Exercise 2.1 · Q4: $\tan^{-1}\left(-\sqrt3\right)$

Basics you need

Solution

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

   Let $y=\tan^{-1}\left(-\sqrt3\right)$⁠. By the definition of the principal value branch, this means both

   $$\tan y=-\sqrt3 \qquad\text{and}\qquad -\frac{\pi}{2}\lt y\lt\frac{\pi}{2}.$$
   $\tan y=-\sqrt3$ on its own has infinitely many solutions, repeating every $\pi$ (Step 6). The range condition picks out the one that $\tan^{-1}$ returns. The inequalities are strict: $\pm\frac{\pi}{2}$ can never be values of $\tan^{-1}$⁠, since tan is not defined there.
   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 open right half of the circle

   As $y$ runs from just above $-\frac{\pi}{2}$ to just below $\frac{\pi}{2}$⁠, the point $(\cos y,\sin y)$ runs up the right half of the circle, with the top and bottom points left out. On this arc $\cos y\gt0$⁠, so the slope $\tan y$ is defined, and it increases steadily: very large negative near the bottom, $0$ at $y=0$⁠, very large positive near the top.

   So every real number is the tangent of exactly one angle in $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$⁠. In particular exactly one $y$ in the range has $\tan y=-\sqrt3$⁠.

   This guarantees the answer exists and is unique, and it says where to look: on the right half, never on the left.
   The graph of $y=\tan^{-1}x$ (Fig 2.5) rises from near $-\frac{\pi}{2}$ to near $\frac{\pi}{2}$ without reaching them. Lesson: graphs of the inverse functions

3. Find the reference angle (ignore the sign)

   Which acute angle $\alpha$ has $\tan\alpha=\sqrt3$⁠?

   $\theta$	$\frac{\pi}{6}$ (30°)	$\frac{\pi}{4}$ (45°)	$\frac{\pi}{3}$ (60°)
   $\tan\theta$	$\frac{1}{\sqrt3}$	$1$	$\sqrt3$

   So $\alpha=\frac{\pi}{3}$ ($60^\circ$⁠). (A steep slope means a large angle: $\sqrt3\approx1.732$ is steeper than the 45° line of slope 1, so $\alpha$ is bigger than $\frac{\pi}{4}$⁠.)

   The steepness of the radius depends only on its tilt from the x-axis, which is the reference angle. Any angle with tangent $\pm\sqrt3$ has reference angle $\frac{\pi}{3}$⁠.
   Tangent increases on $\left(0,\frac{\pi}{2}\right)$⁠, so $\frac{\pi}{3}$ is the only acute angle with this tangent. Basics: special angles and reference angles

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

   We need $\tan y=-\sqrt3$⁠, which is negative. On the principal arc $\cos y\gt0$⁠, so $\tan y$ has the same sign as $\sin y$⁠:

   • for $0\lt y\lt\frac{\pi}{2}$ (above the x-axis) $\tan y\gt0$⁠;
   • at $y=0$⁠, $\tan y=0$⁠;
   • for $-\frac{\pi}{2}\lt y\lt0$ (below the x-axis) $\tan y\lt0$⁠.

   So $y$ lies in $\left(-\frac{\pi}{2},0\right)$⁠, the fourth quadrant. Turning $\frac{\pi}{3}$ clockwise gives

   $$y=-\frac{\pi}{3},\qquad \tan\left(-\frac{\pi}{3}\right)=-\tan\frac{\pi}{3}=-\sqrt3.$$
   On the full circle a negative tangent occurs in the second and the fourth quadrants. The principal range contains the fourth but not the second, so the answer is the fourth-quadrant angle, written as a negative number between $-\frac{\pi}{2}$ and $0$⁠.
   $\tan(-\theta)=-\tan\theta$ (Basic 5). Basics: tan, cot, sec and cosec

5. Check both conditions from Step 1

   Range: $-\frac{\pi}{3}$ is negative and its size $\frac{\pi}{3}$ is less than $\frac{\pi}{2}$⁠, so $-\frac{\pi}{2}\lt -\frac{\pi}{3}\lt\frac{\pi}{2}$⁠. ✓ Equation: $\tan\left(-\frac{\pi}{3}\right)=-\sqrt3$ (Step 4). ✓

   So $\tan^{-1}\left(-\sqrt3\right)=-\frac{\pi}{3}$⁠.

6. Rule out every other angle with tangent $-\sqrt3$

   The line through the centre with slope $-\sqrt3$ meets the circle at two opposite points: the point for $-\frac{\pi}{3}$ and, half a turn away, the point for $-\frac{\pi}{3}+\pi=\frac{2\pi}{3}$⁠. The radius to the opposite point has the same slope, because both $\sin$ and $\cos$ change sign and their ratio does not: $\tan(\theta+\pi)=\tan\theta$⁠. So all solutions are

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

   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?
   $-\frac{7\pi}{3}$	$-\frac{\pi}{3}+n\pi$⁠, $n=-2$	−7.3304	no
   $-\frac{4\pi}{3}$	$-\frac{\pi}{3}+n\pi$⁠, $n=-1$	−4.1888	no
   $-\frac{\pi}{3}$	$-\frac{\pi}{3}+n\pi$⁠, $n=0$	−1.0472	yes
   $\frac{2\pi}{3}$	$-\frac{\pi}{3}+n\pi$⁠, $n=1$	2.0944	no
   $\frac{5\pi}{3}$	$-\frac{\pi}{3}+n\pi$⁠, $n=2$	5.2360	no

   Only $-\frac{\pi}{3}$ is inside. $\frac{2\pi}{3}$ has the same tangent, but its point is on the left half of the circle (second quadrant) and $\frac{2\pi}{3}\gt\frac{\pi}{2}$⁠. Consecutive solutions are $\pi$ apart, and the open interval $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ has length exactly $\pi$⁠, so it can never hold two of them.

   Tangent repeats after half a turn, so no interval longer than $\pi$ can be one-one for it. The principal range has length exactly $\pi$ and is open at both ends, so it cannot contain two numbers $\pi$ apart: it holds at most one solution of any equation $\tan y=c$⁠, and Step 2 showed it holds at least one.
   Tangent has period $\pi$⁠. Lesson: branches

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