Exercise 2.2 · Q13 (MCQ): $\cos^{-1}\left(\cos\frac{7\pi}{6}\right)$

Basics you need

Solution

1. Why the answer is not $\frac{7\pi}{6}$

   The value of $\cos^{-1}$ must lie in $[0, \pi]$⁠. But

   $$\frac{7\pi}{6} = \pi + \frac{\pi}{6} > \pi, \qquad\text{so}\qquad \frac{7\pi}{6} \notin [0, \pi].$$

   So "cancelling" $\cos^{-1}$ against $\cos$ is not allowed here.

   The rule $\cos^{-1}(\cos x) = x$ holds only for $x \in [0, \pi]$⁠. Section 2.3 states the sine version, $\sin^{-1}(\sin x) = x$ on $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$⁠, and says similar results hold for the other functions for suitable values. For cosine the suitable values are exactly the principal range $[0,\pi]$⁠.
   Check the principal range before cancelling. Lesson: sin⁻¹(sin x) and the folding rules

2. Work out the inside: $\cos\frac{7\pi}{6}$

   $\frac{7\pi}{6}$ (210°) is in the third quadrant, where $\cos < 0$⁠, and its reference angle is $\frac{\pi}{6}$⁠. So

   $$\cos\frac{7\pi}{6} = \cos\left(\pi + \frac{\pi}{6}\right) = -\cos\frac{\pi}{6} = -\frac{\sqrt3}{2}.$$

   The expression becomes $\cos^{-1}\left(-\frac{\sqrt3}{2}\right)$⁠.

   Evaluating the inside first reduces the problem to a principal value, which has exactly one answer. The size $\frac{\sqrt3}{2}$ comes from the reference angle; the minus sign comes from the quadrant.
   $\cos(\pi + \phi) = -\cos\phi$⁠: adding $\pi$ moves to the opposite point of the circle, which flips both coordinates. Basics: reference angles

3. Find the principal value of $\cos^{-1}\left(-\frac{\sqrt3}{2}\right)$

   We need $y \in [0,\pi]$ with $\cos y = -\frac{\sqrt3}{2}$⁠. A negative cosine in $[0,\pi]$ means the second quadrant, with reference angle $\frac{\pi}{6}$⁠:

   $$\cos\frac{5\pi}{6} = \cos\left(\pi - \frac{\pi}{6}\right) = -\cos\frac{\pi}{6} = -\frac{\sqrt3}{2}.$$

   Since $0 \le \frac{5\pi}{6} \le \pi$⁠, and $\cos$ takes each value only once on $[0,\pi]$⁠, this is the only answer:

   $$\cos^{-1}\left(\cos\frac{7\pi}{6}\right) = \cos^{-1}\left(-\frac{\sqrt3}{2}\right) = \frac{5\pi}{6}.$$
   Both tests are passed: $\frac{5\pi}{6}$ has the right cosine and lies in the principal range. Uniqueness comes from $\cos$ being strictly decreasing on $[0,\pi]$ (Basic 1), so there is no second candidate to worry about.
   A useful shortcut: $\cos^{-1}(-x) = \pi - \cos^{-1}x$ for $x\in[-1,1]$⁠. Here $\pi - \frac{\pi}{6} = \frac{5\pi}{6}$⁠. Lesson: principal values

4. The folding argument: reflect in the $x$-axis

   Instead of computing the cosine, we can fold the angle directly. For every $\theta$⁠, $\cos(2\pi - \theta) = \cos\theta$⁠, and

   $$2\pi - \frac{7\pi}{6} = \frac{12\pi - 7\pi}{6} = \frac{5\pi}{6} \in [0, \pi].$$

   So $\frac{5\pi}{6}$ has the same cosine as $\frac{7\pi}{6}$ and lies in the principal range. In general,

   $$\cos^{-1}(\cos x) = 2\pi - x \quad\text{for } \pi \le x \le 2\pi.$$
   Every value of $\cos$ on the lower half of the circle is repeated by its mirror image on the upper half. $\cos^{-1}$ can only return the upper-half copy, so for an angle in $[\pi, 2\pi]$ we reflect it: $x \mapsto 2\pi - x$⁠. This is the cosine version of the folding rule; for $\tan$ the fold is a shift by $\pi$ instead, because $\tan$ has period $\pi$⁠.
   $\cos(2\pi - \theta) = \cos(-\theta) = \cos\theta$⁠: $\cos$ has period $2\pi$ and is even. Basics: the unit circle

5. Examine every option

   Option	In $[0,\pi]$⁠?	Its cosine	Verdict
   (A) $\frac{7\pi}{6}$	No, $\frac{7\pi}{6} > \pi$	$-\frac{\sqrt3}{2}$	Right cosine, but outside the range: cannot be a value of $\cos^{-1}$
   (B) $\frac{5\pi}{6}$	Yes	$-\frac{\sqrt3}{2}$	Correct
   (C) $\frac{\pi}{3}$	Yes	$\frac12$	Wrong cosine
   (D) $\frac{\pi}{6}$	Yes	$\frac{\sqrt3}{2}$	Right size, wrong sign: this is the reference angle

   Only (B) $\frac{5\pi}{6}$ passes both tests.

   An answer must pass two tests: lie in $[0,\pi]$ and have cosine $-\frac{\sqrt3}{2}$⁠. (A) is the "cancel" trap and fails the first test. (D) is the "forget the sign" trap and fails the second. (C) has neither the right size nor sign.

Every fact used, and where it comes from