Question

If $\alpha +\beta ={90}^o$, find the maximum and minimum values of $\sin \alpha \sin \beta$.

Answer 1

Let $y=\sin \alpha \sin \beta$. Then,

$y=\frac{1}{2}\left(2\sin \alpha \sin \beta \right)=\frac{1}{2}\left[\cos \right(\alpha -\beta )-\cos (\alpha +\beta \left)\right]$

$=\frac{1}{2}\left[\cos \right(\alpha -\beta )-\cos {90}^o]=\frac{1}{2}\cos (\alpha -\beta )$

we know,

$-1\le \cos (\alpha -\beta )\le 1\Rightarrow \frac{-1}{2}\le \frac{1}{2}\cos (\alpha -\beta )\le \frac{1}{2}$

$\Rightarrow \frac{-1}{2}\le y\le \frac{1}{2}\Rightarrow \frac{-1}{2}\le \sin \alpha \sin \beta \le \frac{1}{2}$

Hence, $\frac{-1}{2}$ and $\frac{1}{2}$ are respectively minimum and maximum values of $\sin \alpha \sin \beta$.