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Exercise 3.4 · Q67

Q.Prove the following: sin⁡x−sin⁡3x+sin⁡5x−sin⁡7xcos⁡x−cos⁡3x−cos⁡5x+cos⁡7x=cot⁡2x\dfrac{\sin x-\sin3x+\sin5x-\sin7x}{\cos x-\cos3x-\cos5x+\cos7x}=\cot2x

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Step 1: Group the numerator as (sin⁡x+sin⁡5x)−(sin⁡3x+sin⁡7x)=2sin⁡3xcos⁡2x−2sin⁡5xcos⁡2x=2cos⁡2x(sin⁡3x−sin⁡5x)(\sin x+\sin5x)-(\sin3x+\sin7x)=2\sin3x\cos2x-2\sin5x\cos2x=2\cos2x(\sin3x-\sin5x).

Step 2: sin⁡3x−sin⁡5x=−2cos⁡4xsin⁡x\sin3x-\sin5x=-2\cos4x\sin x, so the numerator is 2cos⁡2x(−2cos⁡4xsin⁡x)=−4cos⁡2xcos⁡4xsin⁡x2\cos2x(-2\cos4x\sin x)=-4\cos2x\cos4x\sin x.

Step 3: Group the denominator as (cos⁡x−cos⁡3x)−(cos⁡5x−cos⁡7x)=2sin⁡2xsin⁡x−2sin⁡6xsin⁡x=2sin⁡x(sin⁡2x−sin⁡6x)(\cos x-\cos3x)-(\cos5x-\cos7x)=2\sin2x\sin x-2\sin6x\sin x=2\sin x(\sin2x-\sin6x).

Step 4: sin⁡2x−sin⁡6x=−2cos⁡4xsin⁡2x\sin2x-\sin6x=-2\cos4x\sin2x, so the denominator is 2sin⁡x(−2cos⁡4xsin⁡2x)=−4sin⁡xcos⁡4xsin⁡2x2\sin x(-2\cos4x\sin2x)=-4\sin x\cos4x\sin2x. …

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