Question

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

Answer 1

$\sin \theta +\cos \theta =x$ [ Given ]

$(\sin \theta +\cos \theta {)}^2=x^2$ [ Squaring both sides ]

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

$\Rightarrow$ $({\sin }^2\theta +{\cos }^2\theta )+2\sin \theta \cos \theta =x^2$

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

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

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

$\Rightarrow$ $(\sin \theta \cos \theta {)}^2=\frac{(x^2-1{)}^2}{4}$ [ Squaring both sides ]

$\Rightarrow$ ${\sin }^2\theta {\cos }^2\theta =\frac{(x^2-1{)}^2}{4}$

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

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

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

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

$=\frac{4-3(x^2-1{)}^2}{4}$ [ Hence proved ]

$\therefore$ ${\sin }^6\theta +{\cos }^6\theta ==\frac{4-3(x^2-1{)}^2}{4}$