Question

If $\sin \theta +\cos \theta =\frac{a}{2},$ then the value of ${\sin }^6\theta +{\cos }^6\theta$ is

• A. $\frac{32-3(a^2-2{)}^2}{32}$
  
• B. $\frac{64-3(a^2-4{)}^2}{64}$
  
• C. $\frac{64-(a^2-4)}{64}$
  
• D. $\frac{(a^2-4{)}^2}{64}$
  

Answer 1

The correct option is B $\frac{64-3(a^2-4{)}^2}{64}$


Given: $\sin \theta +\cos \theta =\frac{a}{2}$

Squaring on both sides,

$(\sin \theta +\cos \theta {)}^2=\frac{a^2}{4}$

$\Rightarrow {\sin }^2\theta +{\cos }^2\theta +2\sin \theta \cos \theta =\frac{a^2}{4}$

$\Rightarrow 1+2\sin \theta \cos \theta =\frac{a^2}{4}$

$\Rightarrow 2\sin \theta \cos \theta =\frac{a^2}{4}-1$

$\Rightarrow \sin \theta \cos \theta =\frac{1}{2}(\frac{a^2}{4}-1)=\frac{a^2-4}{8}$ ...(i)

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

$=({\sin }^2\theta +{\cos }^2\theta )({\sin }^4\theta +{\cos }^4\theta -{\sin }^2\theta {\cos }^2\theta )[\because a^3+b^3=(a+b\left)\right(a^2+b^2-ab\left)\right]$

$=1\times \left[\right({\sin }^2\theta +{\cos }^2\theta {)}^2-3{\sin }^2\theta {\cos }^2\theta ]$ $(\because {\sin }^2\theta +{\cos }^2\theta =1)$

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

$=1-3{\left(\frac{a^2-4}{8}\right)}^2$ [From (i)]

$=1-3\frac{(a^2-4{)}^2}{64}$

$\Rightarrow {\sin }^6\theta +{\cos }^6\theta =\frac{64-3(a^2-4{)}^2}{64}$

Hence, the correct answer is option (b).