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Question: Period of \(4\cos 3\theta\)is....

Period of 4cos⁡3θ4\cos 3\thetais.

A

2π3\frac{2\pi}{3}

B

π3\frac{\pi}{3}

C

f(x)=sin⁡4x+cos⁡4xf(x) = \sin^{4}x + \cos^{4}x

D

(sin⁡2x+cos⁡2x)2−2sin⁡2xcos⁡2x(\sin^{2}x + \cos^{2}x)^{2} - 2\sin^{2}x\cos^{2}x

Answer

(sin⁡2x+cos⁡2x)2−2sin⁡2xcos⁡2x(\sin^{2}x + \cos^{2}x)^{2} - 2\sin^{2}x\cos^{2}x

Explanation

Solution

Period of =12k3[sin⁡2A(sin⁡2B+sin⁡2C)+sin⁡2B(sin⁡2C+sin⁡2A)= \frac{1}{2}k^{3}\lbrack\sin^{2}A(\sin 2B + \sin 2C) + \sin^{2}B(\sin 2C + \sin 2A) is +sin⁡2C(sin⁡2A+sin⁡2B)]+ \sin^{2}C(\sin 2A + \sin 2B)\rbrack and period of =k3[sin⁡Asin⁡B(sin⁡Acos⁡B+cos⁡Asin⁡B)= k^{3}\lbrack\sin A\sin B(\sin A\cos B + \cos A\sin B) is +sin⁡Bsin⁡C(sin⁡Bcos⁡C+cos⁡Bsin⁡C)+ \sin B\sin C(\sin B\cos C + \cos B\sin C). Therefore period of the expression is +sin⁡Csin⁡A(sin⁡Ccos⁡A+cos⁡Csin⁡A)]+ \sin C\sin A(\sin C\cos A + \cos C\sin A)\rbrack.