If \[ \sin\alpha=\frac{4}{5} \] and \[ \cos\beta=\frac{5}{13}, \] prove that \[ \cos\frac{\alpha-\beta}{2} = \frac{8}{\sqrt{65}} \]
Question
If \[ \sin\alpha=\frac{4}{5} \] and \[ \cos\beta=\frac{5}{13}, \] prove that \[ \cos\frac{\alpha-\beta}{2} = \frac{8}{\sqrt{65}}. \]
Solution
Given,
\[ \sin\alpha=\frac{4}{5} \]
Using
\[ \sin^2\alpha+\cos^2\alpha=1 \]
\[ \cos\alpha = \sqrt{1-\left(\frac{4}{5}\right)^2} \]
\[ = \sqrt{1-\frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5} \]
Also,
\[ \cos\beta=\frac{5}{13} \]
Using
\[ \sin^2\beta+\cos^2\beta=1 \]
\[ \sin\beta = \sqrt{1-\left(\frac{5}{13}\right)^2} \]
\[ = \sqrt{1-\frac{25}{169}} = \sqrt{\frac{144}{169}} = \frac{12}{13} \]
Now use the identity
\[ \cos(\alpha-\beta) = \cos\alpha\cos\beta+\sin\alpha\sin\beta \]
Substituting the values,
\[ \cos(\alpha-\beta) = \frac{3}{5}\times\frac{5}{13} + \frac{4}{5}\times\frac{12}{13} \]
\[ = \frac{15}{65} + \frac{48}{65} = \frac{63}{65} \]
Using the half-angle identity,
\[ \cos\frac{\alpha-\beta}{2} = \sqrt{ \frac{1+\cos(\alpha-\beta)}{2} } \]
\[ = \sqrt{ \frac{ 1+\frac{63}{65} }{2} } \]
\[ = \sqrt{ \frac{ \frac{128}{65} }{2} } \]
\[ = \sqrt{ \frac{64}{65} } = \frac{8}{\sqrt{65}} \]
Hence proved.
Final Answer
\[ \boxed{ \cos\frac{\alpha-\beta}{2} = \frac{8}{\sqrt{65}} } \]