Question

please solve this question . It's urgent.

Answer 1

$\sin ^{2} ( \frac{ \alpha - \beta }{2} ) = \frac{2 \times \sin( \frac{ \alpha - \beta }{2} ) \times \sin( \frac{ \alpha - \beta }{2} ) }{2} \\ \: using \: the \: formula \: \: \: \: 2 \sin( \alpha ) \sin( \beta ) = \cos( \alpha - \beta ) - \cos( \alpha + \beta ) \\ = > 2 \times \sin( \frac{ \alpha - \beta}{2}) \times \sin( \frac{ \alpha - \beta}{2})= \cos(0) - \cos( \alpha - \beta ) \\ = 1 - \cos( \alpha - \beta ) \\ = 1 - (\cos( \alpha ) \cos( \beta ) + \sin( \alpha ) \sin( \beta ) ) \\ \\ given \: that \: ...\cos( \alpha ) = \frac{11}{61} \\ \sin( \alpha ) = \sqrt{1 - { \cos^{2}( \alpha ) } } \: = \sqrt{1 - \frac{121}{3721} } = \sqrt{ \frac{3600}{3721} } = \frac{60}{61} \\ \sin( \beta ) = \frac{4}{5} \\ \cos( \beta ) = \sqrt{1 - \sin^{2} ( \beta ) } = > \cos( \beta ) = \frac{3}{5} \\ now \: plug \: in \: the \: values \: \\ = 1 - (\frac{11}{61} \times \frac{3}{5} + \frac{60}{61} \times \frac{4}{5} )\: \\ = 1 - \frac{273}{305} \\ = \frac{32}{305} \div 2 = > \frac{16}{305}$