Exercise 2.1 · Q3: $\operatorname{cosec}^{-1}(2)$

Basics you need

Solution

1. Say exactly what $\operatorname{cosec}^{-1}(2)$ means

   The input is allowed, because $2\ge1$ puts it in $\R-(-1,1)$⁠. Let $y=\operatorname{cosec}^{-1}(2)$⁠. By the definition of the principal value branch,

   $$\begin{aligned} &\operatorname{cosec}y=2,\\ &-\frac{\pi}{2}\le y\le\frac{\pi}{2},\quad y\ne0. \end{aligned}$$
   The equation alone has infinitely many solutions; the range condition picks one. The textbook chooses $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]-\{0\}$ because on that set cosec is one-one and takes every value in $\R-(-1,1)$⁠.
   Principal value = the value lying in the range of the principal branch (Note 3, Section 2.2). Lesson: principal values

2. Turn the cosec equation into a sine equation

   Since $\operatorname{cosec}y=\frac{1}{\sin y}$⁠,

   $$\frac{1}{\sin y}=2 \iff \sin y=\frac12.$$

   The step is reversible in both directions: if $\sin y=\frac12$⁠, then $\sin y\ne0$⁠, so $\operatorname{cosec}y$ is defined and equals $\frac{1}{1/2}=2$⁠.

   We know the special values of sine by heart, not of cosec. Taking reciprocals turns the question into a familiar one. We check both directions so that no solution is lost or gained: a reciprocal can only be taken when the number is not zero, and $\frac12\ne0$⁠.
   $\operatorname{cosec}$ and $\sin$ are reciprocals wherever both are defined. Basics: tan, cot, sec and cosec

3. See the principal range on 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 circle, and its height rises steadily from $-1$ to $1$⁠. The point $(1,0)$⁠, at $y=0$⁠, is removed, because its height is $0$ and cosec is not defined there.

   Since the heights never repeat on this arc, the cosecants $\frac{1}{\sin y}$ never repeat either: each value with $|\text{value}|\ge1$ is reached exactly once. So exactly one $y$ in the range has $\operatorname{cosec}y=2$⁠, that is, $\sin y=\frac12$⁠.

   The removed point does not affect this question, since $\sin0=0\ne\frac12$⁠, but stating it keeps the range exactly as the textbook defines it.
   Compare the graph of $y=\operatorname{cosec}^{-1}x$ in Fig 2.3. Lesson: graphs of the inverse functions

4. Reference angle, then the sign

   Which acute angle has sine $\frac12$⁠?

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

   The reference angle is $\alpha=\frac{\pi}{6}$⁠. The value $\frac12$ is positive, and on the principal arc the height is positive exactly for $0\lt y\le\frac{\pi}{2}$ (above the x-axis). There the angle equals its reference angle:

   $$y=\frac{\pi}{6}.$$
   A positive sine occurs in the first and the second quadrants of the full circle. The principal arc contains the first but not the second, so the first-quadrant angle is the one we want.
   Reference angle fixes the size; the sign fixes the side. Basics: special angles and reference angles

5. Check both conditions from Step 1

   Range: $-\frac{\pi}{2}\le\frac{\pi}{6}\le\frac{\pi}{2}$ (as $\frac{\pi}{6}\lt\frac{3\pi}{6}$⁠) and $\frac{\pi}{6}\ne0$⁠. ✓ Equation:

   $$\operatorname{cosec}\frac{\pi}{6}=\frac{1}{\sin\frac{\pi}{6}}=\frac{1}{1/2}=2.$$

   ✓ So $\operatorname{cosec}^{-1}(2)=\frac{\pi}{6}$⁠.

6. Rule out every other angle with cosec 2

   By Step 2, the solutions of $\operatorname{cosec}y=2$ are exactly the solutions of $\sin y=\frac12$⁠. On one turn, the line at height $\frac12$ meets the circle at the point for $\frac{\pi}{6}$ and its mirror image in the y-axis, the point for $\pi-\frac{\pi}{6}=\frac{5\pi}{6}$⁠. Adding whole turns gives all of them:

   $$\begin{aligned} y&=\frac{\pi}{6}+2n\pi \quad\text{or}\\ y&=\frac{5\pi}{6}+2n\pi,\qquad n\in\Z. \end{aligned}$$

   The ones nearest the range $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]-\{0\}$⁠, roughly $[-1.5708,\,1.5708]$ without $0$⁠:

   Angle $y$	From the family	Decimal	In the principal range?
   $-\frac{11\pi}{6}$	$\frac{\pi}{6}+2n\pi$⁠, $n=-1$	−5.7596	no
   $-\frac{7\pi}{6}$	$\frac{5\pi}{6}+2n\pi$⁠, $n=-1$	−3.6652	no
   $\frac{\pi}{6}$	$\frac{\pi}{6}+2n\pi$⁠, $n=0$	0.5236	yes
   $\frac{5\pi}{6}$	$\frac{5\pi}{6}+2n\pi$⁠, $n=0$	2.6180	no
   $\frac{13\pi}{6}$	$\frac{\pi}{6}+2n\pi$⁠, $n=1$	6.8068	no
   $\frac{17\pi}{6}$	$\frac{5\pi}{6}+2n\pi$⁠, $n=1$	8.9012	no

   Only $\frac{\pi}{6}$ is inside. $\frac{5\pi}{6}$ has the same height, so the same cosec, but its point is on the left half (second quadrant) and $\frac{5\pi}{6}\gt\frac{\pi}{2}$⁠. $\frac{13\pi}{6}$ is the same point as $\frac{\pi}{6}$ but a different number, outside the range. In general, cosec is one-one on the range (Step 3), so it cannot hold a second solution.

   The two rival families come from the two points at height $\frac12$⁠. The principal arc is the right half, so it contains only the right-hand point, and only one number for it.
   Adding $2\pi$ returns to the same point, so it never changes sine or cosec. Lesson: branches

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