Question

Given that $\sin \alpha =\frac{\sqrt{3}}{2},\cos \beta =\frac{\sqrt{3}}{2}$ find the value of $(\alpha -\beta )$.

Answer 1

$\sin \alpha =\frac{\sqrt{3}}{2},\cos \beta =\frac{\sqrt{3}}{2}weknowthat\Rightarrow \sin \alpha =\sin \frac{\pi }{3}\therefore \alpha =\pi /3\Rightarrow andsimilarly\Rightarrow \cos \beta =\frac{\sqrt{3}}{2}\Rightarrow \cos \beta =\frac{\pi }{6}\therefore \beta =\pi /6Then,valueof\alpha -\beta =\frac{\pi }{3}-\frac{-\pi }{6}=\frac{\pi }{6}Ans.$