Question

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

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

Answer 1

Answer: (C)

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

$si{n}^6\alpha +co{s}^6\alpha$
$=(si{n}^2\alpha {)}^3+(co{s}^2\alpha {)}^3$
$=(si{n}^2\alpha +co{s}^2\alpha )(si{n}^4\alpha +co{s}^4\alpha -si{n}^2\alpha co{s}^2\alpha )$
Now
$si{n}^2\alpha +co{s}^2\alpha =1$
Hence
$(si{n}^2\alpha +co{s}^2\alpha )(si{n}^4\alpha +co{s}^4\alpha -si{n}^2\alpha co{s}^2\alpha )$
$=(si{n}^4\alpha +co{s}^4\alpha -si{n}^2\alpha co{s}^2\alpha )$
$=\left(\right(si{n}^2\alpha +co{s}^2\alpha {)}^2-2si{n}^2\alpha co{s}^2\alpha -si{n}^2\alpha co{s}^2\alpha )$
$=(1-3si{n}^2\alpha co{s}^2\alpha )$ ....(i)
Now it is given that
$sin\alpha +cos\alpha =m$
$(si{n}^2\alpha +co{s}^2\alpha )=1$
$si{n}^2\alpha +co{s}^2\alpha +2sin\alpha .cos\alpha =m^2$
$1+2sin\alpha .cos\alpha =m^2$
$m^2-1=2sin\alpha .cos\alpha$
$\frac{m^2-1}{2}=sin\alpha .cos\alpha$
$(sin\alpha .cos\alpha {)}^2=\frac{(1-m^2{)}^2}{4}$
Hence from i
$(1-3si{n}^2\alpha co{s}^2\alpha )$
$=1-\frac{3(1-m^2{)}^2}{4}$
$=\frac{4-3(1-m^2{)}^2}{4}$