Exercise 2.1 · Q5: $\cos^{-1}\left(-\frac12\right)$

Basics you need

Solution

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

   The input is allowed, since $-1\le-\frac12\le1$⁠. Let $y=\cos^{-1}\left(-\frac12\right)$⁠. By the definition of the principal value branch, this means both

   $$\cos y=-\frac12 \qquad\text{and}\qquad 0\le y\le\pi.$$
   The equation alone has infinitely many solutions (Step 6). The condition $0\le y\le\pi$ picks one, and it immediately tells us something useful: the answer is not negative, even though the input is.
   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 $y$ goes from $0$ to $\pi$⁠, the point $(\cos y,\sin y)$ moves anticlockwise over the upper half of the circle, from $(1,0)$ through $(0,1)$ to $(-1,0)$⁠. Its x-coordinate falls steadily from $1$ to $-1$⁠, never repeating. So exactly one $y$ in $[0,\pi]$ has $\cos y=-\frac12$⁠.

   The negative x-coordinates on this arc all belong to its left part, the second quadrant. That is where we will find the answer.
   Compare the graph of $y=\cos^{-1}x$ (Fig 2.2): it is above the line $y=\frac{\pi}{2}$ exactly for negative $x$⁠. Lesson: graphs of the inverse functions

3. Find the reference angle (ignore the sign)

   Which acute angle $\alpha$ has $\cos\alpha=\frac12$⁠?

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

   So $\alpha=\frac{\pi}{3}$ ($60^\circ$⁠).

   Every angle whose cosine is $\pm\frac12$ has reference angle $\frac{\pi}{3}$⁠. The sign, handled next, decides the quadrant.
   On $\left(0,\frac{\pi}{2}\right)$ cosine decreases, so $\frac{\pi}{3}$ is the only acute angle with cosine $\frac12$⁠. Basics: special angles and reference angles

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

   $-\frac12$ is negative. On the upper half, $\cos y\gt0$ for $0\le y\lt\frac{\pi}{2}$⁠, $\cos\frac{\pi}{2}=0$⁠, and $\cos y\lt0$ for $\frac{\pi}{2}\lt y\le\pi$⁠. So $y$ is in $\left(\frac{\pi}{2},\pi\right]$⁠: the second quadrant, together with its end point $\pi$ on the negative x-axis (where the cosine is $-1$⁠, not $-\frac12$⁠).

   A second-quadrant angle with reference angle $\alpha$ sits $\alpha$ short of $\pi$⁠, so

   $$y=\pi-\frac{\pi}{3}=\frac{2\pi}{3}.$$

   Its cosine, by the mirror rule $\cos(\pi-\theta)=-\cos\theta$⁠:

   $$\cos\frac{2\pi}{3}=\cos\left(\pi-\frac{\pi}{3}\right)=-\cos\frac{\pi}{3}=-\frac12.$$
   A negative cosine occurs in the second and the third quadrants of the full circle. The range $[0,\pi]$ contains the second but not the third, so the answer must be the second-quadrant angle. The tempting $-\frac{\pi}{3}$ is wrong twice over: it is negative, and $\cos\left(-\frac{\pi}{3}\right)=+\frac12$⁠, not $-\frac12$⁠.
   This is the same $\pi-\theta$ move the textbook uses in Example 2 for $\cot^{-1}\left(\frac{-1}{\sqrt3}\right)=\frac{2\pi}{3}$⁠. Example 2

5. Check both conditions from Step 1

   Range: $0\le\frac{2\pi}{3}\le\pi$⁠, since $\frac{2\pi}{3}\lt\frac{3\pi}{3}=\pi$⁠. ✓ Equation: $\cos\frac{2\pi}{3}=-\frac12$ (Step 4). ✓

   So $\cos^{-1}\left(-\frac12\right)=\frac{2\pi}{3}$⁠.

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

   On one turn, the vertical line $x=-\frac12$ meets the circle at the point for $\frac{2\pi}{3}$ (above the x-axis) and its mirror image in the x-axis, the point for $-\frac{2\pi}{3}$ (below it), since $\cos(-\theta)=\cos\theta$⁠. Adding whole turns gives all solutions:

   $$y=\pm\frac{2\pi}{3}+2n\pi,\qquad n\in\Z.$$

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

   Angle $y$	From the family	Decimal	In the principal range?
   $-\frac{8\pi}{3}$	$-\frac{2\pi}{3}+2n\pi$⁠, $n=-1$	−8.3776	no
   $-\frac{4\pi}{3}$	$\frac{2\pi}{3}+2n\pi$⁠, $n=-1$	−4.1888	no
   $-\frac{2\pi}{3}$	$-\frac{2\pi}{3}+2n\pi$⁠, $n=0$	−2.0944	no
   $\frac{2\pi}{3}$	$\frac{2\pi}{3}+2n\pi$⁠, $n=0$	2.0944	yes
   $\frac{4\pi}{3}$	$-\frac{2\pi}{3}+2n\pi$⁠, $n=1$	4.1888	no
   $\frac{8\pi}{3}$	$\frac{2\pi}{3}+2n\pi$⁠, $n=1$	8.3776	no

   Only $\frac{2\pi}{3}$ is inside. $-\frac{2\pi}{3}$ and $\frac{4\pi}{3}$ are the same third-quadrant point, on the lower half: one is negative, the other bigger than $\pi$⁠. $\frac{8\pi}{3}$ is the answer's own point plus a full turn, bigger than $\pi$⁠. Since cosine is one-one on $[0,\pi]$ (Step 2), no other solution can be in the range.

   The x-axis splits the circle into two halves, and the two points of the vertical line are mirror images across it. The principal range is the whole upper half, so for any value strictly between $-1$ and $1$ it contains exactly one of the two points. (For $\pm1$ the two points merge into one, on the x-axis.)
   Cosine has period $2\pi$⁠. Lesson: branches

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