Exercise 2.1 · Q8: Principal value of $\cot^{-1}\left(\sqrt3\right)$

Basics you need

Solution

1. Name the unknown and its range

   Let $y = \cot^{-1}(\sqrt3)$⁠. By the definition of the principal value branch, this means exactly two things:

   $$\cot y = \sqrt3 \qquad\text{and}\qquad 0 < y < \pi.$$
   The range condition is not decoration; it is half of the definition. Without it, "the angle whose cot is $\sqrt3$" would have infinitely many answers. The textbook fixes the branch $(0,\pi)$ for $\cot^{-1}$ in Section 2.2, and the Note (point 2) says the principal branch is meant whenever no branch is named.
   $\cot^{-1} : \R \to (0,\pi)$⁠: domain all real numbers, range the open interval $(0,\pi)$⁠. The ends are excluded because $\cot 0$ and $\cot\pi$ are undefined. Lesson: principal values

2. Find the reference angle

   Ignore the sign for a moment and ask: which acute angle $\alpha$ has $\cot\alpha = \sqrt3$⁠? From the special angles,

   $$\cot\frac{\pi}{6} = \frac{\cos\frac{\pi}{6}}{\sin\frac{\pi}{6}} = \frac{\sqrt3/2}{1/2} = \sqrt3.$$

   So the reference angle is $\alpha = \frac{\pi}{6}$⁠.

   Every trig value of an angle is, up to sign, the value of an acute angle (its reference angle). So we always first identify the acute angle from the special-angle table, then let the sign decide where the true answer sits. For comparison, $\cot\frac{\pi}{3} = \frac{1}{\sqrt3}$ and $\cot\frac{\pi}{4} = 1$⁠, so $\frac{\pi}{6}$ is the only special angle that fits.
   $\sin\frac{\pi}{6} = \frac12$⁠, $\cos\frac{\pi}{6} = \frac{\sqrt3}{2}$⁠. Basics: special angles and reference angles

3. Use the sign to place the answer

   On $(0,\pi)$ the sine is positive, so the sign of $\cot y = \frac{\cos y}{\sin y}$ is the sign of $\cos y$⁠:

   • on $\left(0, \frac{\pi}{2}\right)$⁠: $\cos y > 0$⁠, so $\cot y > 0$⁠;
   • at $\frac{\pi}{2}$⁠: $\cot y = 0$⁠;
   • on $\left(\frac{\pi}{2}, \pi\right)$⁠: $\cos y < 0$⁠, so $\cot y < 0$⁠.

   Our value $\sqrt3$ is positive, so $y$ lies in the first-quadrant part $\left(0,\frac{\pi}{2}\right)$⁠, where an angle equals its own reference angle. Hence

   $$y = \alpha = \frac{\pi}{6}.$$
   This is the sign rule for $\cot^{-1}$⁠: a positive input gives an acute answer $\alpha$⁠; a negative input gives the second-quadrant answer $\pi - \alpha$⁠, because $\cot(\pi-\alpha) = -\cot\alpha$⁠. Example 2 of the textbook is the negative case: $\cot^{-1}\left(-\frac{1}{\sqrt3}\right) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$⁠.
   Quadrant signs: cos is positive in the first quadrant and negative in the second; sin is positive in both. Basics: sin and cos on the unit circle

4. Check it, and exclude every other candidate

   Check. $0 < \frac{\pi}{6} < \pi$ ✓, and $\cot\frac{\pi}{6} = \sqrt3$ ✓ (Step 2). So $\frac{\pi}{6}$ satisfies both conditions of Step 1.

   Other angles with the same cot. Cot has period $\pi$⁠, so every solution of $\cot y = \sqrt3$ is

   $$y = \frac{\pi}{6} + n\pi, \qquad n \in \Z.$$

   For $n = 0$ this is $\frac{\pi}{6}$⁠, inside $(0,\pi)$⁠. For $n = 1$ it is $\frac{7\pi}{6} > \pi$⁠, and for $n = -1$ it is $-\frac{5\pi}{6} < 0$⁠; every other $n$ is further out. So $\frac{\pi}{6}$ is the only candidate in the principal range.

   Two angles have the same cot exactly when their points on the unit circle lie on the same line through $O$⁠, because the ratio $x : y$ is the same. A line through $O$ cuts the circle at two opposite points, one in the upper half and one in the lower half. The upper half is the principal range, so exactly one of the two survives. That is why $\cot^{-1}$ is a genuine function on this branch.
   Cot repeats every $\pi$⁠: $\cot(y+\pi) = \cot y$⁠. On $(0,\pi)$ it is one-one, so it can be inverted there. Lesson: branches

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