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Miscellaneous Exercise 2 · Q82

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

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Step 1. (sin⁡2θ+cos⁡2θ)2=12=1(\sin^2\theta+\cos^2\theta)^2 = 1^2 = 1.

Step 2. Expand the left side: sin⁡4θ+2sin⁡2θcos⁡2θ+cos⁡4θ=1\sin^4\theta+2\sin^2\theta\cos^2\theta+\cos^4\theta=1. …

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