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Exercise 3.1 · Q11

Q.Prove that: cos⁡(x+y)⋅cos⁡(x−y)=cos⁡2y−sin⁡2x\cos(x+y)\cdot\cos(x-y)=\cos^2y-\sin^2x

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Step 1: cos⁡(x+y)cos⁡(x−y)=(cos⁡xcos⁡y−sin⁡xsin⁡y)(cos⁡xcos⁡y+sin⁡xsin⁡y)=cos⁡2xcos⁡2y−sin⁡2xsin⁡2y\cos(x+y)\cos(x-y)=(\cos x\cos y-\sin x\sin y)(\cos x\cos y+\sin x\sin y)=\cos^2x\cos^2y-\sin^2x\sin^2y.

Step 2: Replace cos⁡2x=1−sin⁡2x\cos^2x=1-\sin^2x and sin⁡2y=1−cos⁡2y\sin^2y=1-\cos^2y: =(1−sin⁡2x)cos⁡2y−sin⁡2x(1−cos⁡2y)=(1-\sin^2x)\cos^2y-\sin^2x(1-\cos^2y). …

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