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$\cos \frac{\pi}{12}=$
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Verified Answer
The correct answer is:
$\frac{\sqrt{2}+\sqrt{6}}{4}$
Let $\theta=\frac{\pi}{12}$
$\because \cos 2 \theta=2 \cos ^2 \theta-1$
$\begin{aligned} & \Rightarrow \cos \left(\frac{\pi}{6}\right)=2 \cos ^2 \theta-1 \Rightarrow 2 \cos ^2 \theta-1=\frac{\sqrt{3}}{2} \\ & \Rightarrow \cos ^2 \theta=\frac{\sqrt{3}+2}{4}=\frac{8+4 \sqrt{3}}{16}\end{aligned}$
$=\frac{2+6+2 \sqrt{2.6}}{16}=\frac{(\sqrt{2}+\sqrt{6})^2}{4^2}$
$\therefore \cos \theta=\cos \frac{\pi}{12}=\frac{\sqrt{2}+\sqrt{6}}{4}$
$\because \cos 2 \theta=2 \cos ^2 \theta-1$
$\begin{aligned} & \Rightarrow \cos \left(\frac{\pi}{6}\right)=2 \cos ^2 \theta-1 \Rightarrow 2 \cos ^2 \theta-1=\frac{\sqrt{3}}{2} \\ & \Rightarrow \cos ^2 \theta=\frac{\sqrt{3}+2}{4}=\frac{8+4 \sqrt{3}}{16}\end{aligned}$
$=\frac{2+6+2 \sqrt{2.6}}{16}=\frac{(\sqrt{2}+\sqrt{6})^2}{4^2}$
$\therefore \cos \theta=\cos \frac{\pi}{12}=\frac{\sqrt{2}+\sqrt{6}}{4}$
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