$\sin \theta + \cos \theta = {1 \over 2}$ ${\sin ^2}\theta + {\cos ^2}\theta + 2\sin \theta \cos \theta = {1 \over 4}$ $\sin 2\theta = - {3 \over 4}$ Now : $\cos 4\theta = 1 - 2{\sin ^2}2\theta$ $= 1 - 2{\left( { - {3 \over 4}} \right)^2}$ $= 1 - 2 \times {9 \over {16}} = - {1 \over 8}$ $\sin 6\theta = 3\sin 2\theta - 4{\sin ^3}2\theta$ $= (3 - 4{\sin ^2}2\theta ).\sin 2\theta$ $= \left[ {3 - 4\left( {{9 \over {16}}} \right)} \right].\left( { - {3 \over 4}} \right)$ $\Rightarrow \left[ {{3 \over 4}} \right] \times \left( { - {3 \over 4}} \right) = - {9 \over {16}}$ $16[\sin 2\theta + \cos 4\theta + \sin 6\theta ]$ = $16\left( { - {3 \over 4} - {1 \over 8} - {9 \over {16}}} \right) = - 23$