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