Question

If \( \theta \in \left[\dfrac{\pi}{2}, \dfrac{3\pi}{2}\right] \), then \( \sin^{-1}(\sin\theta) \) is equal to

• A. \( \theta \)
• B. \( \theta-\pi \)
• C. \( \theta-\dfrac{\pi}{2} \)
• D. \( \pi-\theta \)
• E. \( \theta+\dfrac{\pi}{2} \)

Hint:

For inverse trigonometric functions, always remember the principal value range first. Then rewrite the angle into an equivalent angle lying in that principal interval.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The function \(f(x) = \sin^{-1}(\sin x)\) is not equal to \(x\) for all \(x\). The range of the principal value of the inverse sine function, \(\sin^{-1}(y)\), is \([-\frac{\pi}{2}, \frac{\pi}{2}]\). Therefore, the expression \(\sin^{-1}(\sin\theta)\) must yield a value within this range. We need to find an angle \(\alpha\) in the range \([-\frac{\pi}{2}, \frac{\pi}{2}]\) such that \(\sin(\alpha) = \sin(\theta)\).
Step 2: Key Formula or Approach:
The identity \(\sin(\pi - x) = \sin(x)\) is key. We need to check if for a given \(\theta\) in the interval \([\frac{\pi}{2}, \frac{3\pi}{2}]\), the value \(\pi - \theta\) falls within the principal range \([-\frac{\pi}{2}, \frac{\pi}{2}]\).
Step 3: Detailed Explanation:
The given interval for \(\theta\) is \([\frac{\pi}{2}, \frac{3\pi}{2}]\). This covers the second and third quadrants.
Let \(\alpha = \sin^{-1}(\sin\theta)\). We know that \(\sin(\alpha) = \sin(\theta)\) and \(-\frac{\pi}{2} \le \alpha \le \frac{\pi}{2}\).
We use the identity \(\sin(\theta) = \sin(\pi - \theta)\).
Let's check the range of \(\pi - \theta\) for the given interval of \(\theta\).
Given: \(\frac{\pi}{2} \le \theta \le \frac{3\pi}{2}\)
Multiply by -1 (reversing the inequalities):
\(-\frac{3\pi}{2} \le -\theta \le -\frac{\pi}{2}\)
Add \(\pi\) to all parts:
\(\pi - \frac{3\pi}{2} \le \pi - \theta \le \pi - \frac{\pi}{2}\)
\(-\frac{\pi}{2} \le \pi - \theta \le \frac{\pi}{2}\)
This shows that the angle \(\pi - \theta\) lies exactly in the principal value range for \(\sin^{-1}\).
Since \(\sin(\pi - \theta) = \sin(\theta)\) and \(\pi - \theta\) is in the required range, we have:
\[ \sin^{-1}(\sin\theta) = \pi - \theta \] Step 4: Final Answer:
For the given interval of \(\theta\), \(\sin^{-1}(\sin\theta) = \pi - \theta\). Therefore, option (D) is correct.