Exercise 2.1 · Q1: $\sin^{-1}\left(-\frac12\right)$

Basics you need

Solution

1. Say exactly what $\sin^{-1}\left(-\frac12\right)$ means

   First, the input is allowed: $-1\le-\frac12\le1$⁠, and $\sin^{-1}$ is defined on $[-1,1]$⁠. Now give the unknown angle a name. Let

   $$y=\sin^{-1}\left(-\tfrac12\right).$$

   By the definition of the principal value branch, this statement means two things together:

   $$\sin y=-\frac12 \qquad\text{and}\qquad -\frac{\pi}{2}\le y\le\frac{\pi}{2}.$$
   The first condition alone has infinitely many solutions (we list some in Step 6), so it cannot define a function. The textbook makes $\sin^{-1}$ a function by keeping only the branch with range $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$⁠. Whenever no branch is named, the principal branch is meant, and its value is called the principal value. Forgetting the second condition is the most common way to get this question wrong.
   Principal value = the value that lies in the range of the principal branch (Note 3, Section 2.2). Lesson: principal values

2. See the principal range as an arc of the unit circle

   As $y$ runs from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$⁠, the point $(\cos y,\sin y)$ runs up the right half of the unit circle, from the bottom point $(0,-1)$ through $(1,0)$ to the top point $(0,1)$⁠. Both end points are included. Along the way its height $\sin y$ rises steadily from $-1$ to $1$ and never repeats.

   So on this arc every height between $-1$ and $1$ occurs at exactly one point. In particular there is exactly one $y$ in the range with $\sin y=-\frac12$⁠. The figure after the steps shows this arc in blue.

   This step tells us the answer exists and is unique before we look for it. It also tells us where to look: on the right half of the circle, and nowhere else.
   The graph of $y=\sin^{-1}x$ (Fig 2.1) is this same fact drawn differently: it rises from $\left(-1,-\frac{\pi}{2}\right)$ to $\left(1,\frac{\pi}{2}\right)$⁠. Lesson: graphs of the inverse functions

3. Find the reference angle (ignore the sign for now)

   Drop the minus sign and ask: which acute angle $\alpha$ has $\sin\alpha=\frac12$⁠? From the special values:

   $\theta$	$\frac{\pi}{6}$ (30°)	$\frac{\pi}{4}$ (45°)	$\frac{\pi}{3}$ (60°)
   $\sin\theta$	$\frac12$	$\frac{1}{\sqrt2}$	$\frac{\sqrt3}{2}$

   So the reference angle is $\alpha=\frac{\pi}{6}$ (that is, $30^\circ$⁠).

   The size of a sine depends only on how far the radius is tilted from the x-axis, which is the reference angle. Any angle whose sine is $\pm\frac12$ has reference angle $\frac{\pi}{6}$⁠. Separating "how big" (the reference angle) from "which sign" (the quadrant) turns one hard question into two easy ones.
   On $\left(0,\frac{\pi}{2}\right)$ sine increases, so $\frac{\pi}{6}$ is the only acute angle with sine $\frac12$⁠. Basics: special angles and reference angles

4. Use the sign to choose the side, then name the angle

   We need $\sin y=-\frac12$⁠, which is negative. On the principal arc:

   • for $0\lt y\le\frac{\pi}{2}$ the point is above the x-axis, so $\sin y\gt 0$⁠;
   • at $y=0$ the point is $(1,0)$⁠, so $\sin y=0$⁠;
   • for $-\frac{\pi}{2}\le y\lt 0$ the point is below the x-axis, so $\sin y\lt 0$⁠.

   So $y$ must lie in $\left[-\frac{\pi}{2},0\right)$⁠, the part of the arc in the fourth quadrant. The angle there with reference angle $\frac{\pi}{6}$ is reached by turning $\frac{\pi}{6}$ clockwise:

   $$y=-\frac{\pi}{6}.$$

   Its sine follows from the mirror rule $\sin(-\theta)=-\sin\theta$⁠:

   $$\sin\left(-\frac{\pi}{6}\right)=-\sin\frac{\pi}{6}=-\frac12.$$
   The sign rule is what keeps us inside the principal range. In the full circle a negative sine happens in two quadrants, the third and the fourth. The principal arc contains only one of them, the fourth, so the answer must be the fourth-quadrant angle, written as a negative angle between $-\frac{\pi}{2}$ and 0.
   $\sin(-\theta)=-\sin\theta$ because the points for $\theta$ and $-\theta$ are mirror images in the x-axis. Basics: sin and cos on the unit circle

5. Check both conditions from Step 1

   Range: writing every angle over 6,

   $$-\frac{3\pi}{6}\le-\frac{\pi}{6}\le\frac{3\pi}{6},$$

   so $-\frac{\pi}{2}\le-\frac{\pi}{6}\le\frac{\pi}{2}$⁠. ✓ Equation: $\sin\left(-\frac{\pi}{6}\right)=-\frac12$ (Step 4). ✓

   Both conditions hold, so $\sin^{-1}\left(-\frac12\right)=-\frac{\pi}{6}$⁠.

   An answer to a principal-value question is only finished when it passes both tests. Checking the range explicitly is what separates $-\frac{\pi}{6}$ from impostors such as $\frac{11\pi}{6}$⁠, which passes the equation test but fails the range test.

6. Rule out every other angle with sine $-\frac12$

   On one full turn of the circle, the horizontal line at height $-\frac12$ meets the circle at exactly two points: the fourth-quadrant point for $-\frac{\pi}{6}$ and its mirror image in the y-axis, the third-quadrant point for $\pi-\left(-\frac{\pi}{6}\right)=\frac{7\pi}{6}$⁠. (The mirror rule $\sin(\pi-\theta)=\sin\theta$ confirms $\sin\frac{7\pi}{6}=-\frac12$⁠.) Every other angle with sine $-\frac12$ differs from one of these by whole turns, so all solutions of $\sin y=-\frac12$ are

   $$\begin{aligned} y&=-\frac{\pi}{6}+2n\pi \quad\text{or}\\ y&=\frac{7\pi}{6}+2n\pi,\qquad n\in\Z. \end{aligned}$$

   The ones nearest the range $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$⁠, roughly $[-1.5708,\,1.5708]$⁠:

   Angle $y$	From the family	Decimal	In the principal range?
   $-\frac{13\pi}{6}$	$-\frac{\pi}{6}+2n\pi$⁠, $n=-1$	−6.8068	no
   $-\frac{5\pi}{6}$	$\frac{7\pi}{6}+2n\pi$⁠, $n=-1$	−2.6180	no
   $-\frac{\pi}{6}$	$-\frac{\pi}{6}+2n\pi$⁠, $n=0$	−0.5236	yes
   $\frac{7\pi}{6}$	$\frac{7\pi}{6}+2n\pi$⁠, $n=0$	3.6652	no
   $\frac{11\pi}{6}$	$-\frac{\pi}{6}+2n\pi$⁠, $n=1$	5.7596	no
   $\frac{19\pi}{6}$	$\frac{7\pi}{6}+2n\pi$⁠, $n=1$	9.9484	no

   Only $-\frac{\pi}{6}$ is inside. In particular:

   • $\frac{7\pi}{6}$ has the right sine but its point is on the left half of the circle (third quadrant), and $\frac{7\pi}{6}\gt\frac{\pi}{2}$⁠.
   • $\frac{11\pi}{6}$ is the same point as $-\frac{\pi}{6}$⁠, but it is a different number, and $\frac{11\pi}{6}\gt\frac{\pi}{2}$⁠. A principal value is a number in the range, not just a point on the circle.
   • $-\frac{5\pi}{6}$ is negative like the answer, but $-\frac{5\pi}{6}\lt-\frac{\pi}{2}$⁠: it is the third-quadrant point reached clockwise.

   The table only shows nearby angles, but the conclusion holds for all of them: sine takes each value only once on the range (Step 2), so the range cannot contain a second solution.

   Showing that the rival angles fail is what makes the answer convincing rather than a memorised rule. The general fact behind it is that the principal range is one complete, non-repeating sweep of sine values: going any further either way would start repeating heights already used.
   Adding $2\pi$ (a full turn) returns to the same point, so it does not change sine. Lesson: branches

Every fact used, and where it comes from