Solving trigonometric equations: Double angle formula

In a nutshell

The addition and double angle formula can help you solve a wide range of trigonometric equations.

Addition formulae

The trigonometric addition and difference formulae can be used to reduce certain equations to simple trigonometric equations.

The equations are as follows:

$\sin(\alpha + \beta) \equiv \sin(\alpha)\cos(\beta) + \cos(\alpha)\sin(\beta)\\\sin(\alpha - \beta) \equiv \sin(\alpha)\cos(\beta) - \cos(\alpha)\sin(\beta)$

$\cos(\alpha + \beta) \equiv \cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta)\\\cos(\alpha - \beta) \equiv \cos(\alpha)\cos(\beta) + \sin(\alpha)\sin(\beta)$

$\tan(\alpha + \beta) \equiv \dfrac{\tan(\alpha) + \tan(\beta)}{1 - \tan(\alpha)\tan(\beta)}\\\tan(\alpha - \beta ) \equiv \dfrac{\tan(\alpha) - \tan(\beta)}{1 + \tan(\alpha)\tan(\beta)}$

Example 1

Find all solutions to the equation $\sin\left(\alpha + \dfrac{\pi}{6}\right) = 4\cos(\alpha)$ in the interval $0 \le \alpha \le 2\pi$ . Give your answers to $3\space d.p.$ 

Use the addition formula for $\sin$ :

$\begin{aligned}\sin\left(\alpha + \dfrac{\pi}{6}\right) &= 4\cos(\alpha)\\\sin(\alpha)\cos\left(\dfrac{\pi}{6}\right) + \cos(\alpha)\sin\left(\dfrac{\pi}{6}\right) &= 4\cos(\alpha)\\\dfrac{\sqrt{3}}{2}\sin(\alpha) &= \dfrac{7}{2}\cos(\alpha)\\\tan(\alpha) &= \dfrac{7\sqrt{3}}{3}\end{aligned}$

Using either the graph of $\tan(\alpha)$ or otherwise, find that the only solutions to the equation $\tan(\alpha) = \dfrac{7\sqrt{3}}{3}$ are $\alpha = 1.3282\dots$ and $\alpha = 4.4698\dots$

Therefore, to $3\space d.p.$ , the solutions to the equation $\sin\left(\alpha + \dfrac{\pi}{6}\right) = 4\cos(\alpha)$ in the interval $0 \le \alpha \le 2\pi$ are $\underline{\alpha = 1.328}$ and $\underline{\alpha = 4.470}$ radians.

Double angle formulae

The double angle formulae can similarly be used to reduce certain equations to simple trigonometric equations.

The equations are as follows:

$\sin(2\alpha) \equiv 2\sin(\alpha)\cos(\alpha)$

$\cos(2\alpha) \equiv \cos^2(\alpha) - \sin^2(\alpha) \equiv 2\cos^2(\alpha) - 1 \equiv 1 - 2\sin^2(\alpha)$

$\tan(2\alpha) = \dfrac{2\tan(\alpha)}{1 - \tan^2(\alpha)}$

Example 2

Find all solutions to the equation $2\cos(2\alpha) =2\cos(\alpha) -1$ in the interval $0 \le \alpha \le 2\pi$ .

Use the double anglue formula for $\cos$ which only features $\cos(\alpha)$ , and solve the quadratic in $\cos(\alpha)$ :

$\begin{aligned}2\cos(2\alpha) &= 2\cos(\alpha) - 1\\2(2\cos^2(\alpha) - 1) &= 2\cos(\alpha) - 1\\4\cos^2(\alpha) - 2\cos(\alpha) - 1 &= 0\\\cos(\alpha) &= \dfrac{1 \pm \sqrt{5}}{4}\end{aligned}$

The solutions to the equation $\cos(\alpha) = \dfrac{1 + \sqrt{5}}{4}$ over the interval $0 \le \alpha \le 2\pi$ are $\alpha = \dfrac{\pi}{5}, \alpha = \dfrac{9\pi}{5}$ .

The solutions to the equation $\cos(\alpha) = \dfrac{1-\sqrt{5}}{4}$ over the interval $0 \le \alpha \le 2\pi$ are $\alpha = \dfrac{3\pi}{5}, \alpha = \dfrac{7\pi}{5}$ .

Therefore, the solutions to the equation $2\cos(2\alpha) = 2\cos(\alpha) - 1$ over the interval $0 \le \alpha \le 2\pi$ are $\underline{\alpha = \dfrac{\pi}{5}, \alpha = \dfrac{3\pi}{5}, \alpha = \dfrac{7\pi}{5}, \alpha = \dfrac{9\pi}{5}}$ .