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Miscellaneous Exercise 2 · Q85

Q.Prove that sin⁡4θ+2sin⁡2θcos⁡2θ=1−cos⁡4θ\sin^4\theta + 2\sin^2\theta\cos^2\theta = 1 − \cos^4\theta.

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Step 1. RHS =1−cos⁡4θ=(1−cos⁡2θ)(1+cos⁡2θ)=sin⁡2θ(1+cos⁡2θ)=1-\cos^4\theta = (1-\cos^2\theta)(1+\cos^2\theta) = \sin^2\theta(1+\cos^2\theta). …

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