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Solving trigonometric equations: Double angle formula

Solving trigonometric equations: Double angle formula

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Summary

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⁡(α+β)≡sin⁡(α)cos⁡(β)+cos⁡(α)sin⁡(β)sin⁡(α−β)≡sin⁡(α)cos⁡(β)−cos⁡(α)sin⁡(β)\sin(\alpha + \beta) \equiv \sin(\alpha)\cos(\beta) + \cos(\alpha)\sin(\beta)\\\sin(\alpha - \beta) \equiv \sin(\alpha)\cos(\beta) - \cos(\alpha)\sin(\beta)sin(α+β)≡sin(α)cos(β)+cos(α)sin(β)sin(α−β)≡sin(α)cos(β)−cos(α)sin(β)​​


​cos⁡(α+β)≡cos⁡(α)cos⁡(β)−sin⁡(α)sin⁡(β)cos⁡(α−β)≡cos⁡(α)cos⁡(β)+sin⁡(α)sin⁡(β)\cos(\alpha + \beta) \equiv \cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta)\\\cos(\alpha - \beta) \equiv \cos(\alpha)\cos(\beta) + \sin(\alpha)\sin(\beta)cos(α+β)≡cos(α)cos(β)−sin(α)sin(β)cos(α−β)≡cos(α)cos(β)+sin(α)sin(β)​​


​tan⁡(α+β)≡tan⁡(α)+tan⁡(β)1−tan⁡(α)tan⁡(β)tan⁡(α−β)≡tan⁡(α)−tan⁡(β)1+tan⁡(α)tan⁡(β)\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)}tan(α+β)≡1−tan(α)tan(β)tan(α)+tan(β)​tan(α−β)≡1+tan(α)tan(β)tan(α)−tan(β)​​​


​

Example 1

Find all solutions to the equation sin⁡(α+π6)=4cos⁡(α)\sin\left(\alpha + \dfrac{\pi}{6}\right) = 4\cos(\alpha)sin(α+6π​)=4cos(α) in the interval 0≤α≤2π0 \le \alpha \le 2\pi0≤α≤2π. Give your answers to 3 d.p.3\space d.p.3 d.p.​


Use the addition formula for sin⁡\sinsin:


sin⁡(α+π6)=4cos⁡(α)sin⁡(α)cos⁡(π6)+cos⁡(α)sin⁡(π6)=4cos⁡(α)32sin⁡(α)=72cos⁡(α)tan⁡(α)=733\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}sin(α+6π​)sin(α)cos(6π​)+cos(α)sin(6π​)23​​sin(α)tan(α)​=4cos(α)=4cos(α)=27​cos(α)=373​​​​​


Using either the graph of tan⁡(α)\tan(\alpha)tan(α) or otherwise, find that the only solutions to the equation tan⁡(α)=733\tan(\alpha) = \dfrac{7\sqrt{3}}{3}tan(α)=373​​ are α=1.3282…\alpha = 1.3282\dotsα=1.3282… and α=4.4698…\alpha = 4.4698\dotsα=4.4698…​


Therefore, to 3 d.p.3\space d.p.3 d.p., the solutions to the equation sin⁡(α+π6)=4cos⁡(α)\sin\left(\alpha + \dfrac{\pi}{6}\right) = 4\cos(\alpha)sin(α+6π​)=4cos(α) in the interval 0≤α≤2π0 \le \alpha \le 2\pi0≤α≤2π are α=1.328‾\underline{\alpha = 1.328}α=1.328​ and α=4.470‾\underline{\alpha = 4.470}α=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⁡(2α)≡2sin⁡(α)cos⁡(α)\sin(2\alpha) \equiv 2\sin(\alpha)\cos(\alpha)sin(2α)≡2sin(α)cos(α)​​


​cos⁡(2α)≡cos⁡2(α)−sin⁡2(α)≡2cos⁡2(α)−1≡1−2sin⁡2(α)\cos(2\alpha) \equiv \cos^2(\alpha) - \sin^2(\alpha) \equiv 2\cos^2(\alpha) - 1 \equiv 1 - 2\sin^2(\alpha)cos(2α)≡cos2(α)−sin2(α)≡2cos2(α)−1≡1−2sin2(α)​​


​tan⁡(2α)=2tan⁡(α)1−tan⁡2(α)\tan(2\alpha) = \dfrac{2\tan(\alpha)}{1 - \tan^2(\alpha)}tan(2α)=1−tan2(α)2tan(α)​​​



Example 2

Find all solutions to the equation 2cos⁡(2α)=2cos⁡(α)−12\cos(2\alpha) =2\cos(\alpha) -12cos(2α)=2cos(α)−1 in the interval 0≤α≤2π0 \le \alpha \le 2\pi0≤α≤2π.


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


2cos⁡(2α)=2cos⁡(α)−12(2cos⁡2(α)−1)=2cos⁡(α)−14cos⁡2(α)−2cos⁡(α)−1=0cos⁡(α)=1±54\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}2cos(2α)2(2cos2(α)−1)4cos2(α)−2cos(α)−1cos(α)​=2cos(α)−1=2cos(α)−1=0=41±5​​​​​

​

The solutions to the equation cos⁡(α)=1+54\cos(\alpha) = \dfrac{1 + \sqrt{5}}{4}cos(α)=41+5​​ over the interval 0≤α≤2π0 \le \alpha \le 2\pi0≤α≤2π are α=π5,α=9π5\alpha = \dfrac{\pi}{5}, \alpha = \dfrac{9\pi}{5}α=5π​,α=59π​.


The solutions to the equation cos⁡(α)=1−54\cos(\alpha) = \dfrac{1-\sqrt{5}}{4}cos(α)=41−5​​ over the interval 0≤α≤2π0 \le \alpha \le 2\pi0≤α≤2π are α=3π5,α=7π5\alpha = \dfrac{3\pi}{5}, \alpha = \dfrac{7\pi}{5}α=53π​,α=57π​.


Therefore, the solutions to the equation 2cos⁡(2α)=2cos⁡(α)−12\cos(2\alpha) = 2\cos(\alpha) - 12cos(2α)=2cos(α)−1 over the interval 0≤α≤2π0 \le \alpha \le 2\pi0≤α≤2π are α=π5,α=3π5,α=7π5,α=9π5‾\underline{\alpha = \dfrac{\pi}{5}, \alpha = \dfrac{3\pi}{5}, \alpha = \dfrac{7\pi}{5}, \alpha = \dfrac{9\pi}{5}}α=5π​,α=53π​,α=57π​,α=59π​​.



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FAQs - Frequently Asked Questions

How can addition formulae help me solve trigonometric equations?

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

How can double angle formulae help me solve trigonometric equations?

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

What is the formula for tan(a-b)?

(tan(a)-tan(b))/(1 + tan(a)tan(b))

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