$If \sin \theta +\cos \theta =\frac{1}{2}, then 16\left(\sin \right(2\theta )+\cos (4\theta )+\sin (6\theta \left)\right) is equal to$

$\begin{array}{lll} & & 16(\sin 2\mathrm{\theta }+\cos 4\mathrm{\theta }+\sin 6\mathrm{\theta })\\ & & 16(2\sin 4\mathrm{\theta } \sin 2\mathrm{\theta }+\cos 4\mathrm{\theta })\\ & & 16(2\sin 4\mathrm{\theta } \cos 2\mathrm{\theta }+\cos 4\mathrm{\theta })\\ & & \begin{array}{rcl}\mathrm{sin\theta }+\mathrm{cos\theta }& =& \frac{1}{2} \Rightarrow 1+\sin 2\mathrm{\theta }=\frac{1}{4}\end{array}\\ & & \begin{array}{rcl}\sin 2\mathrm{\theta }& =& \frac{-3}{4} \cos 2\mathrm{\theta }=\frac{\sqrt{7}}{4}\end{array}\\ & & 16(4\sin 2\mathrm{\theta } {\cos }^2\mathrm{\theta }+2{\cos }^{2}2\mathrm{\theta }-1)\\ & & 16\left(4\left(\frac{-3}{4}\right)·\frac{7}{16}+2\left(\frac{7}{16}\right)-1\right)\\ & & \begin{array}{lll}16\left(\frac{-21+14}{16}-1\right)& =& 16\left(\frac{-23}{16}\right)=-23\end{array}\end{array}$