Example 13: Where $\sin x+\cos x$ increases and decreases on $[0,2\pi]$

Basics you need

Solution

1. Differentiate

   Using Basic 2 and the sum rule,

   $$f'(x) = \cos x - \sin x.$$
   Sine and cosine are differentiable everywhere, so $f$ is continuous on $[0,2\pi]$ and differentiable inside it: Theorem 1 applies to every closed piece of the domain.

2. Solve $f'(x) = 0$ in $[0, 2\pi]$

   $f'(x) = 0$ means $\sin x = \cos x$⁠. At such a point $\cos x \ne 0$⁠: if $\cos x$ were 0, then $\sin x = \pm1$⁠, which is not equal to $\cos x = 0$⁠. So we may divide by $\cos x$⁠:

   $$\tan x = 1 \iff x = \frac{\pi}{4} + n\pi.$$

   In $[0, 2\pi]$ this gives $n = 0$ and $n = 1$⁠:

   $$x = \frac{\pi}{4} \quad\text{and}\quad x = \frac{5\pi}{4}.$$

   ($n = -1$ gives $-\frac{3\pi}{4} < 0$ and $n = 2$ gives $\frac{9\pi}{4} > 2\pi$⁠.) These two points split $[0, 2\pi]$ into $\left[0, \frac{\pi}{4}\right)$⁠, $\left(\frac{\pi}{4}, \frac{5\pi}{4}\right)$ and $\left(\frac{5\pi}{4}, 2\pi\right]$⁠.

   Dividing by $\cos x$ is only allowed when $\cos x \ne 0$⁠; checking this first guarantees that no solution of $\sin x = \cos x$ is lost by the division.

3. Rewrite $f'$ as a single cosine

   By the compound-angle formula (Basic 3),

   $$\begin{aligned} \sqrt2\cos\left(x + \tfrac{\pi}{4}\right) &= \sqrt2\left(\cos x\cdot\tfrac{1}{\sqrt2} - \sin x\cdot\tfrac{1}{\sqrt2}\right) \\ &= \cos x - \sin x. \end{aligned}$$

   So $f'(x) = \sqrt2\cos\left(x + \frac{\pi}{4}\right)$⁠, and since $\sqrt2 > 0$⁠, $f'(x)$ has the sign of $\cos\left(x + \frac{\pi}{4}\right)$.

   A difference of a cosine and a sine changes sign at places that are hard to see; a single cosine changes sign only where its angle passes $\frac{\pi}{2}$ or $\frac{3\pi}{2}$⁠. This is the same idea as Example 12: reduce to one trigonometric function of one angle.
   The identity also confirms Step 2: $\cos\left(x + \frac{\pi}{4}\right) = 0$ when $x + \frac{\pi}{4} = \frac{\pi}{2}$ or $\frac{3\pi}{2}$⁠, that is $x = \frac{\pi}{4}$ or $\frac{5\pi}{4}$⁠.

4. The sign of $f'$ on each piece

   Write $\theta = x + \frac{\pi}{4}$⁠. As $x$ runs over $[0, 2\pi]$⁠, $\theta$ runs over $\left[\frac{\pi}{4}, \frac{9\pi}{4}\right]$⁠.

   • $0 \le x < \frac{\pi}{4}$⁠: $\frac{\pi}{4} \le \theta < \frac{\pi}{2}$⁠, first quadrant, $\cos\theta > 0$⁠, so $f'(x) > 0$⁠.
   • $\frac{\pi}{4} < x < \frac{5\pi}{4}$⁠: $\frac{\pi}{2} < \theta < \frac{3\pi}{2}$⁠, second and third quadrants, $\cos\theta < 0$⁠, so $f'(x) < 0$⁠.
   • $\frac{5\pi}{4} < x \le 2\pi$⁠: $\frac{3\pi}{2} < \theta \le \frac{9\pi}{4}$⁠, fourth quadrant and then (past $2\pi$⁠) the first quadrant again, $\cos\theta > 0$⁠, so $f'(x) > 0$⁠.

   Test values: $f'(0) = 1 - 0 = 1$⁠, $f'\!\left(\frac{\pi}{2}\right) = 0 - 1 = -1$⁠, $f'\!\left(\frac{3\pi}{2}\right) = 0 - (-1) = 1$⁠.

   The angle $\frac{9\pi}{4}$ is one full turn plus $\frac{\pi}{4}$⁠, so the last stretch of $\theta$ wraps round into the first quadrant, where cosine is still positive. There is no further sign change.
   Quadrant signs of cosine: $+$⁠, $-$⁠, $-$⁠, $+$⁠. Basics: signs of sin and cos

5. Conclude with Theorem 1

   $f$ is continuous on each closed piece and $f'$ has a fixed strict sign inside it, so by Theorem 1 (and, for strictness, the Mean Value Theorem argument of its proof):

   • $f$ is increasing on $\left[0, \frac{\pi}{4}\right]$ and on $\left[\frac{5\pi}{4}, 2\pi\right]$;
   • $f$ is decreasing on $\left[\frac{\pi}{4}, \frac{5\pi}{4}\right]$.

   The book states the same answer with the critical points left out: increasing in $\left[0, \frac{\pi}{4}\right)$ and $\left(\frac{5\pi}{4}, 2\pi\right]$⁠, decreasing in $\left(\frac{\pi}{4}, \frac{5\pi}{4}\right)$⁠. Both versions are correct.

   Note that $f$ is not increasing on the union $\left[0, \frac{\pi}{4}\right) \cup \left(\frac{5\pi}{4}, 2\pi\right]$⁠: for example $\frac{\pi}{6} < \frac{3\pi}{2}$ but $f\!\left(\frac{\pi}{6}\right) = \frac12 + \frac{\sqrt3}{2}$ $\approx 1.36603$ is larger than $f\!\left(\frac{3\pi}{2}\right) = -1$⁠.

   The two increasing pieces are separated by a long decreasing stretch, so a point early in the first piece can be higher than a point in the last piece.
   Theorem 1, Section 6.3. Lesson: increasing and decreasing functions

6. See it on the graph

   By the same compound-angle idea (with $\sin(A+B) = \sin A\cos B + \cos A\sin B$⁠), $f(x) = \sqrt2\sin\left(x + \frac{\pi}{4}\right)$⁠. So $f$ is a sine wave of height $\sqrt2$⁠: it starts at $f(0) = 1$⁠, rises to $f\!\left(\frac{\pi}{4}\right) = \sqrt2$⁠, falls to $f\!\left(\frac{5\pi}{4}\right) = -\sqrt2$⁠, and rises back to $f(2\pi) = 1$⁠.

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