# Question #02a6f

Feb 6, 2017

To be contd.............

#### Explanation:

We use, ${a}^{3} + {b}^{3} = \left(a + b\right) \left({a}^{2} - a b + {b}^{2}\right)$ to get,

${\left(\cos \theta\right)}^{6} + {\left(\sin \theta\right)}^{6} = {\cos}^{6} \theta + {\sin}^{6} \theta$

$= \left({\cos}^{2} \theta + {\sin}^{2} \theta\right) \left({\cos}^{4} \theta - {\cos}^{2} \theta {\sin}^{2} \theta + {\sin}^{4} \theta\right)$

$= \left(1\right) \left\{{\left({\cos}^{2} \theta + {\sin}^{2} \theta\right)}^{2} - 2 {\sin}^{2} \theta {\cos}^{2} \theta - {\cos}^{2} \theta {\sin}^{2} \theta\right\}$

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

$= 1 - \frac{3}{4} {\left(2 \sin \theta \cos \theta\right)}^{2}$

$= 1 - \frac{3}{4} {\left(\sin 2 \theta\right)}^{2}$

$= 1 - \frac{3}{4} \left({\sin}^{2} \left(2 \theta\right)\right)$

Recall that, $1 - \cos 2 A = 2 {\sin}^{2} A$.

Hence, the Exp.$= 1 - \frac{3}{8} \left\{2 {\sin}^{2} \left(2 \theta\right)\right\}$

$= 1 - \frac{3}{8} \left(1 - \cos 4 \theta\right)$

$= 1 - \frac{3}{8} + \frac{3}{8} \cos 4 \theta$

$= \frac{5}{8} + \frac{3}{8} \cos 4 \theta$