Question

If cos (alpha+beta)=0 then prove that sin (alpha-beta) = cos 2beta.​

Answer 1

$Answer \: \\ \\ GIVEN \: \: Question \: \: Is \: \: \\ \\ \cos( \alpha + \beta ) = 0 \: \: \: \: then \: \: prove \: that \\ \sin( \alpha - \beta ) = \cos(2 \beta ) \\ \\ \cos( \alpha + \beta ) = 0 \\ \\ \cos( \alpha + \beta ) = \cos( \frac{\pi}{2} ) \\ \\ \alpha + \beta = \frac{\pi}{2} \\ \\ \alpha = \frac{\pi}{2} - \beta \\ lhs \\ \\ \sin( \alpha - \beta ) \\ \\ \sin( \frac{\pi}{2} - \beta - \beta ) \\ \\ \sin( \frac{\pi}{2} - 2 \beta ) \\ \\ \cos(2 \beta ) \: \: \: \: hence \: proved \: \\ \\ Note \: \: \\ \\ \sin( \frac{\pi}{2} - 2x) = \cos(2x)$

Answer 2

Step-by-step explanation:

$\begin{lgathered} \\ \\ \cos( \alpha + \beta ) = 0 \: \: \: \: \\ \\ Then \: \: Prove \: That \\ \\ \sin( \alpha - \beta ) = \cos(2 \beta ) \\ \\ \\ \cos( \alpha + \beta ) = 0 \\ \\ \\ \cos( \alpha + \beta ) = \cos( \frac{\pi}{2} ) \\ \\ \\ \alpha + \beta = \frac{\pi}{2} \\ \\ \\ \alpha = \frac{\pi}{2} - \beta \\ \\ LHS \\ \\ \\ = \sin( \alpha - \beta ) \\ \\ = \sin( \frac{\pi}{2} - \beta - \beta ) \\ \\ = \sin( \frac{\pi}{2} - 2 \beta ) \\ \\ =\cos(2 \beta ) \: \: \: \: \\ \\ Hence \: Proved \: \\ \\ \end{lgathered}$