The greatest and least values of sin x cos x | vedclass

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(B) Let $f(x) = \sin x \cos x$. Multiplying and dividing by $2$,we get $f(x) = \frac{2 \sin x \cos x}{2} = \frac{1}{2} \sin(2x)$. We know that the ran

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$\frac{1}{2}, -\frac{1}{2}$

Vedclass · Answer

(B) Let $f(x) = \sin x \cos x$. Multiplying and dividing by $2$,we get $f(x) = \frac{2 \sin x \cos x}{2} = \frac{1}{2} \sin(2x)$. We know that the range of the sine function is $[-1, 1]$,so $-1 \le \sin(2x) \le 1$. Multiplying the entire inequality by $\frac{1}{2}$,we get $-\frac{1}{2} \le \frac{1}{2} \sin(2x) \le \frac{1}{2}$. Therefore,the greatest value is $\frac{1}{2}$ and the least value is $-\frac{1}{2}$.

The greatest and least values of $\sin x \cos x$ are

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