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 $f'(x) < 0$. 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 $(\frac{\pi}{2}, \pi)$.