$2{\sin }^2\left(8\frac{1^0}{2}\right)+4\cos {16}^o\sin \left(7\frac{1^0}{2}\right)\sin \left(8\frac{1^0}{2}\right)+\cos {32}^o=$

$$\begin{gathered} 2{\sin }^2\left(8\frac{1^0}{2}\right)+4\cos {16}^o\sin \left(7\frac{1^0}{2}\right)\sin \left(8\frac{1^0}{2}\right)+\cos {32}^o \\ =1-\cos 17°+2\cos {16}^o\left(2\sin \left(8\frac{1^0}{2}\right)\sin \left(7\frac{1^0}{2}\right)\right)+\cos {32}^o \\ =1-\cos 17°+2\cos {16}^o\left(\cos 1°-\cos 16°\right)+\cos {32}^o \\ =1-\cos 17°+2\cos {16}^o\cos 1°-2{\cos }^2{16}^o+\cos 32° \\ =1-\cos 17°+\cos 17°+\cos 15°-(1+\cos 32°)+\cos 32° \\ =\cos 15°=\frac{\sqrt{3}+1}{2\sqrt{2}} \end{gathered}$$