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Miscellaneous Exercise 3 (II) · Q105

Q.Prove the following: 3(sin⁡x−cos⁡x)4+6(sin⁡x+cos⁡x)2+4(sin⁡6x+cos⁡6x)=133(\sin x-\cos x)^4+6(\sin x+\cos x)^2+4(\sin^6x+\cos^6x)=13

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Step 1: (sin⁡x−cos⁡x)2=1−2sin⁡xcos⁡x=1−sin⁡2x(\sin x-\cos x)^2=1-2\sin x\cos x=1-\sin2x, so (sin⁡x−cos⁡x)4=(1−sin⁡2x)2(\sin x-\cos x)^4=(1-\sin2x)^2.

Step 2: (sin⁡x+cos⁡x)2=1+sin⁡2x(\sin x+\cos x)^2=1+\sin2x.

Step 3: sin⁡6x+cos⁡6x=1−3sin⁡2xcos⁡2x=1−34sin⁡22x\sin^6x+\cos^6x=1-3\sin^2x\cos^2x=1-\dfrac34\sin^22x (using a3+b3a^3+b^3 factoring with a2+b2=1a^2+b^2=1, and sin⁡2xcos⁡2x=14sin⁡22x\sin^2x\cos^2x=\frac14\sin^22x).

Step 4: Substitute all three: 3(1−sin⁡2x)2+6(1+sin⁡2x)+4(1−34sin⁡22x)3(1-\sin2x)^2+6(1+\sin2x)+4\left(1-\dfrac34\sin^22x\right). …

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