Using the angle addition formulae

​​In a nutshell

The addition formulae for trigonometric functions, and the exact values of trigonometric functions evaluated at certain angles, can be used to compute the exact values of trigonometric functions evaluated at new angles.

Addition formulae

The addition formulae for the trigonometric functions are:

​$$\begin{aligned}\sin(\alpha + \beta) &\equiv \sin(\alpha)\cos(\beta) + \cos(\alpha)\sin(\beta)\\\sin(\alpha - \beta) &\equiv \sin(\alpha)\cos(\beta) - \cos(\alpha)\sin(\beta)\\\\\cos(\alpha + \beta) &\equiv \cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta)\\\cos(\alpha - \beta) &\equiv \cos(\alpha)\cos(\beta) + \sin(\alpha)\sin(\beta)\\\\\tan(\alpha + \beta) &\equiv \dfrac{\tan(\alpha) + \tan(\beta)}{1 - \tan(\alpha)\tan(\beta)}\\\tan(\alpha - \beta) &\equiv \dfrac{\tan(\alpha) - \tan(\beta)}{1 + \tan(\alpha)\tan(\beta)}\end{aligned}$$​​

Given the exact values of trigonometric functions evaluated at angles $$\alpha$$ and $$\beta$$​, you can use these identities to compute the exact value of trigonometric functions evaluated at $$\alpha + \beta$$​ or $$\alpha - \beta$$​.

Example 1

Compute the exact value of $$\tan\left(\dfrac{7\pi}{12}\right)$$.

You have that $$\dfrac{7\pi}{12} = \dfrac{\pi}{4} + \dfrac{\pi}{3}$$.

$$\tan\left(\dfrac{\pi}{4}\right) = 1$$, and $$\tan\left(\dfrac{\pi}{3}\right) = \sqrt{3}$$. Use the addition formula for $$\tan$$:

$$\begin{aligned}\tan\left(\dfrac{7\pi}{12}\right) &= \tan\left(\dfrac{\pi}{4} + \dfrac{\pi}{3}\right)\\&= \dfrac{\tan\left(\dfrac{\pi}{4}\right) + \tan\left(\dfrac{\pi}{3}\right)}{1 - \tan\left(\dfrac{\pi}{4}\right)\tan\left(\dfrac{\pi}{3}\right)}\\&= \dfrac{1 + \sqrt{3}}{1 - \sqrt{3}}\\&= \dfrac{(1 + \sqrt{3})^2}{(1 - \sqrt{3})(1 + \sqrt{3})}\\&= \dfrac{4 + 2\sqrt{3}}{-2}\\&= - 2 - \sqrt{3}\end{aligned}$$​​

Therefore, the exact value of $$\tan\left(\dfrac{7\pi}{12}\right)$$ is $$\underline{-2 - \sqrt{3}}$$.

​

Example 2

Given that $$\cos(\alpha) = \dfrac{20}{29}$$, with $$\alpha$$ acute, and $$\sin(\beta) = \dfrac{24}{25}$$, with $$\beta$$ obtuse, compute the exact value of $$\sin(\alpha + \beta)$$.

If $$\cos(\alpha) = \dfrac{20}{29}$$, then $$\sin(\alpha) = \pm \sqrt{1- \cos^2(\alpha)} = \pm\dfrac{21}{29}$$.

However, $$\alpha$$ is acute, therefore $$\sin(\alpha) = \dfrac{21}{29}$$.

If $$\sin(\beta) = \dfrac{24}{25}$$, then $$\cos(\beta) = \pm\sqrt{1 - \sin^2(\beta)} = \pm \dfrac{7}{25}$$.

However, $$\beta$$ is obtuse, therefore $$\cos(\beta) = -\dfrac{7}{25}$$.

Put these together via the addition formula for $$\sin$$:

$$\begin{aligned}\sin(\alpha + \beta) &= \sin(\alpha)\cos(\beta) + \cos(\alpha)\sin(\beta)\\&= \left(\dfrac{21}{29}\right)\left(-\dfrac{7}{25}\right) + \left(\dfrac{20}{29}\right)\left(\dfrac{24}{25}\right)\\&= \dfrac{333}{725}\end{aligned}$$​​

Therefore, the exact value of $$\sin(\alpha + \beta)$$ is $$\underline{\dfrac{333}{725}}$$.