Question

What is the maximum value of $\sin \alpha \sin \beta$

• A. ${\sin }^2\left(c/2\right)$
• B. ${\cos }^2\left(c/2\right)$
• C. ${\sin }^2\left(c\right)$
• D. None of these
  

Answer 1

The correct option is A ${\sin }^2\left(c/2\right)$

Let $y=\sin a.\sin \beta =\sin a.\sin (c-a)=\sin a(\sin c\cos a-\sin a\cos c)$
$\Rightarrow y=1/2\sin c.\sin 2a-\cos c.{\sin }^{2}a$
For maximum value of $y$,
$\frac{dy}{da}=0\Rightarrow \sin c.\cos 2a-\cos c.\sin 2a=0\Rightarrow \sin (2a-c)=0\Rightarrow c=2a$
$\therefore y_{max}=\sin \left(c/2\right).\sin \left(c/2\right)={\sin }^2\left(c/2\right)$