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The greatest and least value of $\sin x \cos x$ are
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Verified Answer
The correct answer is:
$\frac{1}{2},-\frac{1}{2}$
Let $f(x)=\sin x \cos x=\frac{1}{2} \sin 2 x$
We know $-1 \leq \sin 2 x \leq 1 \Rightarrow-\frac{1}{2} \leq \frac{1}{2} \sin 2 x \leq \frac{1}{2}$
Thus the greatest and least value of f(x) are $\frac{1}{2}$ and $-\frac{1}{2}$ respectively.
We know $-1 \leq \sin 2 x \leq 1 \Rightarrow-\frac{1}{2} \leq \frac{1}{2} \sin 2 x \leq \frac{1}{2}$
Thus the greatest and least value of f(x) are $\frac{1}{2}$ and $-\frac{1}{2}$ respectively.
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