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  4. If α+β=90°, then the maximum value of sin α sin β is

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Q. If $\alpha+\beta=90^{\circ}$, then the maximum value of $\sin \alpha \sin \beta$ is

Trigonometric Functions

Solution:

Let $y=\sin \alpha \sin \beta$
Now $y=\sin \alpha \sin \left(90^{\circ}-\alpha\right)=\sin \alpha \cos \alpha=\frac{1}{2}(2 \sin \alpha \cos \alpha)$
$=\frac{1}{2} \sin ^{2} \alpha$
Now $\sin 2 \alpha \leq 1$
$\Rightarrow \frac{1}{2} \sin 2 \alpha \leq \frac{1}{2}$