Exercise 2.1 · Q2: $\cos^{-1}\left(\frac{\sqrt3}{2}\right)$

Basics you need

Solution

1. Say exactly what $\cos^{-1}\left(\frac{\sqrt3}{2}\right)$ means

   The input is allowed: $\sqrt3\approx1.732$⁠, so $\frac{\sqrt3}{2}\approx0.866$ lies in $[-1,1]$⁠. Let

   $$y=\cos^{-1}\left(\frac{\sqrt3}{2}\right).$$

   By the definition of the principal value branch, this means both

   $$\cos y=\frac{\sqrt3}{2} \qquad\text{and}\qquad 0\le y\le\pi.$$
   The equation $\cos y=\frac{\sqrt3}{2}$ alone has infinitely many solutions (Step 6 lists several). The range condition is what picks out one of them and makes $\cos^{-1}$ a function. With no branch named, the principal branch is meant.
   Principal value = the value lying in the range of the principal branch (Note 3, Section 2.2). Lesson: principal values

2. See the principal range as the upper half of the circle

   As $y$ runs from $0$ to $\pi$⁠, the point $(\cos y,\sin y)$ runs anticlockwise over the upper half of the unit circle: from $(1,0)$ through $(0,1)$ to $(-1,0)$⁠, both ends included. Its x-coordinate $\cos y$ falls steadily from $1$ to $-1$ and never repeats.

   So every value between $-1$ and $1$ is the cosine of exactly one angle in $[0,\pi]$⁠, and there is exactly one $y$ in the range with $\cos y=\frac{\sqrt3}{2}$⁠. The blue arc in the figure after the steps is this half.

   Notice this is a different half from the one used for $\sin^{-1}$⁠. For sine the right half works because heights do not repeat there; for cosine we need a half on which x-coordinates do not repeat, and the upper half is the textbook's choice.
   The graph of $y=\cos^{-1}x$ falls from $(-1,\pi)$ down to $(1,0)$⁠: the same fact read off a graph (Fig 2.2). Lesson: graphs of the inverse functions

3. Find the reference angle

   Which acute angle $\alpha$ has $\cos\alpha=\frac{\sqrt3}{2}$⁠? From the special values:

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

   So $\alpha=\frac{\pi}{6}$ ($30^\circ$⁠).

   Any angle whose cosine is $\pm\frac{\sqrt3}{2}$ has reference angle $\frac{\pi}{6}$⁠, because the x-coordinate's size depends only on how far the radius is tilted from the x-axis. Here the number is already positive, so there is no sign to drop.
   On $\left(0,\frac{\pi}{2}\right)$ cosine decreases, so $\frac{\pi}{6}$ is the only acute angle with cosine $\frac{\sqrt3}{2}$⁠. Basics: special angles and reference angles

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

   $\frac{\sqrt3}{2}$ is positive. On the upper half:

   • for $0\le y\lt\frac{\pi}{2}$ the point is right of the y-axis, so $\cos y\gt 0$⁠;
   • at $y=\frac{\pi}{2}$ the point is $(0,1)$⁠, so $\cos y=0$⁠;
   • for $\frac{\pi}{2}\lt y\le\pi$ the point is left of the y-axis, so $\cos y\lt 0$⁠.

   So $y$ lies in $\left[0,\frac{\pi}{2}\right)$⁠, the first quadrant, where an angle equals its own reference angle:

   $$y=\frac{\pi}{6},\qquad \cos\frac{\pi}{6}=\frac{\sqrt3}{2}.$$
   A positive cosine occurs in two quadrants of the full circle, the first and the fourth. The principal range $[0,\pi]$ contains the first but not the fourth, so the answer is the first-quadrant angle.

5. Check both conditions from Step 1

   Range: $0\le\frac{\pi}{6}\le\pi$⁠, since $\frac{\pi}{6}$ is positive and smaller than $\frac{6\pi}{6}=\pi$⁠. ✓ Equation: $\cos\frac{\pi}{6}=\frac{\sqrt3}{2}$⁠. ✓

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

   Passing both tests is what "principal value" means. The rival $-\frac{\pi}{6}$ passes the equation test but fails the range test, which is the subject of the next step.

6. Rule out every other angle with cosine $\frac{\sqrt3}{2}$

   On one full turn, the vertical line $x=\frac{\sqrt3}{2}$ meets the circle at exactly two points: the point for $\frac{\pi}{6}$ (above the x-axis) and its mirror image in the x-axis, the point for $-\frac{\pi}{6}$ (below it), because $\cos(-\theta)=\cos\theta$⁠. Adding whole turns gives all solutions:

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

   The ones nearest the range $[0,\pi]\approx[0,\,3.1416]$⁠:

   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 right cosine, but it is negative, and its point is on the lower half of the circle.
   • $\frac{11\pi}{6}$ is the same point as $-\frac{\pi}{6}$ and is bigger than $\pi$⁠.
   • $\frac{13\pi}{6}$ is the same point as $\frac{\pi}{6}$⁠, but as a number it is bigger than $\pi$⁠. The principal value is a number in the range, not just a point.

   Beyond the table nothing changes: cosine takes each value only once on $[0,\pi]$ (Step 2), so no second solution can sit in the range.

   The two points of the vertical line are mirror images across the x-axis. The principal range for $\cos^{-1}$ is the upper of the two halves that the x-axis cuts the circle into, so for any value strictly between $-1$ and $1$ it contains exactly one of the two mirror points. (For $\pm1$ the two points merge into one, on the x-axis.)
   A full turn ($2\pi$⁠) brings you back to the same point, so it never changes cosine. Lesson: branches

Every fact used, and where it comes from