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Exercise 3.3 · Q56

Q.Prove the following: 2cos⁡4x+12cos⁡x+1=(2cos⁡x−1)(2cos⁡2x−1)\dfrac{2\cos4x+1}{2\cos x+1}=(2\cos x-1)(2\cos2x-1)

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Step 1: Expand RHS: (2cos⁡x−1)(2cos⁡2x−1)=4cos⁡xcos⁡2x−2cos⁡x−2cos⁡2x+1(2\cos x-1)(2\cos2x-1)=4\cos x\cos2x-2\cos x-2\cos2x+1.

Step 2: 4cos⁡xcos⁡2x=2(cos⁡x+cos⁡3x)4\cos x\cos2x=2(\cos x+\cos3x) (product-to-sum), so RHS =2cos⁡x+2cos⁡3x−2cos⁡x−2cos⁡2x+1=2cos⁡3x−2cos⁡2x+1=2\cos x+2\cos3x-2\cos x-2\cos2x+1=2\cos3x-2\cos2x+1.

Step 3: Multiply RHS by (2cos⁡x+1)(2\cos x+1) and expand using 4cos⁡3xcos⁡x=2cos⁡2x+2cos⁡4x4\cos3x\cos x=2\cos2x+2\cos4x and 4cos⁡2xcos⁡x=2cos⁡x+2cos⁡3x4\cos2x\cos x=2\cos x+2\cos3x; after collecting terms every cos⁡2x,cos⁡3x,cos⁡x\cos2x,\cos3x,\cos x term cancels in pairs, leaving 2cos⁡4x+12\cos4x+1. …

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