Question

If $\sin \alpha +\cos \alpha =m,m^2\le 2$, then ${\sin }^6\alpha +{\cos }^6\alpha =$

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

Answer 1

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

$\sin \alpha +\cos \alpha =m$
squaring both sides
${\sin }^2\alpha +{\cos }^2\alpha +2\sin \alpha \cos \alpha =m^2$

$\sin \alpha \cos \alpha =\frac{m^2-1}{2}$
Now,

$si{n}^6\alpha +co{s}^6\alpha =(si{n}^2\alpha +co{s}^2\alpha )({\sin }^4\alpha +co{s}^4\alpha -si{n}^2\alpha co{s}^2\alpha )$

$=(si{n}^2\alpha +{\cos }^2\alpha {)}^2-3{\sin }^2\alpha {\cos }^2\alpha$

$=1-3{\sin }^2\alpha {\cos }^2\alpha$

Substitute the value of $\sin \alpha \cos \alpha$

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