For $0\le x\le 2\pi$, $\sin x$ and $\cos x$ are both decreasing functions in the interval______.

Understanding Decreasing Trigonometric Functions

A function is decreasing on an interval if its derivative is negative within that interval.

For a function $f(x)$, it is decreasing when $f^{\prime }(x)<0$. We need to find the interval where both $\sin x$ and $\cos x$ satisfy this condition for $0\le x\le 2\pi$.

**Analyzing $\sin x$ Decreasing Interval**

The derivative of $\sin x$ is $\cos x$. Thus, $\sin x$ is decreasing when its derivative, $\cos x$, is negative.

In the interval $0\le x\le 2\pi$, $\cos x<0$ holds true for the interval $x\in (\frac{\pi }{2},\frac{3\pi }{2})$.

**Analyzing $\cos x$ Decreasing Interval**

The derivative of $\cos x$ is $-\sin x$. Thus, $\cos x$ is decreasing when its derivative, $-\sin x$, is negative.

The condition $-\sin x<0$ simplifies to $\sin x>0$.

In the interval $0\le x\le 2\pi$, $\sin x>0$ holds true for the interval $x\in (0,\pi )$.

Finding the Common Interval

We need the interval where both $\sin x$ and $\cos x$ are decreasing. This requires finding the intersection of the intervals derived above:

- $\sin x$ decreasing interval: $(\frac{\pi }{2},\frac{3\pi }{2})$
- $\cos x$ decreasing interval: $(0,\pi )$

The intersection of these two intervals is the region where both conditions are met simultaneously.

Intersection: $(\frac{\pi }{2},\frac{3\pi }{2})\cap (0,\pi )=(\frac{\pi }{2},\pi )$.

Therefore, both $\sin x$ and $\cos x$ are decreasing functions in the interval $(\frac{\pi }{2},\pi )$.