回復 1# thankyou 的帖子
\(\begin{align}
& 8{{\sin }^{2}}\theta +\sqrt{2}\cos \theta =7 \\
& 8{{\cos }^{2}}\theta -\sqrt{2}\cos \theta -1=0 \\
& \cos \theta =\frac{\sqrt{2}+\sqrt{34}}{16} \\
& \cos {{\theta }_{1}}=\cos {{\theta }_{2}}=\frac{\sqrt{2}+\sqrt{34}}{16} \\
& 32{{\cos }^{2}}\frac{{{\theta }_{1}}}{2}{{\cos }^{2}}\frac{{{\theta }_{2}}}{2}=32\times {{\left( \frac{1+\cos {{\theta }_{1}}}{2} \right)}^{2}}=\frac{73+8\sqrt{2}+\sqrt{17}+8\sqrt{34}}{8} \\
\end{align}\)
