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: Recall the principal value range of \( \sin^{-1}x \).
The principal value range of inverse sine is \[ \sin^{-1}x \in \left[-\frac{\pi}{2},\frac{\pi}{2}\right] \] This means whatever the input is, the output of \( \sin^{-1}(\sin\theta) \) must always lie in this interval.

Step 2: Note the given interval for \( \theta \).
We are given \[ \theta \in \left[\frac{\pi}{2},\frac{3\pi}{2}\right] \] This interval covers the second quadrant and the third quadrant.
Now \( \sin^{-1}(\sin\theta) \) will not always be equal to \( \theta \), because \( \theta \) may lie outside the principal value range of \( \sin^{-1} \).

Step 3: Use the standard identity for second quadrant angles.
For angles in the interval \[ \left[\frac{\pi}{2},\pi\right] \] we use the identity \[ \sin(\pi-\theta)=\sin\theta \] and since \[ \pi-\theta \in \left[0,\frac{\pi}{2}\right] \] which lies in the principal value range, we get \[ \sin^{-1}(\sin\theta)=\pi-\theta \]

Step 4: Check the option pattern.
Among the given options, the expression that correctly represents the principal value corresponding to the given interval is \[ \pi-\theta \] This is the standard result used for such questions.

Step 5: Verify with a sample value.
Take \[ \theta=\frac{2\pi}{3} \] Then \[ \sin\theta=\sin\frac{2\pi}{3}=\frac{\sqrt{3}}{2} \] Now, \[ \sin^{-1}\left(\frac{\sqrt{3}}{2}\right)=\frac{\pi}{3} \] Also, \[ \pi-\theta=\pi-\frac{2\pi}{3}=\frac{\pi}{3} \] So the result matches perfectly.

Step 6: Why \( \theta \) itself is not correct.
If we take \[ \theta=\frac{2\pi}{3} \] then \( \theta=\frac{2\pi}{3} \notin \left[-\frac{\pi}{2},\frac{\pi}{2}\right] \).
But the output of \( \sin^{-1} \) must lie in the principal range.
Hence \( \sin^{-1}(\sin\theta) \neq \theta \) in general on this interval.

Step 7: Final conclusion.
Therefore, \[ \boxed{\sin^{-1}(\sin\theta)=\pi-\theta} \] Hence, the correct option is \[ \boxed{(4)\ \pi-\theta} \]