Exercise 2.1 · Q11: Value of $\tan^{-1}1+\cos^{-1}\left(-\frac12\right)+\sin^{-1}\left(-\frac12\right)$

Basics you need

Solution

1. First term: $\tan^{-1}(1)$

   Let $a = \tan^{-1}(1)$⁠. Then $\tan a = 1$ and $-\frac{\pi}{2} < a < \frac{\pi}{2}$⁠.

   Reference angle: $\tan\frac{\pi}{4} = \frac{\sin(\pi/4)}{\cos(\pi/4)} = 1$⁠, so $\alpha = \frac{\pi}{4}$⁠. Sign: on $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$⁠, tan is positive exactly on $\left(0,\frac{\pi}{2}\right)$⁠, and $1 > 0$⁠, so $a = \alpha$⁠:

   $$a = \tan^{-1}(1) = \frac{\pi}{4}.$$

   Exclusion: all solutions of $\tan a = 1$ are $\frac{\pi}{4} + n\pi$⁠. The neighbours $-\frac{3\pi}{4}$ and $\frac{5\pi}{4}$ are outside $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$⁠, so $\frac{\pi}{4}$ is the only one in range.

   Tan has period $\pi$⁠, and the principal range is an open interval of length exactly $\pi$⁠, so it catches exactly one solution. Points with equal tan lie on one line through $O$⁠, and only the right-half end of that line is in the range (left panel of the figure).
   $\tan^{-1} : \R \to \left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ (Section 2.2). Basics: tan, cot, sec and cosec

2. Second term: $\cos^{-1}\left(-\frac12\right)$

   Let $b = \cos^{-1}\left(-\frac12\right)$⁠. Then $\cos b = -\frac12$ and $0 \le b \le \pi$⁠.

   Reference angle: $\cos\frac{\pi}{3} = \frac12$⁠, so $\alpha = \frac{\pi}{3}$⁠. Sign: on $[0,\pi]$⁠, cos is negative exactly on $\left(\frac{\pi}{2},\pi\right]$⁠, the second quadrant, so $b = \pi - \alpha$⁠. Indeed $\cos\left(\pi - \frac{\pi}{3}\right) = -\cos\frac{\pi}{3} = -\frac12$⁠. Hence

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

   Exclusion: all solutions of $\cos b = -\frac12$ are $\pm\frac{2\pi}{3} + 2n\pi$⁠. The nearest others, $-\frac{2\pi}{3}$ and $\frac{4\pi}{3}$⁠, are outside $[0,\pi]$⁠.

   The range of $\cos^{-1}$ has no negative angles, so the answer cannot be $-\frac{\pi}{3}$ (whose cosine is $+\frac12$ anyway). A negative cosine pushes the answer past $\frac{\pi}{2}$⁠, into the second quadrant. This is the same value as Q5.
   $\cos(\pi - \alpha) = -\cos\alpha$⁠. Basics: sin and cos on the unit circle

3. Third term: $\sin^{-1}\left(-\frac12\right)$

   Let $c = \sin^{-1}\left(-\frac12\right)$⁠. Then $\sin c = -\frac12$ and $-\frac{\pi}{2} \le c \le \frac{\pi}{2}$⁠.

   Reference angle: $\sin\frac{\pi}{6} = \frac12$⁠, so $\alpha = \frac{\pi}{6}$⁠. Sign: on $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$⁠, sin is negative exactly on $\left[-\frac{\pi}{2},0\right)$⁠, so $c = -\alpha$⁠. Indeed $\sin\left(-\frac{\pi}{6}\right) = -\sin\frac{\pi}{6} = -\frac12$⁠. Hence

   $$c = \sin^{-1}\left(-\frac12\right) = -\frac{\pi}{6}.$$

   Exclusion: all solutions of $\sin c = -\frac12$ are $-\frac{\pi}{6} + 2n\pi$ and $\frac{7\pi}{6} + 2n\pi$⁠. The nearest others, $-\frac{5\pi}{6}$ and $\frac{7\pi}{6}$⁠, are outside $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$⁠.

   Compare with Step 2: the input $-\frac12$ is the same, but the function and its range differ, so the sign rule differs. The sine range is symmetric about $0$⁠, so a negative input simply gives a negative angle. This is the same value as Q1.
   $\sin(-\alpha) = -\sin\alpha$⁠. Basics: domain and range of sin and cos

4. Add the three values

   Now the expression is a sum of three ordinary numbers. Use the common denominator $12$⁠:

   $$\begin{aligned} a + b + c &= \frac{\pi}{4} + \frac{2\pi}{3} - \frac{\pi}{6}\\ &= \frac{3\pi}{12} + \frac{8\pi}{12} - \frac{2\pi}{12}\\ &= \frac{9\pi}{12} = \frac{3\pi}{4}. \end{aligned}$$
   The least common multiple of $4, 3, 6$ is $12$⁠, so every term becomes a whole number of twelfths and the addition is exact. The total $\frac{3\pi}{4}$ is just a number; there is no requirement that a sum of principal values lie in any principal range.
   $\frac{\pi}{4} = \frac{3\pi}{12}$⁠, $\frac{2\pi}{3} = \frac{8\pi}{12}$⁠, $\frac{\pi}{6} = \frac{2\pi}{12}$⁠. Basics: angles and radians

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