If x+y=frac{pi}{2},then the maximum value of | vedclass

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Question

(A) Given $x+y=\frac{\pi}{2}$,we have $y=\frac{\pi}{2}-x$. Substituting this into the expression $\sin x \cdot \sin y$,we get: $\sin x \cdot \sin \lef

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(A) Given $x+y=\frac{\pi}{2}$,we have $y=\frac{\pi}{2}-x$. Substituting this into the expression $\sin x \cdot \sin y$,we get: $\sin x \cdot \sin \left(\frac{\pi}{2}-x\right) = \sin x \cdot \cos x$. Multiplying and dividing by $2$,we get: $\frac{2 \sin x \cdot \cos x}{2} = \frac{\sin 2x}{2}$. We know that the range of $\sin 2x$ is $[-1, 1]$. Therefore,the range of $\frac{\sin 2x}{2}$ is $\left[\frac{-1}{2}, \frac{1}{2}\right]$. Thus,the maximum value is $\frac{1}{2}$.

If $x+y=\frac{\pi}{2}$,then the maximum value of $\sin x \cdot \sin y$ is

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