Example 26: $\sin x+\cos x$ is not one-one on $[0,\pi/2]$

Basics you need

Solution

1. Set up: two distinct points of the interval

   To prove $f$ and $g$ one-one we must handle every pair of distinct inputs. So let $x_1, x_2$ be any two different points of $\left[0,\frac{\pi}{2}\right]$⁠, and name them so that $x_1 < x_2$⁠. Then

   $$0 \le x_1 < x_2 \le \frac{\pi}{2}.$$

   From this we get two facts about the half-difference and the half-sum, which are the angles that appear in the sum-to-product formulas:

   $$0 < \frac{x_2 - x_1}{2} \le \frac{\pi}{4}, \qquad 0 < \frac{x_1 + x_2}{2} < \frac{\pi}{2}.$$

   For the first: $x_2 - x_1 > 0$⁠, and at most $\frac{\pi}{2} - 0$⁠. For the second: $x_1 + x_2 > 0$ because $x_2 > x_1 \ge 0$⁠, and $x_1 + x_2 < \pi$ because $x_1 < x_2 \le \frac{\pi}{2}$ makes $x_1 < \frac{\pi}{2}$⁠.

   The textbook simply states that $\sin x_1 \ne \sin x_2$ and $\cos x_1 \ne \cos x_2$ for distinct points. To prove it we will look at the differences $\sin x_2 - \sin x_1$ and $\cos x_1 - \cos x_2$ and show they are strictly positive. The two inequalities above are exactly what we need to decide those signs. Naming the smaller point $x_1$ costs nothing, because any two different numbers have a smaller one.
   To prove a "for all" statement, work with arbitrary elements, not chosen examples (Basic 2). Both bounds on the half-sum are strict, which matters in the next step.

2. $f(x) = \sin x$ is one-one

   By the sum-to-product formula (Basic 3) with $A = x_2,\ B = x_1$⁠:

   $$\sin x_2 - \sin x_1 = 2\cos\frac{x_2 + x_1}{2}\,\sin\frac{x_2 - x_1}{2}.$$

   By Step 1, $\frac{x_1+x_2}{2}$ lies strictly between 0 and $\frac{\pi}{2}$⁠, so its cosine is positive; and $\frac{x_2-x_1}{2}$ lies in $\left(0,\frac{\pi}{4}\right]$⁠, so its sine is positive. The product of 2 and two positive numbers is positive:

   $$\sin x_2 - \sin x_1 > 0, \qquad\text{that is}\qquad \sin x_1 < \sin x_2.$$

   So distinct inputs give distinct outputs (in fact $\sin$ is strictly increasing here), and $f$ is one-one.

   A difference is easiest to judge when it is written as a product, because then its sign is just the product of the signs of the factors. The sum-to-product formula does exactly that. The strict inequality $\frac{x_1+x_2}{2} < \frac{\pi}{2}$ from Step 1 is what keeps $\cos\frac{x_1+x_2}{2}$ from being 0.
   In the first quadrant, $0<\theta<\frac{\pi}{2}$⁠, sine and cosine are both positive (Basic 4). A strictly increasing function is one-one (Basic 2).

3. $g(x) = \cos x$ is one-one

   By the second formula (Basic 3) with $A = x_1,\ B = x_2$⁠:

   $$\begin{aligned} \cos x_1 - \cos x_2 &= -2\sin\frac{x_1 + x_2}{2}\,\sin\frac{x_1 - x_2}{2} \\ &= 2\sin\frac{x_1 + x_2}{2}\,\sin\frac{x_2 - x_1}{2}. \end{aligned}$$

   (The second form uses $\sin(-\theta) = -\sin\theta$ to absorb the minus sign.) By Step 1 both angles $\frac{x_1+x_2}{2}$ and $\frac{x_2-x_1}{2}$ lie strictly between 0 and $\frac{\pi}{2}$⁠, so both sines are positive. Hence

   $$\cos x_1 - \cos x_2 > 0, \qquad\text{that is}\qquad \cos x_1 > \cos x_2.$$

   So $\cos$ is strictly decreasing on the interval, distinct inputs give distinct outputs, and $g$ is one-one.

   Same method, same reason: write the difference as a product and read off its sign. Here we compare $\cos x_1 - \cos x_2$ (larger minus smaller, as it turns out) so that the result comes out positive.
   $\sin(-\theta) = -\sin\theta$ (sine is an odd function). A strictly decreasing function is one-one too (Basic 2).

4. $f+g$ is not one-one: one counterexample

   The sum is $(f+g)(x) = \sin x + \cos x$⁠. Evaluate it at the two ends of the interval:

   $$\begin{aligned} (f+g)(0) &= \sin 0 + \cos 0 = 0 + 1 = 1,\\ (f+g)\left(\tfrac{\pi}{2}\right) &= \sin\tfrac{\pi}{2} + \cos\tfrac{\pi}{2} = 1 + 0 = 1. \end{aligned}$$

   The inputs $0$ and $\frac{\pi}{2}$ are different, but the outputs are equal. So $f+g$ is not one-one.

   "Not one-one" is the failure of a "for all" statement, so a single pair of distinct inputs with equal outputs settles it (Basic 1). The end points are a natural place to look: $\sin$ and $\cos$ swap their values 0 and 1 there, so their sums agree.
   $\sin 0 = 0,\ \cos 0 = 1,\ \sin\frac{\pi}{2} = 1,\ \cos\frac{\pi}{2} = 0$⁠.

5. Why the sum fails: it rises and then falls

   Using $\sin\frac{\pi}{4} = \cos\frac{\pi}{4} = \frac{1}{\sqrt2}$ and the formula for $\sin(A+B)$⁠,

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

   As $x$ runs from 0 to $\frac{\pi}{2}$⁠, the angle $x + \frac{\pi}{4}$ runs from $\frac{\pi}{4}$ to $\frac{3\pi}{4}$⁠, passing through $\frac{\pi}{2}$ where sine is largest. So $f+g$ climbs to its maximum $\sqrt2$ at $x = \frac{\pi}{4}$ and then comes back down. It is also symmetric about $\frac{\pi}{4}$⁠:

   $$\begin{aligned} (f+g)\left(\tfrac{\pi}{2} - x\right) &= \sin\left(\tfrac{\pi}{2}-x\right) + \cos\left(\tfrac{\pi}{2}-x\right) \\ &= \cos x + \sin x = (f+g)(x). \end{aligned}$$

   So every value strictly between 1 and $\sqrt2$ is taken twice, at $x$ and at $\frac{\pi}{2}-x$⁠. For example, at $\frac{\pi}{6}$ and $\frac{\pi}{3}$⁠:

   $$\sin\tfrac{\pi}{6} + \cos\tfrac{\pi}{6} = \tfrac12 + \tfrac{\sqrt3}{2} = \tfrac{1+\sqrt3}{2} = \cos\tfrac{\pi}{3} + \sin\tfrac{\pi}{3}.$$
   The counterexample in Step 4 already proves the claim; this step explains it. An increasing function plus a decreasing function can do anything: here the increase of $\sin$ wins for $x<\frac{\pi}{4}$ and the decrease of $\cos$ wins for $x>\frac{\pi}{4}$⁠, so the sum turns around, and a function that turns around repeats values.
   $\sin(A+B) = \sin A\cos B + \cos A\sin B$⁠, and $\sin\left(\frac{\pi}{2}-x\right) = \cos x$⁠, $\cos\left(\frac{\pi}{2}-x\right) = \sin x$ (Class XI, Trigonometric Functions).

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