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Miscellaneous Exercise 2 · Q84

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

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Step 1. cos⁡4θ−sin⁡4θ=(cos⁡2θ−sin⁡2θ)(cos⁡2θ+sin⁡2θ)=cos⁡2θ−sin⁡2θ\cos^4\theta-\sin^4\theta = (\cos^2\theta-\sin^2\theta)(\cos^2\theta+\sin^2\theta) = \cos^2\theta-\sin^2\theta (using sin²θ+cos²θ=1).

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

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