Exercise 6.2 · Q3: $\sin x$ rises on $\left(0,\frac\pi2\right)$⁠, falls on $\left(\frac\pi2,\pi\right)$

Basics you need

Solution

1. Differentiate

   $\sin x$ is differentiable (hence continuous) at every real $x$⁠, and

   $$f'(x) = \frac{d}{dx}(\sin x) = \cos x.$$
   Theorem 1 needs two things on an interval $[a, b]$⁠: continuity on $[a, b]$ and differentiability inside. Sine has both everywhere, so the only work left is the sign of $\cos x$⁠.
   $\frac{d}{dx}(\sin x) = \cos x$ (Table 5.3).

2. (a) On $\left(0, \frac{\pi}{2}\right)$⁠: $\cos x > 0$⁠, so sine is increasing

   For $0 < x < \frac{\pi}{2}$ the point at angle $x$ on the unit circle is in the first quadrant, so its $x$-coordinate $\cos x$ is positive:

   $$f'(x) = \cos x > 0 \quad\text{for every } x \in \left(0, \tfrac{\pi}{2}\right).$$

   Sine is continuous on $\left[0, \frac{\pi}{2}\right]$⁠, so Theorem 1 (a) gives: $f$ is increasing on $\left[0, \frac{\pi}{2}\right]$⁠, and therefore on the smaller interval $\left(0, \frac{\pi}{2}\right)$⁠. Because $f' > 0$ (not just $\ge 0$⁠), the proof of Theorem 1 even gives strictly increasing.

   If $f$ is increasing on an interval, it is increasing on every interval inside it: the condition "$x_1 < x_2 \Rightarrow f(x_1) \le f(x_2)$" for all pairs of the big interval includes all pairs of the small one. That is why the closed-interval result also covers the open interval in the question.
   First quadrant: both $\sin x$ and $\cos x$ are positive. Basics: signs of sin and cos

3. (b) On $\left(\frac{\pi}{2}, \pi\right)$⁠: $\cos x < 0$⁠, so sine is decreasing

   For $\frac{\pi}{2} < x < \pi$ the point is in the second quadrant, so its $x$-coordinate is negative:

   $$f'(x) = \cos x < 0 \quad\text{for every } x \in \left(\tfrac{\pi}{2}, \pi\right).$$

   Sine is continuous on $\left[\frac{\pi}{2}, \pi\right]$⁠, so Theorem 1 (b) gives: $f$ is decreasing on $\left[\frac{\pi}{2}, \pi\right]$⁠, hence on $\left(\frac{\pi}{2}, \pi\right)$ (strictly, by the proof, since $f' < 0$⁠).

   Same reasoning as (a), with the sign reversed. At the single point $x = \frac{\pi}{2}$ we have $\cos x = 0$⁠, which is exactly where the behaviour switches from rising to falling.
   Second quadrant: $\sin x > 0$ but $\cos x < 0$⁠. Basics: signs of sin and cos

4. (c) On $(0, \pi)$⁠: one pair rules out "decreasing", another rules out "increasing"

   Take three points of $(0, \pi)$⁠: $\frac{\pi}{6} < \frac{\pi}{2} < \frac{5\pi}{6}$⁠, with

   $$\sin\tfrac{\pi}{6} = \tfrac12,\quad \sin\tfrac{\pi}{2} = 1,\quad \sin\tfrac{5\pi}{6} = \tfrac12.$$
   • Not decreasing: $\frac{\pi}{6} < \frac{\pi}{2}$ but $f\left(\frac{\pi}{6}\right) = \frac12 < 1 = f\left(\frac{\pi}{2}\right)$⁠. Decreasing would need $\frac12 \ge 1$⁠, which is false.
   • Not increasing: $\frac{\pi}{2} < \frac{5\pi}{6}$ but $f\left(\frac{\pi}{2}\right) = 1 > \frac12 = f\left(\frac{5\pi}{6}\right)$⁠. Increasing would need $1 \le \frac12$⁠, which is false.

   So on $(0, \pi)$⁠, $f$ is neither increasing nor decreasing.

   Parts (a) and (b) alone do not prove (c) by themselves; they describe the two halves. "Neither" is a statement about the whole interval, and the definition makes it easy: each property is a "for all pairs" claim, so a single failing pair destroys it. We picked points where the sine values are exact. (The book argues Example 9 (c) for cosine the same way, "by (a) and (b)": the pairs above are what that phrase means.)
   $\sin\frac{\pi}{6} = \frac12$⁠, $\sin\frac{\pi}{2} = 1$⁠, and $\sin\frac{5\pi}{6} = \sin\left(\pi - \frac{\pi}{6}\right) = \sin\frac{\pi}{6} = \frac12$⁠.

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