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The maximum value of $\sin(\cos x)$ is equal to

$(a)\;\sin 1\qquad(b)\;1\qquad(c)\;\sin\large\big(\frac{1}{\sqrt 2}\big)$$\qquad(d)\;\sin\big(\large\frac{\sqrt 3}{2}\big)$

1 Answer

$\cos x\in [-1,1]\forall x\in R$ and $\sin x$ increasing in $\bigg[\large\frac{-\pi}{2},\frac{\pi}{2}\bigg]$
$\Rightarrow $ Maximum values of $\sin x(\cos x)=\sin 1$
Hence (a) is the correct option.
answered Oct 7, 2013 by sreemathi.v
 

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