Solveeit Logo
Login

Question

Question: \[\sin^{6}\theta + \cos^{6}\theta + 3\sin^{2}\theta\cos^{2}\theta =\]...

sin⁡6θ+cos⁡6θ+3sin⁡2θcos⁡2θ=\sin^{6}\theta + \cos^{6}\theta + 3\sin^{2}\theta\cos^{2}\theta =

A

0

B

–1

C

1

D

None of these

Answer

1

Explanation

Solution

sin⁡6θ+cos⁡6θ+3sin⁡2θcos⁡2θ\sin^{6}\theta + \cos^{6}\theta + 3\sin^{2}\theta\cos^{2}\theta

=(sin⁡2θ+cos⁡2θ)3−3sin⁡2θcos⁡2θ+3sin⁡2θcos⁡2θ=1.= (\sin^{2}\theta + \cos^{2}\theta)^{3} - 3\sin^{2}\theta\cos^{2}\theta + 3\sin^{2}\theta\cos^{2}\theta = 1.

Trick : Put θ=0o,\theta = 0^{o}, we get the value of expression equal to 1. Again put θ=45o,\theta = 45^{o}, the value remains 1, it means that the expression is independent of θ and is equal to 1.