If $\sin \alpha=\frac{4}{5}$ and $\cos \beta=\frac{5}{13}$, prove that $\cos \frac{\alpha-\beta}{2}=\frac{8}{\sqrt{65}}$
Given:
$\sin \alpha=\frac{4}{5}$
$\cos \beta=\frac{5}{13}$
Now,
$\cos \alpha=\sqrt{1-\sin ^{2} \alpha}=\sqrt{1-\left(\frac{4}{5}\right)^{2}}=\frac{3}{5}$
And,
$\sin \beta=\sqrt{1-\cos ^{2} \alpha}=\sqrt{1-\left(\frac{5}{13}\right)^{2}}=\frac{12}{13}$
Now,
$\cos (\alpha-\beta)=\cos \alpha \times \cos \beta+\sin \alpha \times \sin \beta$
$\Rightarrow \cos (\alpha-\beta)=\frac{3}{5} \times \frac{5}{13} \times \frac{4}{5} \times \frac{12}{13}=\frac{63}{65}$
Thus,
$\cos \frac{\alpha-\beta}{2}=\sqrt{\frac{1+\cos (\alpha-\beta)}{2}}$
$=\sqrt{\frac{1+\frac{63}{65}}{2}}$
$=\frac{8}{\sqrt{65}}$
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