Understanding the Principal Value of Inverse Sine

The question asks for the principal value of \(\sin^{-1}\left(\frac{1}{\sqrt{2}}\right)\). Inverse trigonometric functions, also known as arcsin, arccos, arctan, etc., help us find the angle when the trigonometric ratio is known. For \(\sin^{-1}(x)\), it gives us the angle whose sine is \(x\).

However, sine is a periodic function, meaning multiple angles can have the same sine value. To make the inverse function unique, we define a specific range for its output. This unique output within the defined range is called the principal value.

Range of the Principal Value of Inverse Sine

For the function \(\sin^{-1}(x)\), the principal value is defined to be in the range \([-\frac{\pi}{2}, \frac{\pi}{2}]\). This range corresponds to the right half of the unit circle, specifically Quadrants I and IV.

• If \(x > 0\), the principal value \(\sin^{-1}(x)\) is in Quadrant I (\(0 < \sin^{-1}(x) \le \frac{\pi}{2}\)).
• If \(x < 0\), the principal value \(\sin^{-1}(x)\) is in Quadrant IV (\(-\frac{\pi}{2} \le \sin^{-1}(x) < 0\)).
• If \(x = 0\), the principal value is \(0\).

In our case, we need to find the principal value of \(\sin^{-1}\left(\frac{1}{\sqrt{2}}\right)\). Since \(\frac{1}{\sqrt{2}}\) is positive, the principal value will be in Quadrant I, within the range \((0, \frac{\pi}{2}]\).

Finding the Principal Value

Let \(\theta = \sin^{-1}\left(\frac{1}{\sqrt{2}}\right)\). This means \(\sin(\theta) = \frac{1}{\sqrt{2}}\).

We need to find the angle \(\theta\) such that:

1. \(\sin(\theta) = \frac{1}{\sqrt{2}}\)
2. \(\theta\) is in the principal value range \([-\frac{\pi}{2}, \frac{\pi}{2}]\).

We know the standard trigonometric values for common angles. For example:

From the table, we see that \(\sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}\).

Now we check if this angle \(\frac{\pi}{4}\) is within the principal value range \([-\frac{\pi}{2}, \frac{\pi}{2}]\). The range in degrees is \([-90^\circ, 90^\circ]\). The angle \(\frac{\pi}{4}\) is \(45^\circ\), which is indeed within this range.

Therefore, the principal value of \(\sin^{-1}\left(\frac{1}{\sqrt{2}}\right)\) is \(\frac{\pi}{4}\).

Evaluating the Options

Let's look at the given options:

1. \(\frac{\pi}{4}\): We found that \(\sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}\) and \(\frac{\pi}{4}\) is in the range \([-\frac{\pi}{2}, \frac{\pi}{2}]\). This is the principal value.
2. \(\frac{3\pi}{4}\): We know that \(\sin\left(\frac{3\pi}{4}\right) = \sin\left(\pi - \frac{\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}\). However, \(\frac{3\pi}{4}\) (\(135^\circ\)) is not in the principal value range \([-\frac{\pi}{2}, \frac{\pi}{2}]\). So, this is not the principal value.
3. \(\frac{5\pi}{4}\): We know that \(\sin\left(\frac{5\pi}{4}\right) = \sin\left(\pi + \frac{\pi}{4}\right) = -\sin\left(\frac{\pi}{4}\right) = -\frac{1}{\sqrt{2}}\). The sine value is negative, not \(\frac{1}{\sqrt{2}}\). Also, \(\frac{5\pi}{4}\) (\(225^\circ\)) is not in the principal value range.
4. None of these: Since option 1 is the correct principal value, this option is incorrect.

The unique value in the range \([-\frac{\pi}{2}, \frac{\pi}{2}]\) for which \(\sin(\theta) = \frac{1}{\sqrt{2}}\) is \(\frac{\pi}{4}\). This confirms it as the principal value of inverse sine.

Thus, the principal value of \(\sin^{-1}\left(\frac{1}{\sqrt{2}}\right)\) is \(\frac{\pi}{4}\).