Example 9: $\cos x$⁠: decreasing on $(0,\pi)$⁠, increasing on $(\pi,2\pi)$

Basics you need

Solution

1. Differentiate

   $$f'(x) = \frac{d}{dx}(\cos x) = -\sin x .$$

   So $f'(x)$ has the opposite sign to $\sin x$ at every $x$⁠.

   Theorem 1 links the direction of $f$ to the sign of $f'$⁠. Because of the minus sign, cosine falls exactly where sine is positive and rises exactly where sine is negative.
   Derivative of $\cos x$⁠, Section 5.3. The minus sign is easy to drop: $\frac{d}{dx}\sin x = \cos x$ but $\frac{d}{dx}\cos x = -\sin x$⁠.

2. Part (a): decreasing on $(0, \pi)$

   For every $x \in (0, \pi)$⁠, $\sin x > 0$⁠, so $f'(x) = -\sin x < 0$⁠.

   Now take any $x_1 < x_2$ in $(0, \pi)$⁠. On $[x_1, x_2]$ the cosine is continuous, it is differentiable on $(x_1, x_2)$⁠, and $f'(x) < 0$⁠, so $f'(x) \le 0$⁠, there (because $(x_1, x_2)$ lies inside $(0, \pi)$⁠). By Theorem 1(b), $f$ is decreasing in $[x_1, x_2]$⁠, so $f(x_1) \ge f(x_2)$⁠. Hence $\cos x$ is decreasing on $(0, \pi)$.

   Theorem 1 is stated for a closed interval, while $(0, \pi)$ is open. Definition 1 compares only two points at a time, and every pair $x_1 < x_2$ of $(0, \pi)$ spans a closed interval $[x_1, x_2]$ inside $(0, \pi)$⁠, where the theorem applies. With $f' < 0$ strictly, the Mean Value Theorem argument in the book's proof of Theorem 1 even gives $f(x_1) > f(x_2)$⁠: strictly decreasing.
   Sine is positive in the first and second quadrants. Basics: signs of sin and cos

3. Part (b): increasing on $(\pi, 2\pi)$

   For every $x \in (\pi, 2\pi)$⁠, $\sin x < 0$⁠, so $f'(x) = -\sin x > 0$⁠.

   For any $x_1 < x_2$ in $(\pi, 2\pi)$⁠, Theorem 1(a) on $[x_1, x_2]$ gives $f(x_1) \le f(x_2)$ (in fact $<$⁠). Hence $\cos x$ is increasing on $(\pi, 2\pi)$.

   The argument is the mirror image of part (a): the sign of sine flips at $\pi$⁠, so the sign of $f'$ flips, and so does the direction of the graph.
   Sine is negative in the third and fourth quadrants. Basics: signs of sin and cos

4. Part (c): neither increasing nor decreasing on $(0, 2\pi)$

   Not increasing. $\frac{\pi}{2}$ and $\pi$ are in $(0, 2\pi)$ with $\frac{\pi}{2} < \pi$⁠, but

   $$\cos\tfrac{\pi}{2} = 0 \;>\; -1 = \cos\pi .$$

   So the implication $x_1 < x_2 \Rightarrow f(x_1) \le f(x_2)$ fails for this pair, and $f$ is not increasing on $(0, 2\pi)$⁠.

   Not decreasing. $\pi$ and $\frac{3\pi}{2}$ are in $(0, 2\pi)$ with $\pi < \frac{3\pi}{2}$⁠, but

   $$\cos\pi = -1 \;<\; 0 = \cos\tfrac{3\pi}{2} .$$

   So $x_1 < x_2 \Rightarrow f(x_1) \ge f(x_2)$ fails for this pair, and $f$ is not decreasing on $(0, 2\pi)$⁠. Hence $\cos x$ is neither increasing nor decreasing on $(0, 2\pi)$.

   "Increasing on I" is a claim about every pair of points of I, so one pair in the wrong order is enough to disprove it. The book says "clearly by (a) and (b)": inside $(0, \pi)$ the function strictly falls, which already rules out "increasing" on the larger interval, and inside $(\pi, 2\pi)$ it strictly rises, which rules out "decreasing". The two pairs above make that concrete. (Knowing only the non-strict "decreasing" of part (a) would not be enough, since a constant function is both increasing and decreasing.)
   $\cos\frac{\pi}{2} = 0$⁠, $\cos\pi = -1$⁠, $\cos\frac{3\pi}{2} = 0$⁠. Basics: signs of sin and cos

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