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.