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Miscellaneous Exercise 2 · Q86

Q.Prove that sin⁡3θ+cos⁡3θsin⁡θ+cos⁡θ+sin⁡3θ−cos⁡3θsin⁡θ−cos⁡θ=2\dfrac{\sin^3\theta+\cos^3\theta}{\sin\theta+\cos\theta} + \dfrac{\sin^3\theta−\cos^3\theta}{\sin\theta−\cos\theta} = 2.

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Step 1. sin⁡3θ+cos⁡3θ=(sin⁡θ+cos⁡θ)(sin⁡2θ−sin⁡θcos⁡θ+cos⁡2θ)=(sin⁡θ+cos⁡θ)(1−sin⁡θcos⁡θ)\sin^3\theta+\cos^3\theta = (\sin\theta+\cos\theta)(\sin^2\theta-\sin\theta\cos\theta+\cos^2\theta) = (\sin\theta+\cos\theta)(1-\sin\theta\cos\theta), so sin⁡3θ+cos⁡3θsin⁡θ+cos⁡θ=1−sin⁡θcos⁡θ\dfrac{\sin^3\theta+\cos^3\theta}{\sin\theta+\cos\theta} = 1-\sin\theta\cos\theta. …

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