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If $x+y=\frac{\pi}{2}$, then the maximum value of $\sin x$.siny is
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Verified Answer
The correct answer is:
$\frac{1}{2}$
(C)
$x+y=\frac{\pi}{2} \quad \Rightarrow \quad y=\frac{\pi}{2}-x$
$\begin{array}{l}
\sin x \cdot \sin y=\sin x \cdot \sin \left(\frac{\pi}{2}-x\right)=\sin x \cos x=\frac{2 \sin x \cdot \cos x}{2}=\frac{\sin 2 x}{2} \\
\text { We know, }-1 \leq \sin 2 x \leq 1 \Rightarrow \frac{-1}{2} \leq \frac{\sin 2 x}{2} \leq \frac{1}{2}
\end{array}$
So maximum value is $\frac{1}{2}$
$x+y=\frac{\pi}{2} \quad \Rightarrow \quad y=\frac{\pi}{2}-x$
$\begin{array}{l}
\sin x \cdot \sin y=\sin x \cdot \sin \left(\frac{\pi}{2}-x\right)=\sin x \cos x=\frac{2 \sin x \cdot \cos x}{2}=\frac{\sin 2 x}{2} \\
\text { We know, }-1 \leq \sin 2 x \leq 1 \Rightarrow \frac{-1}{2} \leq \frac{\sin 2 x}{2} \leq \frac{1}{2}
\end{array}$
So maximum value is $\frac{1}{2}$
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