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Exercise 3.3 · Q49

Q.Prove the following: 16sin⁡θcos⁡θcos⁡2θcos⁡4θcos⁡8θ=sin⁡16θ16\sin\theta\cos\theta\cos2\theta\cos4\theta\cos8\theta=\sin16\theta

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Step 1: 16sin⁡θcos⁡θcos⁡2θcos⁡4θcos⁡8θ=8(2sin⁡θcos⁡θ)cos⁡2θcos⁡4θcos⁡8θ=8sin⁡2θcos⁡2θcos⁡4θcos⁡8θ16\sin\theta\cos\theta\cos2\theta\cos4\theta\cos8\theta=8(2\sin\theta\cos\theta)\cos2\theta\cos4\theta\cos8\theta=8\sin2\theta\cos2\theta\cos4\theta\cos8\theta.

Step 2: =4(2sin⁡2θcos⁡2θ)cos⁡4θcos⁡8θ=4sin⁡4θcos⁡4θcos⁡8θ=2(2sin⁡4θcos⁡4θ)cos⁡8θ=2sin⁡8θcos⁡8θ=4(2\sin2\theta\cos2\theta)\cos4\theta\cos8\theta=4\sin4\theta\cos4\theta\cos8\theta=2(2\sin4\theta\cos4\theta)\cos8\theta=2\sin8\theta\cos8\theta. …

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