Question

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

Answer 1

Given:
$$\begin{gathered} \sin \alpha =\frac{4}{5} \\ \cos \beta =\frac{5}{13} \end{gathered}$$.
Now,
$$\begin{gathered} \cos \alpha =\sqrt{1-{\sin }^2\mathrm{\alpha }}=\sqrt{1-{\left(\frac{4}{5}\right)}^2}=\frac{3}{5} \\ \mathrm{And}, \\ \mathrm{sin\beta }=\sqrt{1-{\cos }^2\mathrm{\alpha }}=\sqrt{1-{\left(\frac{5}{13}\right)}^2}=\frac{12}{13} \end{gathered}$$

Now,

$$\begin{gathered} \cos \left(\mathrm{\alpha }-\mathrm{\beta }\right)=\mathrm{cos\alpha }\times \mathrm{cos\beta }+\mathrm{sin\alpha }\times \mathrm{sin\beta } \\ \Rightarrow \cos \left(\mathrm{\alpha }-\mathrm{\beta }\right)=\frac{3}{5}\times \frac{5}{13}\times \frac{4}{5}\times \frac{12}{13}=\frac{63}{65} \\ \mathrm{Thus}, \\ \cos \frac{\mathrm{\alpha }-\mathrm{\beta }}{2}=\sqrt{\frac{1+\cos \left(\mathrm{\alpha }-\mathrm{\beta }\right)}{2}} \\ =\sqrt{\frac{1+{\displaystyle \frac{63}{65}}}{2}} \\ =\frac{8}{\sqrt{65}} \end{gathered}$$