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MHT CET · Maths · Application of Derivatives

If \(x+y=\frac{\pi}{2}\), then the maximum value of \(\sin x\).siny is

  1. A \(\frac{1}{2}\)
  2. B \(\frac{-1}{2}\)
  3. C \(\frac{-1}{\sqrt{2}}\)
  4. D \(\frac{1}{\sqrt{2}}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\frac{1}{2}\)

Step-by-step Solution

Detailed explanation

(C)
\(x+y=\frac{\pi}{2} \quad \Rightarrow \quad y=\frac{\pi}{2}-x\)
\(\sin x \cdot \sin y=\sin x \cdot \sin \left(\frac{\pi}{2}-x\right)=\sin x \cos x=\)\(\frac{2 \sin x \cdot \cos x}{2}=\frac{\sin 2 x}{2} \)
\( \text {We know, }-1 \leq \sin 2 x \leq 1 \Rightarrow \frac{-1}{2} \leq \frac{\sin 2 x}{2} \leq \frac{1}{2}\)
So maximum value is \(\frac{1}{2}\)