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If $\cos \theta=\frac{3}{5}$ and $\cos \phi=\frac{4}{5}$, where $\theta$ and $\phi$ are positive acute angles, then $\cos \frac{\theta-\phi}{2}=$
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The correct answer is:
$\frac{7}{5 \sqrt{2}}$
We have $\cos \theta=\frac{3}{5}$ and $\cos \phi=\frac{4}{5}$
$\begin{aligned}
& \begin{aligned}
\text {Therefore } \cos (\theta-\phi) & =\cos \theta \cos \phi+\sin \theta \sin \phi \\
& =\frac{3}{5} \cdot \frac{4}{5}+\frac{4}{5} \cdot \frac{3}{5}=\frac{24}{25}
\end{aligned} \\
& \text { But } 2 \cos ^2\left(\frac{\theta-\phi}{2}\right)=1+\cos (\theta-\phi)=1+\frac{24}{25}=\frac{49}{50} \\
& \therefore \cos ^2\left(\frac{\theta-\phi}{2}\right)=\frac{49}{50} \text {. Hence, } \cos \left(\frac{\theta-\phi}{2}\right)=\frac{7}{5 \sqrt{2}}
\end{aligned}$
$\begin{aligned}
& \begin{aligned}
\text {Therefore } \cos (\theta-\phi) & =\cos \theta \cos \phi+\sin \theta \sin \phi \\
& =\frac{3}{5} \cdot \frac{4}{5}+\frac{4}{5} \cdot \frac{3}{5}=\frac{24}{25}
\end{aligned} \\
& \text { But } 2 \cos ^2\left(\frac{\theta-\phi}{2}\right)=1+\cos (\theta-\phi)=1+\frac{24}{25}=\frac{49}{50} \\
& \therefore \cos ^2\left(\frac{\theta-\phi}{2}\right)=\frac{49}{50} \text {. Hence, } \cos \left(\frac{\theta-\phi}{2}\right)=\frac{7}{5 \sqrt{2}}
\end{aligned}$
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