Question

lf in a $\Delta ABC,\angle C={90}^0$, then the maximum value of $\sin A\sin B$ is

• A. $\frac{1}{2}$
• B. $1$
• C. $2$
• D. $\frac{1}{\sqrt{2}}$

Answer 1

The correct option is A $\frac{1}{2}$

we know that

$2\sin A\sin B=$ $\cos (A-B)+\cos (A+B)$

$\sin A\sin B=\frac{1}{2}\left[\cos \right(A-B)-\cos (A+B\left)\right]$
$=\frac{1}{2}\left[\cos \right(A-B)-0]$.......... $[A+B={90}^o]$
$=\frac{\cos (A-B)}{2}$

$\cos \left(\theta \right)$ lies in the interval $[-1,1]$ $\Rightarrow \cos \left(\theta \right)$ has the maximum value of $1$

$\Rightarrow \frac{cos(A-B)}{2}$ has the maximum value of $\frac{1}{2}$
So, max of $\sin A\sin B=\frac{1}{2}$