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:

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

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

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

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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 $$\cos(\alpha)$$​:

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

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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 $$\underline{\alpha = \dfrac{\pi}{5}, \alpha = \dfrac{3\pi}{5}, \alpha = \dfrac{7\pi}{5}, \alpha = \dfrac{9\pi}{5}}$$.