Q.Show that the function given by $f(x) = \sin x$ is

Concept: Monotonicity of a function is determined by the sign of its derivative. If $f'(x) > 0$ on an interval, $f$ is increasing there; if $f'(x) < 0$, $f$ is decreasing.

Reasoning:

1. Compute the derivative: $f'(x) = \cos x$.

2. On $(0, \frac{\pi}{2})$, $\cos x > 0$, so $f'(x) > 0$ — hence $f$ is increasing.

3. On $(\frac{\pi}{2}, \pi)$, $\cos x < 0$, so $f'(x) < 0$ — hence $f$ is decreasing.

4. Since $f$ increases on the first half and decreases on the second half of $(0, \pi)$, it is neither purely increasing nor purely decreasing over the whole interval.

The core idea here is simple: a function is increasing where its derivative is positive, decreasing where its derivative is negative, and neither if the derivative changes sign over the interval. For $f(x) = \sin x$, the derivative is $f'(x) = \cos x$. So the entire problem reduces to asking: where is $\cos x$ positive, where is it negative, and does it stay the same sign throughout $(0, \pi)$?

Let’s walk through each part.

1. Part (a): Increasing in $(0, \frac{\pi}{2})$

   On the open interval $(0, \frac{\pi}{2})$, the cosine function is positive. You can see this from the unit circle: for angles between $0$ and $\frac{\pi}{2}$ (first quadrant), the $x$-coordinate (which is $\cos x$) is positive.

   Since $f'(x) = \cos x > 0$ for every $x$ in $(0, \frac{\pi}{2})$, the function $f(x) = \sin x$ is strictly increasing on this interval.

   Tip

   A quick mental check: at $x = 0$, $\sin 0 = 0$; at $x = \frac{\pi}{2}$, $\sin \frac{\pi}{2} = 1$. The value goes up, confirming the derivative’s story.

2. Part (b): Decreasing in $(\frac{\pi}{2}, \pi)$

   On $(\frac{\pi}{2}, \pi)$, we are in the second quadrant. Here, the $x$-coordinate (cosine) becomes negative. So $f'(x) = \cos x < 0$ for all $x$ in this interval.

   A negative derivative means the function is strictly decreasing. Indeed, $\sin \frac{\pi}{2} = 1$ and $\sin \pi = 0$, so the value falls from 1 to 0.

3. Part (c): Neither increasing nor decreasing in $(0, \pi)$

   Now consider the whole interval $(0, \pi)$. The derivative $\cos x$ is positive on $(0, \frac{\pi}{2})$ and negative on $(\frac{\pi}{2}, \pi)$. Since the sign of $f'(x)$ changes within the interval, the function cannot be monotonic (purely increasing or purely decreasing) over the entire $(0, \pi)$.

   Watch out

   A common mistake is to think that because $\sin x$ goes from 0 to 1 to 0, it is “increasing then decreasing” — but the question asks about the whole interval at once. A function is increasing on an interval only if for every pair $x_1 < x_2$ in that interval, $f(x_1) \le f(x_2)$. Here, take $x_1 = \frac{\pi}{4}$ and $x_2 = \frac{3\pi}{4}$: $\sin \frac{\pi}{4} = \frac{\sqrt{2}}{2} \approx 0.707$, $\sin \frac{3\pi}{4} = \frac{\sqrt{2}}{2}$ as well — equal, so not strictly increasing. But worse, take $x_1 = \frac{\pi}{6}$ and $x_2 = \frac{2\pi}{3}$: $\sin \frac{\pi}{6} = 0.5$, $\sin \frac{2\pi}{3} \approx 0.866$ — that’s an increase. Yet take $x_1 = \frac{\pi}{3}$ and $x_2 = \frac{5\pi}{6}$: $\sin \frac{\pi}{3} \approx 0.866$, $\sin \frac{5\pi}{6} = 0.5$ — a decrease. So the function is neither consistently increasing nor consistently decreasing across the whole interval.

Method: Determining Monotonicity of a Trigonometric Function Across Sub-intervals

Trigonometric functions are not monotonic over their whole domain — they rise and fall in a repeating pattern. This method finds where a trig function increases or decreases by tracking the sign of its derivative (another, related trig function) across each piece of the interval.

Steps

Step 1: Differentiate the trigonometric function

Recall the standard derivatives: $\frac{d}{dx}\sin x = \cos x$, $\frac{d}{dx}\cos x = -\sin x$, $\frac{d}{dx}\tan x = \sec^2 x$.

Step 2: Determine the sign of the derivative on each given sub-interval separately

Use the unit-circle/quadrant behaviour of the derivative function to decide its sign on each requested piece — e.g. for $\cos x$: positive in the first quadrant, negative in the second. Never assume the sign carries over from one sub-interval to the next; check each one independently.

Step 3: Apply the Increasing/Decreasing Function Test on each piece

Where the derivative is positive on a sub-interval, the function is strictly increasing there; where it is negative, strictly decreasing there.

Step 4 (Applying to a combined/whole interval): Check whether the sign is consistent throughout

If a question also asks about monotonicity over the union of the sub-intervals, check whether the derivative keeps one sign across the entire combined interval. If the sign changes anywhere inside it, the function is neither increasing nor decreasing over that whole interval — even though it is monotonic on each smaller piece.

This piece-by-piece sign tracking generalises directly to $\cos x$ (governed by $-\sin x$) and $\tan x$ (governed by $\sec^2 x$, always positive between consecutive asymptotes).

Common Mistakes

Mistake 1: Calling $\sin x$ simply "increasing" without specifying an interval

Why it's wrong: $\sin x$ is not increasing over its entire domain — it rises and falls periodically. A blanket statement like "$\sin x$ is increasing" is meaningless without naming the interval on which the claim holds. Correct approach: always state monotonicity together with the specific interval, e.g. "increasing on $(0, \frac{\pi}{2})$."

Mistake 2: Misreading the sign of $\cos x$ in the second quadrant

Why it's wrong: Students sometimes assume $\cos x$ stays positive throughout $(0, \pi)$ because $\sin x$ is positive there, confusing the two functions. In fact $\cos x$ turns negative for $x > \frac{\pi}{2}$. Correct approach: check the sign of $\cos x$ (not $\sin x$) quadrant by quadrant using the unit circle.

Mistake 3: Concluding "neither increasing nor decreasing" means the function is constant

Why it's wrong: On $(0,\pi)$, $\sin x$ rises then falls — it is not constant, it simply fails to be monotonic over the whole interval because its derivative changes sign within it. Correct approach: use a concrete counterexample pair $x_1 < x_2$ where $f(x_1) > f(x_2)$ (from the decreasing piece) to show the function is not increasing on the whole interval, rather than claiming it stays flat.