Question

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

• A. $\frac{877}{1024}$
• B. $\frac{77}{1024}$
• C. $\frac{878}{1024}$
• D. $\frac{789}{1024}$
• E. $\frac{878}{1064}$

Answer 1

Answer: (A)

$\frac{877}{1024}$

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