Question

Evaluate: $\sin^{-1}[\sin(-600^\circ)] + \cot^{-1}(-\sqrt{3})$

• A. $\frac{\pi}{6}$
• B. $\frac{\pi}{4}$
• C. $\frac{\pi}{3}$
• D. $\frac{7\pi}{6}$

Hint:

Always respect the principal value branches! For inverse sine, the output must fall between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$. For inverse cotangent, the output must be strictly between $0$ and $\pi$.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We need to find the principal values of two inverse trigonometric expressions and add them together.

Step 2: Detailed Explanation:
First, let's simplify the first term: $\sin^{-1}[\sin(-600^\circ)]$.
Since sine has a period of $360^\circ$, we can add multiples of $360^\circ$ to find an equivalent angle within the principal range:
$$-600^\circ = -600^\circ + 2(360^\circ) = -600^\circ + 720^\circ = 120^\circ$$ So, $\sin(-600^\circ) = \sin(120^\circ)$.
We know that $\sin(120^\circ) = \sin(180^\circ - 60^\circ) = \sin(60^\circ) = \frac{\sqrt{3}}{2}$.
The principal value branch for $\sin^{-1}(x)$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.
$$\sin^{-1}\left(\frac{\sqrt{3}}{2}\right) = 60^\circ = \frac{\pi}{3}$$ Now, let's evaluate the second term: $\cot^{-1}(-\sqrt{3})$.
The principal value branch for $\cot^{-1}(x)$ is $(0, \pi)$.
We know $\cot\left(\frac{\pi}{6}\right) = \sqrt{3}$.
Since the argument is negative, the angle must be in the second quadrant:
$$\cot^{-1}(-\sqrt{3}) = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$$ Finally, add the two results together:
$$\text{Sum} = \frac{\pi}{3} + \frac{5\pi}{6} = \frac{2\pi}{6} + \frac{5\pi}{6} = \frac{7\pi}{6}$$

Step 3: Final Answer:
The total sum evaluates to $\frac{7\pi}{6}$, matching option (D).