Question

For x greater than or equal to zero and less than or equal to $2\pi$, $\sin x$ and $\cos x$ are both decreasing on the intervals

• A. $(0,\frac{\pi }{2})$
• B. $(\frac{\pi }{2},\pi )$
• C. $(\pi ,3\frac{\pi }{2})$
• D. $(\frac{3\pi }{2},2\pi )$

Answer 1

Answer: (B)

$(\frac{\pi }{2},\pi )$

Given that $0\le x\le 2\pi$. Also we know any function $f\left(x\right)$ is increasing or decreasing as its derivative is $\ge 0$ or $\le 0$.

Consider $f\left(x\right)=\sin x$

$f^{\prime }\left(x\right)=\cos x$

To make sure $f$ is decreasing we should have

$\cos x\le 0\Rightarrow x\in [\frac{\pi }{2},\frac{3\pi }{2}]$

Let $g\left(x\right)=\cos x$

$g^{\prime }\left(x\right)=-\sin x$

To make sure $g$ is decreasing we should have

$-\sin \le 0\Rightarrow \sin x\ge 0$

$\Rightarrow x\in [0,\pi ]$

So for $x\in [\frac{\pi }{2},\frac{3\pi }{2}]\sin x$ is deceasing

$x\in [0,\pi ]\cos x$ is decreasing

Intersection will $x\in [\frac{\pi }{2},\pi ]$ where $\sin x$ and $\cos x$ both decreasing.

option B