Exercise 2.2 · Q11: $\tan^{-1}\left(\tan\frac{3\pi}{4}\right)$

Basics you need

Solution

1. Why the answer is not simply $\frac{3\pi}{4}$

   It is tempting to "cancel" $\tan^{-1}$ against $\tan$ and write $\frac{3\pi}{4}$⁠. But the value of $\tan^{-1}$ must lie in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$⁠, and

   $$\frac{3\pi}{4} > \frac{\pi}{2}, \qquad\text{so}\qquad \frac{3\pi}{4} \notin \left(-\frac{\pi}{2}, \frac{\pi}{2}\right).$$

   So $\frac{3\pi}{4}$ can never be an output of $\tan^{-1}$⁠.

   The rule $\tan^{-1}(\tan x) = x$ holds only for $x$ in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$⁠. Section 2.3 states the sine version, $\sin^{-1}(\sin x) = x$ for $x\in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$⁠, and says similar results hold for the other functions "for suitable values". Outside that interval, $\tan$ takes each value many times, and $\tan^{-1}$ can return only the one copy inside its range.
   Always check the principal range before cancelling an inverse against its function. Lesson: sin⁻¹(sin x) and the folding rules

2. Work out the inside: $\tan\frac{3\pi}{4}$

   On the unit circle the angle $\frac{3\pi}{4}$ (that is, 135°) gives the point

   $$\left(\cos\tfrac{3\pi}{4}, \sin\tfrac{3\pi}{4}\right) = \left(-\tfrac{\sqrt2}{2}, \tfrac{\sqrt2}{2}\right).$$

   So

   $$\tan\frac{3\pi}{4} = \frac{\sin\frac{3\pi}{4}}{\cos\frac{3\pi}{4}} = \frac{\sqrt2/2}{-\sqrt2/2} = -1.$$

   The expression becomes $\tan^{-1}(-1)$⁠.

   Evaluating the inner function first turns the problem into an ordinary principal-value question, which has exactly one answer. The sign comes from the quadrant: in the second quadrant $\cos < 0 < \sin$⁠, so the ratio is negative. The size, 1, comes from the reference angle $\frac{\pi}{4}$⁠.
   $\tan\theta = \dfrac{\sin\theta}{\cos\theta}$⁠. In the second quadrant, $\tan(\pi - \phi) = -\tan\phi$⁠. Basics: reference angles

3. Find the principal value of $\tan^{-1}(-1)$

   We need $y$ with $-\frac{\pi}{2} < y < \frac{\pi}{2}$ and $\tan y = -1$⁠. Since $\tan\frac{\pi}{4} = 1$ and $\tan(-\theta) = -\tan\theta$⁠,

   $$\tan\left(-\frac{\pi}{4}\right) = -\tan\frac{\pi}{4} = -1.$$

   And $-\frac{\pi}{4}$ lies in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$⁠. Because $\tan$ is strictly increasing on this interval, no other angle in it has tangent $-1$⁠. Hence

   $$\tan^{-1}\left(\tan\frac{3\pi}{4}\right) = \tan^{-1}(-1) = -\frac{\pi}{4}.$$
   A principal value has to pass two tests: it gives the right tangent, and it lies in the principal range. $\frac{3\pi}{4}$ passes the first test and fails the second; $-\frac{\pi}{4}$ passes both. Uniqueness matters too: since $\tan$ is one-one on $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$⁠, finding one angle that works means we have found the answer.
   $\tan$ is an odd function: $\tan(-\theta) = -\tan\theta$⁠. So $\tan^{-1}(-a) = -\tan^{-1}a$⁠. Lesson: principal values

4. The folding argument: why $-\frac{\pi}{4}$ and $\frac{3\pi}{4}$ share a tangent

   Since $\tan$ has period $\pi$⁠, subtracting $\pi$ does not change the tangent:

   $$\tan\left(\frac{3\pi}{4} - \pi\right) = \tan\frac{3\pi}{4}, \qquad \frac{3\pi}{4} - \pi = -\frac{\pi}{4}.$$

   In general, for any $x$ where $\tan x$ is defined, there is exactly one integer $k$ with $x - k\pi$ in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$⁠, and

   $$\tan^{-1}(\tan x) = x - k\pi.$$

   For $x = \frac{3\pi}{4}$ we need $k = 1$⁠, giving $-\frac{\pi}{4}$ again.

   The principal range of $\tan^{-1}$ has length exactly $\pi$⁠, the period of $\tan$⁠. So every angle can be moved into it, by whole multiples of $\pi$⁠, in exactly one way. That move is the folding: it keeps the tangent and lands the angle where $\tan^{-1}$ can see it. On the unit circle, moving by $\pi$ jumps to the opposite point, and $\tan\theta$ is the slope of the line from the centre to the point, which is the same at both ends.
   $\tan(\theta - \pi) = \tan\theta$⁠: both $\sin$ and $\cos$ change sign when $\pi$ is subtracted, so their ratio is unchanged. Basics: the unit circle

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