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Miscellaneous Exercise 2 · Q83

Q.Prove that 2(sin⁡6θ+cos⁡6θ)−3(sin⁡4θ+cos⁡4θ)+1=02(\sin^6\theta + \cos^6\theta) − 3(\sin^4\theta + \cos^4\theta) + 1 = 0.

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Step 1. From the a³+b³ identity, sin⁡6θ+cos⁡6θ=1−3sin⁡2θcos⁡2θ\sin^6\theta+\cos^6\theta = 1-3\sin^2\theta\cos^2\theta (proved in §2.2.4, Example 5).

Step 2. From squaring sin²θ+cos²θ=1, 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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