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If $ \sin \theta+\cos \theta=x $, prove that $ \sin ^{6} \theta+\cos ^{6} \theta=\frac{4-3\left(x^{2}-1\right)^{2}}{4} $
Given:
\( \sin \theta+\cos \theta=x \)
To do:
We have to prove that \( \sin ^{6} \theta+\cos ^{6} \theta=\frac{4-3\left(x^{2}-1\right)^{2}}{4} \).
Solution:
We know that,
$\sin^2 A+\cos^2 A=1$
$\operatorname{cosec}^2 A-\cot^2 A=1$
$\sec^2 A-\tan^2 A=1$
$\cot A=\frac{\cos A}{\sin A}$
$\tan A=\frac{\sin A}{\cos A}$
$\operatorname{cosec} A=\frac{1}{\sin A}$
$\sec A=\frac{1}{\cos A}$
Therefore,
$\sin \theta+\cos \theta=x$
Squaring on both sides, we get,
$(\sin \theta+\cos \theta)^{2}=x^{2}$
$\Rightarrow \sin ^{2} \theta+\cos ^{2} \theta+2 \sin \theta \cos \theta=x^{2}$
$\Rightarrow 1+2 \sin \theta \cos \theta=x^{2}$
$\Rightarrow 2 \sin \theta \cos \theta=x^{2}-1$
$\Rightarrow \sin \theta \cos \theta=\frac{x^{2}-1}{2}$
This implies,
$\sin ^{6}+\cos ^{6} \theta=(\sin ^{2} \theta+\cos ^{2} \theta)^{3}-3\sin ^{2} \theta \cos ^{2} \theta(\sin ^{2} \theta+\cos ^{2} \theta)$ [since $a^{3}+b^{3}=(a+b)^{3}-3ab(a+b)$]
$=(1)^{3}-3(\frac{x^{2}-1}{2})^{2} (1)$
$=1-3 \frac{(x^{2}-1)^{2}}{4}$
$=\frac{4-3(x^{2}-1)^{2}}{4}$
Hence proved.
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