Exercise 2.1 · Q7: $\sec^{-1}\left(\frac{2}{\sqrt3}\right)$

Basics you need

Solution

1. Check the input and say exactly what the question means

   Domain: $\sqrt3\lt2$ (because $3\lt4$⁠), so $\frac{2}{\sqrt3}\gt1$⁠. Hence $\frac{2}{\sqrt3}\in\R-(-1,1)$ and $\sec^{-1}\left(\frac{2}{\sqrt3}\right)$ is defined. Let $y=\sec^{-1}\left(\frac{2}{\sqrt3}\right)$⁠. By the definition of the principal value branch,

   $$\begin{aligned} &\sec y=\frac{2}{\sqrt3},\\ &0\le y\le\pi,\quad y\ne\frac{\pi}{2}. \end{aligned}$$
   Secant never takes values strictly between $-1$ and $1$⁠, so $\sec^{-1}$ of such a number would not exist. Checking the domain first confirms the question makes sense. The range condition then picks one solution out of infinitely many.
   Principal value = the value lying in the range of the principal branch (Note 3, Section 2.2). Lesson: principal values

2. Turn the secant equation into a cosine equation

   Since $\sec y=\frac{1}{\cos y}$⁠,

   $$\frac{1}{\cos y}=\frac{2}{\sqrt3}\iff\cos y=\frac{\sqrt3}{2}.$$

   Both directions hold: if $\cos y=\frac{\sqrt3}{2}$⁠, then $\cos y\ne0$⁠, so $\sec y$ is defined and equals $\frac{2}{\sqrt3}$⁠.

   We know special values of cosine, not of secant, so we flip both sides. Checking both directions guarantees the two equations have exactly the same solutions.
   Reciprocals: $\sec$ and $\cos$ (Basics 1 and 5). Basics: tan, cot, sec and cosec

3. See the principal range on the unit circle

   As $y$ runs from $0$ to $\pi$⁠, the point $(\cos y,\sin y)$ runs over the upper half of the circle and $\cos y$ falls steadily from $1$ to $-1$⁠. The top point $(0,1)$⁠, at $y=\frac{\pi}{2}$⁠, is removed because its x-coordinate is $0$ and sec is undefined there.

   Because the cosines never repeat on this arc, neither do the secants. So exactly one $y$ in the range has $\cos y=\frac{\sqrt3}{2}$⁠, that is, $\sec y=\frac{2}{\sqrt3}$⁠.

   The removed point does not matter here, as $\cos\frac{\pi}{2}=0\ne\frac{\sqrt3}{2}$⁠, but the range must be stated exactly as the textbook defines it.
   Compare the graph of $y=\sec^{-1}x$ (Fig 2.4). Lesson: graphs of the inverse functions

4. Reference angle, then the sign

   Which acute angle has cosine $\frac{\sqrt3}{2}$⁠?

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

   So $\alpha=\frac{\pi}{6}$⁠. The value is positive, and on the principal arc $\cos y\gt0$ exactly for $0\le y\lt\frac{\pi}{2}$ (right of the y-axis). There the angle equals its reference angle:

   $$y=\frac{\pi}{6}.$$
   A positive cosine occurs in the first and the fourth quadrants. Only the first is inside $[0,\pi]$⁠, so the answer is the first-quadrant angle.
   Reference angle fixes the size; the sign fixes the side. Basics: special angles and reference angles

5. Check both conditions from Step 1

   Range: $0\le\frac{\pi}{6}\le\pi$ and $\frac{\pi}{6}\ne\frac{\pi}{2}$⁠. ✓ Equation:

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

   ✓ So $\sec^{-1}\left(\frac{2}{\sqrt3}\right)=\frac{\pi}{6}$⁠.

6. Rule out every other angle with this secant

   By Step 2 we need all solutions of $\cos y=\frac{\sqrt3}{2}$⁠. The vertical line $x=\frac{\sqrt3}{2}$ meets the circle at the points for $\frac{\pi}{6}$ and $-\frac{\pi}{6}$⁠, mirror images in the x-axis. With whole turns added:

   $$y=\pm\frac{\pi}{6}+2n\pi,\qquad n\in\Z.$$

   The ones nearest the range $[0,\pi]-\left\{\frac{\pi}{2}\right\}$⁠, roughly $[0,\,3.1416]$ without $1.5708$⁠:

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

   Only $\frac{\pi}{6}$ is inside. $-\frac{\pi}{6}$ has the same secant but its point is on the lower half and it is negative; $\frac{11\pi}{6}$ is that same point, and $\frac{13\pi}{6}$ is the answer's point after a full turn, both bigger than $\pi$⁠. Secant is one-one on the range (Step 3), so there is no second solution in it.

   As with $\cos^{-1}$⁠, the range is the upper half of the circle, and of the two mirror points exactly one is on the upper half.
   Cosine, and so secant, has period $2\pi$⁠. Lesson: branches

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