Question
If $\sin\alpha=\frac{4}{5}$ and $\cos\beta=\frac{5}{13},$ prove that $\cos\frac{\alpha-\beta}{2}=\frac{8}{\sqrt{65}}$

Answer

We have,
$\sin\alpha=\frac{4}{5}\ \&\cos\beta=\frac{4}{5}\Rightarrow\cos\alpha=\frac{3}{5}\ \&\cos\alpha=\frac{3}{5}\ \&\sin\beta=\frac{12}{13}$
$\therefore\cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha.\sin\beta$
$=\frac{3}{5}.\frac{5}{13}+\frac{4}{5}.\frac{12}{13}$
$=\frac{15}{65}+\frac{48}{65}=\frac{63}{65}$
Now,
$\cos\Big(\frac{\alpha-\beta}{2}\Big)=\sqrt{\frac{1+\cos(\alpha-\beta)}{2}}$
$=\sqrt{\frac{1+\frac{63}{65}}{2}}$
$=\sqrt{\frac{128}{65\times2}}=\sqrt{\frac{64}{65}}$
$=\pm\frac{8}{\sqrt{65}}$
$\therefore\cos\Big(\frac{\alpha-\beta}{2}\Big)=\frac{8}{\sqrt{65}}$

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