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Worked solution

Reduce a fourth power using double angles twice

Show that cos⁡4θ≡18(3+4cos⁡2θ+cos⁡4θ)\cos^4\theta\equiv\dfrac18(3+4\cos2\theta+\cos4\theta).

CJC · O Level revision · Example 21

Answer

cos⁡4θ≡18(3+4cos⁡2θ+cos⁡4θ)\cos^4\theta\equiv\dfrac18(3+4\cos2\theta+\cos4\theta).

Compare your reasoning

Rearrange the first double-angle identity

cos⁡4θ=(cos⁡2θ)2=(1+cos⁡2θ2)2=14(1+2cos⁡2θ+cos⁡22θ)\begin{aligned}&\cos^4\theta\\ & =(\cos^2\theta)^2\\ & =\left(\frac{1+\cos2\theta}{2}\right)^2\\ & =\frac14(1+2\cos2\theta+\cos^22\theta)\end{aligned}

As the source callout says, make cos²θ the subject of cos2θ=2cos²θ−1.

Reduce the remaining squared cosine

cos⁡4θ=14[1+2cos⁡2θ+(1+cos⁡4θ2)]=18(3+4cos⁡2θ+cos⁡4θ)\begin{aligned}&\cos^4\theta\\ & =\frac14\left[1+2\cos2\theta+\left(\frac{1+\cos4\theta}{2}\right)\right]\\ & =\frac18(3+4\cos2\theta+\cos4\theta)\end{aligned}

Now rearrange cos4θ=2cos²2θ−1, the second source callout.

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