Exercise 2.2 · Q14 (MCQ): $\sin\left(\frac{\pi}{3}-\sin^{-1}\left(-\frac12\right)\right)$

Basics you need

Solution

1. Find $\sin^{-1}\left(-\frac12\right)$

   We need the angle $y$ with $-\frac{\pi}{2} \le y \le \frac{\pi}{2}$ and $\sin y = -\frac12$⁠. Since $\sin\frac{\pi}{6} = \frac12$ and $\sin$ is odd,

   $$\sin\left(-\frac{\pi}{6}\right) = -\sin\frac{\pi}{6} = -\frac12,$$

   and $-\frac{\pi}{6}$ is in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$⁠. As $\sin$ is strictly increasing on that interval, it is the only such angle:

   $$\sin^{-1}\left(-\frac12\right) = -\frac{\pi}{6}.$$
   Other angles also have sine $-\frac12$⁠, such as $\frac{7\pi}{6}$ and $\frac{11\pi}{6}$⁠, but they are outside the principal range, so $\sin^{-1}$ never returns them. When no branch is mentioned, the principal branch is meant (Note 2 after the table in Section 2.2).
   $\sin^{-1}(-x) = -\sin^{-1}x$ for $x\in[-1,1]$⁠, because $\sin$ is odd and the range $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ is symmetric about 0. Lesson: principal values

2. Simplify the angle

   Substitute into the bracket. Subtracting a negative number is adding:

   $$\begin{aligned} \frac{\pi}{3} - \left(-\frac{\pi}{6}\right) &= \frac{\pi}{3} + \frac{\pi}{6} \\ &= \frac{2\pi + \pi}{6} = \frac{\pi}{2}. \end{aligned}$$
   This is where the question tests care: there is a minus inside $\sin^{-1}$ and another in front of it. Writing the bracket $\left(-\frac{\pi}{6}\right)$ explicitly before simplifying keeps both signs visible.
   $a - (-b) = a + b$⁠. Common denominator 6: $\frac{\pi}{3} = \frac{2\pi}{6}$⁠. Basics: angles and radians

3. Take the sine

   $$\begin{aligned} &\sin\left(\frac{\pi}{3} - \sin^{-1}\left(-\frac12\right)\right) \\ &= \sin\frac{\pi}{2} = 1. \end{aligned}$$
   $\frac{\pi}{2}$ is 90°, the top of the unit circle, the point $(0, 1)$⁠. Its $y$-coordinate, the sine, is 1. Note that the outer function is $\sin$⁠, not $\sin^{-1}$⁠, so no range restriction applies to the angle $\frac{\pi}{2}$ here: $\sin$ accepts every real angle.
   $\sin\frac{\pi}{2} = 1$⁠: the maximum value of $\sin$⁠. Basics: the unit circle

4. A second check with the subtraction formula

   With $A = \frac{\pi}{3}$ and $B = -\frac{\pi}{6}$⁠:

   $$\begin{aligned} \sin(A - B) &= \sin A\cos B - \cos A\sin B \\ &= \frac{\sqrt3}{2}\cdot\frac{\sqrt3}{2} - \frac12\cdot\left(-\frac12\right) \\ &= \frac34 + \frac14 = 1. \end{aligned}$$

   Here $\cos B = \cos\left(-\frac{\pi}{6}\right) = \frac{\sqrt3}{2}$ because $\cos$ is even. The same answer, 1.

   The formula route does not need us to notice that the angle is $\frac{\pi}{2}$⁠, so it is an independent check of Steps 2 and 3. Agreement of two different methods is strong evidence that no sign was lost.
   $\cos(-\theta) = \cos\theta$ and $\sin(-\theta) = -\sin\theta$⁠.

5. Examine every option

   Option	Verdict
   (A) $\frac12$	Wrong. This is what you get by dropping the minus sign inside $\sin^{-1}$⁠: $\sin\left(\frac{\pi}{3} - \frac{\pi}{6}\right) = \sin\frac{\pi}{6} = \frac12$⁠.
   (B) $\frac13$	Wrong. No special angle in this problem has sine $\frac13$⁠.
   (C) $\frac14$	Wrong. $\frac14$ is only one of the two terms in Step 4, $\frac12\cdot\frac12$⁠.
   (D) $1$	Correct, by Steps 3 and 4.

   So the answer is (D).

   (A) is the trap the question is designed around, so it is worth seeing exactly which slip produces it. (B) and (C) do not come from any consistent method; the computation gives 1 by two routes.

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