Find the principal of :
(1) `sin ^(-1)((1)/(2))` (2) `tan^(-1)(-sqrt(3))` (3) `cos^(-1)(-(1)/(sqrt(2)))` (4) `sin^(-1)(-(1)/(2))`.

Answer

To find the principal values of the given inverse trigonometric functions, we will follow the standard ranges for each function. Let's solve each part step by step.

### Part 1: Find `sin^(-1)(1/2)`

1. **Set the equation**: Let \( y = \sin^{-1}\left(\frac{1}{2}\right) \).
2. **Apply the sine function**: This implies \( \sin(y) = \frac{1}{2} \).
3. **Find the angle**: The angle \( y \) for which \( \sin(y) = \frac{1}{2} \) in the range of \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \) is \( y = \frac{\pi}{6} \).
4. **Conclusion**: Therefore, \( \sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6} \).

### Part 2: Find `tan^(-1)(-sqrt(3))`

1. **Set the equation**: Let \( y = \tan^{-1}(-\sqrt{3}) \).
2. **Apply the tangent function**: This implies \( \tan(y) = -\sqrt{3} \).
3. **Find the angle**: The angle \( y \) for which \( \tan(y) = -\sqrt{3} \) in the range of \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \) is \( y = -\frac{\pi}{3} \).
4. **Conclusion**: Therefore, \( \tan^{-1}(-\sqrt{3}) = -\frac{\pi}{3} \).

### Part 3: Find `cos^(-1)(-1/sqrt(2))`

1. **Set the equation**: Let \( y = \cos^{-1}\left(-\frac{1}{\sqrt{2}}\right) \).
2. **Apply the cosine function**: This implies \( \cos(y) = -\frac{1}{\sqrt{2}} \).
3. **Find the angle**: The angles \( y \) for which \( \cos(y) = -\frac{1}{\sqrt{2}} \) in the range of \( [0, \pi] \) are \( y = \frac{3\pi}{4} \).
4. **Conclusion**: Therefore, \( \cos^{-1}\left(-\frac{1}{\sqrt{2}}\right) = \frac{3\pi}{4} \).

### Part 4: Find `sin^(-1)(-1/2)`

1. **Set the equation**: Let \( y = \sin^{-1}\left(-\frac{1}{2}\right) \).
2. **Apply the sine function**: This implies \( \sin(y) = -\frac{1}{2} \).
3. **Find the angle**: The angle \( y \) for which \( \sin(y) = -\frac{1}{2} \) in the range of \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \) is \( y = -\frac{\pi}{6} \).
4. **Conclusion**: Therefore, \( \sin^{-1}\left(-\frac{1}{2}\right) = -\frac{\pi}{6} \).

### Summary of Principal Values:
1. \( \sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6} \)
2. \( \tan^{-1}(-\sqrt{3}) = -\frac{\pi}{3} \)
3. \( \cos^{-1}\left(-\frac{1}{\sqrt{2}}\right) = \frac{3\pi}{4} \)
4. \( \sin^{-1}\left(-\frac{1}{2}\right) = -\frac{\pi}{6} \)