If $sinθ+cosθ=x$, prove that $sin^6θ+cos^6θ=\frac{4−3(x^2−1)^2}{4}$.

Given: $sinθ+cosθ=x$

To do: To prove that $sin^6θ+cos^6θ=\frac{4−3(x^2−1)^2}{4}$.

As given, $sin\theta +cos\theta=x$

$\Rightarrow sin^2\theta+cos^2\theta +2sin\theta cos\theta=x^2$

$\Rightarrow 1+2sin\theta cos\theta=x^2$

$\Rightarrow sin\theta cos\theta=\frac{( x^2-1)}{2}$

$sin^6\theta +cos^6\theta =( sin^2\theta )^3+( cos^2\theta )^3$

$=( sin^2\theta +cos^2\theta )( sin^4\theta +cos^4\theta - sin^2\theta cos^2\theta )$

$=((sin^2\theta +cos^2\theta )^2- 2sin^2\theta cos^2\theta - sin^2\theta cos^2\theta )=1- 3sin^2\theta cos^2\theta $

$=1- 3( \frac{( x^2-1)}{2})^2$

$=\frac{4- 3( x^2-1)^2}{4}$