2\theta)$ and $y=1-3\sin(\theta)$ , find the Cartesian equation. (See more on parametric equations).

Solution:

Since there is a $\sin(2x)$ terms, it suggests that we use the identity $\sin(2x)=2\sin(x)\cos(x)$ . Hence, the equation becomes $8\sin(x)\cos(x)=\sin(x)$ . At this point it is tempting to divide both sides by $\sin(x)$ but this is assuming that $\sin(x)\ne 0$ which of course it could be. Rather, subtract $\sin(x)$ from both sides to get $8\sin(x)\cos(x)-\sin(x)=0$ and factorise $\sin(x)\left(8\cos(x)-1\right)=0$ . It follows that the solutions come from solving the two equations $\sin(x)=0$ and $\cos(x)=\frac{1}{8}$ . Hence, the solutions on the given interval are $x=0,\pi$ (exact) and $x=1.45, 4.84$ (to 2 d.p.).
We can rewrite the equations as $\frac{x}{2}=\cos(2\theta)$ and $\frac{1-y}{3}=\sin(\theta)$ . Using the double angle identity $\cos(2\theta)=1-2\sin^2(\theta)$ , we can write $\frac{x}{2}=1-2\sin^2(\theta)=1-2\left(\frac{1-y}{3}\right)^2$ or $x-2=\frac{4}{9}\left(1-y\right)^2$ . This can be left implicitly or we can obtain an explicit equation by making $y$ the subject: $y=1\pm\sqrt{\frac{9}{4}(x-2)}$ .

Express $3\sin(x)+4\cos(x)$ in the form $R\sin(x+\alpha)$ . Hence solve the equation $3\sin(x)+4\cos(x)=2$ on the interval $-\pi\leq x\leq \pi$
Write $\cos(x)+\sqrt{2}\sin(x)$ in the form $R\cos(x-\alpha)$ . Hence, sketch the curve of $y=\cos(x)-\sqrt{2}\sin(x)$ .

Solution:

Setting $R=\sqrt{3^2+4^2}=5$ and $\alpha=\tan^{-1}\left(\frac{4}{3}\right)=0.93$ to 2 decimal places, we can write $3\sin(x)+4\cos(x)=5\sin(x+0.93)$ . Hence, we solve the equation $5\sin(x+0.93)=2$ or $\sin(x+0.93)=\frac{2}{5}$ . It follows that $x+0.93=…-5.87, -3.55, 0.41, 2.73, 6.69,..$ and so the solutions in the given interval are $x=-0.52$ and $x=1.80$ to 2 d.p.
We follow the steps of the notes but this time we look at the identity $\cos(A-B)=\cos(A)\cos(B)+\sin(A)\sin(B)$ . It follows that $R\cos(x-\alpha)=R\cos(x)\cos(\alpha)+R\sin(x)\sin(\alpha)$ . Equating the coefficients gives $R\cos(\alpha)=1$ and $R\sin(\alpha)=\sqrt{2}$ . It follows that $R=\sqrt{1^2+(\sqrt{2})^2}=\sqrt{3}$ and $\alpha=\tan^{-1}\left(\frac{\sqrt{2}}{1}\right)=0.955$ to 3 d.p. Hence, $\cos(x)+\sqrt{2}\sin(x)=\sqrt{3}\cos(x-0.955)$ . It follows that graph of $y=\cos(x)+\sqrt{2}\sin(x)$ is the graph of $\cos(x)$ shifted to the right by 0.955 and stretched in the $y$ -direction by a factor of $\sqrt{3}$ . See more on transformations.