Question

Cos(alpha +betha) =0 then find value of sin(alpha-beta) =¿

Answer 1

We have,

$\cos( \alpha + \beta ) = 0 \\ = > \cos( \alpha + \beta ) = \cos(90) \\ (as \: \cos(90) = 0) \\ = > \alpha + \beta = 90 \\ \alpha = 90 - \beta ..............................(1) \\ now \: \\ \sin(( \alpha + \beta ) = \sin((90 - \beta ) - \beta ))..............using(1) \\ \sin( \alpha - \beta = \sin(90 - 2 \beta) ) \\ = > \sin( \alpha - \beta ) = \cos(2 \beta ) \\ = > as \: \sin(90 - x) = \cos(x) \\ therfore \\ \cos(2 \beta ) \\ proved....$