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Exercise 3.12 · Q8

Q.Let fk(x)=1k(sin⁡kx+cos⁡kx)f_k(x)=\dfrac1k\left(\sin^kx+\cos^kx\right) where x∈Rx\in\mathbb R and k≥1k\ge1. Then f4(x)−f6(x)=f_4(x)-f_6(x)=

(1) 14\dfrac14
(2) 112\dfrac1{12}
(3) 16\dfrac16
(4) 13\dfrac13
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Write sin⁡4x+cos⁡4x\sin^4x+\cos^4x and sin⁡6x+cos⁡6x\sin^6x+\cos^6x in terms of s=sin⁡2xcos⁡2xs=\sin^2x\cos^2x; the ss-terms cancel in the difference.

Step 1. sin⁡4x+cos⁡4x=(sin⁡2x+cos⁡2x)2−2sin⁡2xcos⁡2x=1−2s\sin^4x+\cos^4x=(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x=1-2s, where s=sin⁡2xcos⁡2xs=\sin^2x\cos^2x. So f4(x)=14(1−2s)f_4(x)=\dfrac14(1-2s).

Step 2. sin⁡6x+cos⁡6x=(sin⁡2x+cos⁡2x)3−3sin⁡2xcos⁡2x(sin⁡2x+cos⁡2x)=1−3s\sin^6x+\cos^6x=(\sin^2x+\cos^2x)^3-3\sin^2x\cos^2x(\sin^2x+\cos^2x)=1-3s. So f6(x)=16(1−3s)f_6(x)=\dfrac16(1-3s). …

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