Question

The principal value of sin^-1(sin 3ℼ/4) is?

Answer 1

Let's find the principal value of \( \sin^{-1}(\sin(3\pi/4)) \).

1. Understand the Principal Value:
The principal value of \( \sin^{-1}(x) \), also known as the inverse sine function, is always taken to be in the interval \([- \frac{\pi}{2}, \frac{\pi}{2}]\). This means we need to find an angle \( \theta \) in this interval such that \( \sin(\theta) = \sin(3\pi/4) \).

2. Evaluate \( \sin(3\pi/4) \):
To begin, we need to evaluate \( \sin(3\pi/4) \). First, observe that:

• \( 3\pi/4 \) is in the second quadrant of the unit circle.
• In the second quadrant, the sine function is positive.
• Using the identity \( \sin(\pi - \theta) = \sin(\theta) \), we have: \[ \sin(3\pi/4) = \sin(\pi - \pi/4) = \sin(\pi/4) \]
• \( \sin(\pi/4) = \frac{\sqrt{2}}{2} \). Hence: \[ \sin(3\pi/4) = \frac{\sqrt{2}}{2} \]

3. Find \( \sin^{-1}(\frac{\sqrt{2}}{2}) \):
Now, we need to find the angle \( \theta \) in the interval \([- \frac{\pi}{2}, \frac{\pi}{2}]\) such that \( \sin(\theta) = \frac{\sqrt{2}}{2} \). From trigonometric identities, we know that:

• \( \sin(\pi/4) = \frac{\sqrt{2}}{2} \).
• \( \pi/4 \) lies within the principal value range \([- \frac{\pi}{2}, \frac{\pi}{2}]\).

Therefore, we conclude that: \[ \sin^{-1} \left( \frac{\sqrt{2}}{2} \right) = \pi/4 \]

Final Conclusion:
Hence, the principal value of \( \sin^{-1}(\sin(3\pi/4)) \) is \( \pi/4 \).