Question

Pls answer this trigonometry equation.

Answer 1

Answer:

Step-by-step explanation:

Given that sinθ+cosθ=x

(sinθ+cosθ)2=x2

sin2⁡θ+cos2⁡θ+2sin⁡θcos⁡θ=x2

1+2sinθcosθ=x2

sinθcosθ=(x2−1)/2 .........(1)

∴sin6θ+cos6θ

=(sin2θ)3+(cos2θ)3

=(sin2θ+cos2θ)((sin2θ)2+(cos2θ)2−sin2θcos2θ)

=(1)((sin2θ)2+(cos2θ)2+2sin2θcos2θ−3sin2θcos2θ)

=(sin2⁡θ+cos2⁡θ)2−3sin2⁡θcos2⁡θ

=(1)2−3(sinθcosθ)2

=1−3(x2−1/2)2(from (1), setting sinθcosθ=(x2−1)/2)

=1−{3(x2−1/2)²/4}

=4−3(x2−1)²/4

Proved.