Question
If $\sin\theta+\cos\theta=\text{x},$ prove that $\sin^6\theta+\cos^6\theta=\frac{4-3(\text{x}^2-1)^2}{4}.$

Answer

Given,
$\sin\theta+\cos\theta=\text{x}$
Squaring the given equation, we have
$(\sin\theta+\cos\theta)^2=\text{x}^2$
$\Rightarrow\ \sin^2+2\sin\theta\cos\theta+\cos^2\theta=\text{x}^2$
$\Rightarrow\ (\sin^2\theta+\cos^2\theta)+2\sin\theta\cos\theta=\text{x}^2$
$\Rightarrow\ 1+2\sin\theta\cos\theta=\text{x}^2$
$\Rightarrow\ 2\sin\theta\cos\theta=\text{x}^2-1$
$\Rightarrow \sin\theta\cos\theta=\frac{\text{x}^2-1}{2}$
Squaring the last equation, we have
$(\sin\theta\cos\theta)^2=\frac{(\text{x}^2-1)^2}{4}$
$\Rightarrow \sin^2\theta\cos^2\theta=\frac{(\text{x}^2-1)^2}{4}$
Therefore, we have
$\sin^6\theta+\cos^6\theta=(\sin^2\theta)^3+(\cos^2\theta)^3$
$=(\sin^2\theta+\cos^2\theta)^3-3\sin^2\theta\cos^2\theta(\sin^2\theta+\cos^2\theta)$
$=(1)^3-3\frac{(\text{x}^2-1)^2}{4}(1)$
$=1-3\frac{(\text{x}^2-1)^2}{4}$
$=\frac{4-3(\text{x}^2-1)^2}{4}$
Hence proved.

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