Question

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

Answer 1

We have,

$sin\alpha =\frac{4}{5}\text{and}cos\beta =\frac{5}{13}\Rightarrow cos\alpha =\frac{3}{5}\text{and}sin\beta =\frac{12}{13}\therefore cos(\alpha -\beta )=cos\alpha cos\beta +sin\alpha .sin\beta =\frac{3}{5}.\frac{5}{13}+\frac{4}{5}.\frac{12}{13}=\frac{15}{65}+\frac{48}{65}=\frac{63}{65}$

Now,

$cos\left(\frac{\alpha -\beta }{2}\right)=\sqrt{\frac{1+cos(\alpha -\beta )}{2}}=\sqrt{\frac{1+\frac{63}{65}}{2}}=\sqrt{\frac{128}{65\times 2}}=\sqrt{\frac{64}{65}}=\pm \frac{8}{\sqrt{65}}\therefore cos\left(\frac{\alpha -\beta }{2}\right)=\frac{8}{\sqrt{65}}$