Question

If $cos\alpha +sin\alpha$ =$\frac{3}{4}$,then $si{n}^6\alpha +co{s}^6\alpha$ is equal to

• A. $\frac{877}{1024}$
• B. $\frac{777}{1024}$
• C. $\frac{878}{1024}$
• D. $\frac{789}{1024}$
• E. $\frac{878}{1064}$

Answer 1

Answer: (A)

Given, $\sin \alpha +\cos \alpha =\frac{3}{4}\dots (i)$
Now, ${\sin }^6\alpha +{\cos }^6\alpha$
$={\left({\sin }^2\alpha \right)}^3+{\left({\cos }^2\alpha \right)}^3=\left({\sin }^2\alpha +{\cos }^2\alpha \right)$
$\left({\sin }^4\alpha +{\cos }^4\alpha -{\sin }^2\alpha \cdot {\cos }^2\alpha \right)$
$=1\cdot \left\{{\left({\sin }^2\alpha +{\cos }^2\alpha \right)}^2-3{\sin }^2\alpha \cdot {\cos }^2\alpha \right\}$
$\left(\because {\sin }^2\alpha +{\cos }^2\alpha =1\right)$
$=\left\{1-\frac{3}{4}(\sin 2\alpha {)}^2\right\}\dots (ii)$
On squaring both sides of Eq. (i), we get
$\left({\sin }^2\alpha +{\cos }^2\alpha \right)+2\sin \alpha \cdot \cos \alpha =\frac{9}{16}$
$\Rightarrow \sin 2\alpha =\frac{9}{16}-1=\frac{-7}{16}$
$\Rightarrow (\sin 2\alpha {)}^2=\frac{49}{256}$
On putting this value in Eq. (ii), we get
${\sin }^6\alpha +{\cos }^6\alpha =\left\{1-\frac{3}{4}\times \frac{49}{756}\right\}=\left(1-\frac{147}{1024}\right)$
$=\frac{877}{1024}$