Question

If $\sin \theta +\cos \theta =x,then{\sin }^6\theta +{\cos }^6\theta =$

• A. $\frac{4-3{(x^2-1)}^2}{4}$
• B. $\frac{4-3{(x^2+1)}^2}{4}$
• C. $\frac{3-4{(x^2-1)}^2}{4}$
• D. $\frac{3-4{(x^2+1)}^2}{4}$

Answer 1

The correct option is A $\frac{4-3{(x^2-1)}^2}{4}$

GIven $sin\theta +cos\theta =x$
$\Rightarrow (sin\theta +cos\theta {)}^2=x^2$
OR $Si{n}^2\theta +co{s}^2\theta +2sin\theta cos\theta =x^2$
Or $2sin\theta cos\theta =x^2-1$
Or $sin\theta cos\theta =\frac{1}{2}{x}^2-1$
$si{n}^6\theta +co{s}^6\theta =(si{n}^2\theta {)}^3+(co{s}^2\theta {)}^3$
= $(si{n}^2\theta +co{s}^2\theta {)}^3-3si{n}^2\theta co{s}^2\theta (si{n}^2\theta +co{s}^2\theta )$
=$(si{n}^2\theta +co{s}^2\theta {)}^3-3si{n}^2\theta co{s}^2\theta (si{n}^2\theta +co{s}^2\theta )$
=$1-3(\frac{1-x^2}{2}{)}^2$
=$\frac{4-3(x^2-1{)}^2}{4}$