Question

The value of $\sin^{-1}\left(-\frac{1}{2}\right) + \sin^{-1}\left(-\frac{\sqrt{3}}{2}\right)$ is

• A. $\frac{\pi}{3}$
• B. $-\frac{\pi}{6}$
• C. $-\frac{\pi}{3}$
• D. $-\frac{\pi}{2}$

Hint:

Think of these calculations in terms of degrees to make quick mental addition simpler: $-30^\circ + (-60^\circ) = -90^\circ$. Converting $-90^\circ$ back to radians gives $-\frac{\pi}{2}$ instantly without working through fraction steps!

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The problem requires computing the sum of the principal values of two inverse sine functions with negative arguments.

Step 2: Key Formula or Approach:
Recall that the principal value branch of $\sin^{-1} x$ is strictly restricted to the interval $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. For negative arguments, inverse sine behaves as an odd function: $$\sin^{-1}(-x) = -\sin^{-1}x$$ We will use standard reference angles: $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$ and $\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$.

Step 3: Detailed Explanation:
Apply the negative extraction identity to both terms: $$\sin^{-1}\left(-\frac{1}{2}\right) = -\sin^{-1}\left(\frac{1}{2}\right) = -\frac{\pi}{6}$$ $$\sin^{-1}\left(-\frac{\sqrt{3}}{2}\right) = -\sin^{-1}\left(\frac{\sqrt{3}}{2}\right) = -\frac{\pi}{3}$$ Now add the two principal angles together: $$\text{Sum} = -\frac{\pi}{6} + \left(-\frac{\pi}{3}\right) = -\frac{\pi}{6} - \frac{\pi}{3}$$ Find a common denominator of 6 to combine the fractions: $$\text{Sum} = \frac{-\pi - 2\pi}{6} = \frac{-3\pi}{6} = -\frac{\pi}{2}$$ This matches option (D).

Step 4: Final Answer:
The value of the sum is $-\frac{\pi}{2}$, which corresponds to option (D).