請教一題三角函數

請教一題三角函數,如附件, 謝謝!!

若 \(\displaystyle -90^\circ <\theta_1<\theta_2<90^\circ\) 且\(\theta_1, \theta_2\) 為\(\displaystyle 8\sin^2\theta +\sqrt{2}\cos \theta =7\) 的兩根，求 \(\displaystyle 32\cos^2\frac{\theta _1}{2} \cos^2\frac{\theta _2}{2}=?\)

\(\displaystyle 8\sin^2\theta +\sqrt{2}\cos \theta =7\)

\(\displaystyle 8\left(1-\cos^2\theta\right) +\sqrt{2}\cos \theta =7\)

\(8\cos^2\theta-\sqrt{2}\cos\theta-1=0\)

令 \(t=\cos\theta\)，則 \(8t^2-\sqrt{2} t-1=0\) 的兩根為 \(\cos\theta_1\) 與 \(\cos\theta_2\)

由根與係數關係式，可知 \(\displaystyle\cos\theta_1+\cos\theta_2=\frac{\sqrt{2}}{8}\) 且 \(\displaystyle\cos\theta_1 \cos\theta_2=\frac{-1}{8}\)

所求 \(\displaystyle=32\cos^2\frac{\theta _1}{2} \cos^2\frac{\theta _2}{2}=32\left(\frac{1+\cos\theta_1}{2}\right)\left(\frac{1+\cos\theta_2}{2}\right)=.........\)

-----------------如後續 thepiano 老師的回覆，我上面寫的掉入陷阱了～ＸＤ

因為 \(\theta_1,\theta_2\) 在第一、四象限的關係，所以 \(\displaystyle\cos\theta_1=\cos\theta_2 = \frac{\sqrt{2}+\sqrt{34}}{16}\)

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

這題有陷阱...

thepiano 老師說的對， \(\cos\theta_1, \cos\theta_2\) 皆為正數，所以～我上面掉入陷阱了（後續沒檢查～