Q.Prove that the function given by $f(x) = \cos x$ is

Concept: Monotonicity of Trigonometric Functions — we check the sign of $f'(x)$ in each interval.

1. $f(x) = \cos x \implies f'(x) = -\sin x$.
2. On $(0, \pi)$, $\sin x > 0$, so $f'(x) = -\sin x < 0$ → $f$ is decreasing.
3. On $(\pi, 2\pi)$, $\sin x < 0$, so $f'(x) = -\sin x > 0$ → $f$ is increasing.
4. Since $f$ is decreasing on $(0, \pi)$ and increasing on $(\pi, 2\pi)$, it is neither purely increasing nor purely decreasing over the whole interval $(0, 2\pi)$.

Why the derivative tells the story

To decide whether a function is increasing or decreasing on an interval, we look at the sign of its derivative. If $f'(x) > 0$ for all $x$ in an open interval, the function is strictly increasing there. If $f'(x) < 0$, it is strictly decreasing. If the derivative changes sign, the function is neither.

For $f(x) = \cos x$, we have $f'(x) = -\sin x$. So the monotonicity of $\cos x$ is entirely controlled by the sign of $\sin x$ — but flipped.

Step-by-step reasoning

1. Sign of $\sin x$ on $(0, \pi)$

   On the interval $(0, \pi)$, $\sin x$ is positive (it starts at $0$, rises to $1$ at $\pi/2$, then falls back to $0$ at $\pi$).

   Therefore $f'(x) = -\sin x$ is negative throughout $(0, \pi)$.

   A negative derivative means $f$ is strictly decreasing.

   Hence $\cos x$ is strictly decreasing on $(0, \pi)$.

2. Sign of $\sin x$ on $(\pi, 2\pi)$

   On $(\pi, 2\pi)$, $\sin x$ is negative (it goes from $0$ at $\pi$ to $-1$ at $3\pi/2$, then back to $0$ at $2\pi$).

   So $f'(x) = -\sin x$ becomes positive throughout $(\pi, 2\pi)$.

   A positive derivative means $f$ is strictly increasing.

   Hence $\cos x$ is strictly increasing on $(\pi, 2\pi)$.

3. Behaviour on the full interval $(0, 2\pi)$

   Since $\cos x$ decreases on $(0, \pi)$ and then increases on $(\pi, 2\pi)$, it is not monotonic over the whole interval $(0, 2\pi)$.

   A function that goes down and then up is neither purely increasing nor purely decreasing on the combined interval.

   Therefore $\cos x$ is neither increasing nor decreasing on $(0, 2\pi)$.

Method: Determining Monotonicity of a Trigonometric Function Using Known Quadrant Signs

Because $\sin x$, $\cos x$, and $\tan x$ repeat their sign pattern every period, you rarely need a numerical test point — you can read the sign of the derivative directly from standard quadrant sign rules.

Steps

Step 1: Differentiate the trig function.

Use the standard derivatives:

Step 2: Recall (or sketch) the sign of the resulting trig function over one full period.

For example, $\sin x>0$ on $(0,\pi)$ and $\sin x<0$ on $(\pi,2\pi)$; $\cos x>0$ on $\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$ and negative on $\left(\tfrac{\pi}{2},\tfrac{3\pi}{2}\right)$.

Step 3: Match the given sub-interval to the sign region, and account for any leading negative sign in the derivative.

If the derivative is $-\sin x$, a region where $\sin x>0$ makes the derivative negative (decreasing), and vice versa — don't forget to flip the sign.

Step 4: For an interval spanning both regions, conclude "neither increasing nor decreasing."

If the sub-interval you're asked about contains a point where the derivative's sign changes, the function is not monotonic across all of it — state explicitly that it decreases on part of the interval and increases on the rest, so it is correctly classified as neither purely increasing nor purely decreasing there.

Common Mistakes

Mistake 1: Dropping the negative sign when differentiating $\cos x$.

Why it's wrong: writing $f'(x) = \sin x$ instead of $f'(x) = -\sin x$ flips every sign conclusion that follows, making the increasing and decreasing intervals come out backwards. Correct approach: memorise $\dfrac{d}{dx}\cos x = -\sin x$ precisely, and double-check the sign before reading off the intervals.

Mistake 2: Mixing up which half of $(0, 2\pi)$ has $\sin x$ positive versus negative.

Why it's wrong: $\sin x$ is positive on $(0,\pi)$ (the "upper" half of the unit circle) and negative on $(\pi, 2\pi)$ — reversing this swaps the increasing and decreasing conclusions for $\cos x$. Correct approach: sketch (or recall) the unit circle / sine graph quickly before assigning signs, rather than guessing.