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If $\cos A=\frac{7}{25}$ and $\frac{3 \pi}{2} < A < 2 \pi$, then $\cos \frac{A}{4}+\cos \frac{A}{2}-\cos 2 A=$
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The correct answer is:
$\frac{1}{\sqrt{10}}+\frac{27}{625}$
Given, $\quad \cos A=\frac{7}{25}$ and $\frac{3 \pi}{2} < A < 2 \pi$
Then, $\quad \cos 2 A=2\left(\frac{49}{625}\right)-1=\frac{98-625}{625}=-\frac{527}{625}$ $\cos \frac{A}{2}=-\frac{4}{5}$ and $\cos \frac{A}{4}=\frac{1}{\sqrt{10}}$
So, $\quad \cos \frac{A}{4}+\cos \frac{A}{2}-\cos 2 A=\frac{1}{\sqrt{10}}+\frac{27}{625}$.
Then, $\quad \cos 2 A=2\left(\frac{49}{625}\right)-1=\frac{98-625}{625}=-\frac{527}{625}$ $\cos \frac{A}{2}=-\frac{4}{5}$ and $\cos \frac{A}{4}=\frac{1}{\sqrt{10}}$
So, $\quad \cos \frac{A}{4}+\cos \frac{A}{2}-\cos 2 A=\frac{1}{\sqrt{10}}+\frac{27}{625}$.
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