Porisma

Worked solution

Prove the triple-angle cosine identity

Prove that cos⁡3θ≡4cos⁡3θ−3cos⁡θ\cos3\theta\equiv4\cos^3\theta-3\cos\theta. The formulae for sin⁡2θ\sin2\theta and cos⁡2θ\cos2\theta may be quoted without proof.

CJC · O Level revision · Example 20

Answer

cos⁡3θ≡4cos⁡3θ−3cos⁡θ\cos3\theta\equiv4\cos^3\theta-3\cos\theta.

Compare your reasoning

Expand the sum of angles

cos⁡3θ=cos⁡(2θ+θ)=cos⁡2θcos⁡θ−sin⁡θsin⁡2θ\cos3\theta=\cos(2\theta+\theta)=\cos2\theta\cos\theta-\sin\theta\sin2\theta

Use the source addition-formula callout with A=2θ and B=θ.

Replace both double angles

cos⁡3θ=(2cos⁡2θ−1)cos⁡θ−sin⁡θ(2sin⁡θcos⁡θ)=(2cos⁡2θ−1)cos⁡θ−2sin⁡2θcos⁡θ.\begin{aligned}\cos3\theta&=(2\cos^2\theta-1)\cos\theta-\sin\theta(2\sin\theta\cos\theta)\\&=(2\cos^2\theta-1)\cos\theta-2\sin^2\theta\cos\theta.\end{aligned}

The sine formula returns every angle to θ. Choose cos2θ=2cos²θ−1 to work towards the requested cosine-only expression; the source callout incorrectly says “sine terms only”.

Use the basic identity and collect like terms

cos⁡3θ=(2cos⁡2θ−1)cos⁡θ−2(1−cos⁡2θ)cos⁡θ=2cos⁡3θ−cos⁡θ−2cos⁡θ+2cos⁡3θ=4cos⁡3θ−3cos⁡θ.\begin{aligned}\cos3\theta&=(2\cos^2\theta-1)\cos\theta-2(1-\cos^2\theta)\cos\theta\\&=2\cos^3\theta-\cos\theta-2\cos\theta+2\cos^3\theta\\&=4\cos^3\theta-3\cos\theta.\end{aligned}

Replace sin²θ by 1−cos²θ, retaining each printed expansion stage.

How did this question go?

Solution notes

The native source preserves the “sine terms only” callout. The lesson corrects it to cosine-only terms, matching the formula and target identity.