Find the exact value of each expression, if it is defined.
a) sin^−1 1
b) cos^−1
(
cos 7π/
6
)
c) sin^−1
√2/
2
d) cos^−1 −
√2/
2
e) cos^−1 0
f) sin^−1
(
sin 7π/6)


Let's find the exact values for each of the given expressions. These involve inverse trigonometric functions, so we need to determine angles based on known values of sine and cosine.

### a) \( \sin^{-1} 1 \)

We are looking for the angle \( \theta \) such that \( \sin \theta = 1 \). The sine of \( \frac{\pi}{2} \) (90 degrees) is 1, and since the inverse sine function only takes values from \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \), the answer is:

\[
\sin^{-1}(1) = \frac{\pi}{2}
\]

### b) \( \cos^{-1} \left( \cos \frac{7\pi}{6} \right) \)

The cosine of \( \frac{7\pi}{6} \) is negative, as the angle is in the third quadrant. Specifically, \( \cos \left( \frac{7\pi}{6} \right) = -\frac{\sqrt{3}}{2} \). The inverse cosine will return the principal angle between \( 0 \) and \( \pi \). The reference angle for \( \frac{7\pi}{6} \) is \( \frac{\pi}{6} \), so:

\[
\cos^{-1} \left( \cos \frac{7\pi}{6} \right) = \frac{7\pi}{6}
\]

### c) \( \sin^{-1} \frac{\sqrt{2}}{2} \)

We need to find the angle \( \theta \) such that \( \sin \theta = \frac{\sqrt{2}}{2} \). This corresponds to an angle of \( \frac{\pi}{4} \) (or 45 degrees), and since \( \sin^{-1} \) returns values in \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \), we get:

\[
\sin^{-1} \frac{\sqrt{2}}{2} = \frac{\pi}{4}
\]

### d) \( \cos^{-1} \left( -\frac{\sqrt{2}}{2} \right) \)

We need to find the angle \( \theta \) such that \( \cos \theta = -\frac{\sqrt{2}}{2} \). This corresponds to an angle of \( \frac{3\pi}{4} \) (135 degrees), because \( \cos \) is negative in the second quadrant. Thus:

\[
\cos^{-1} \left( -\frac{\sqrt{2}}{2} \right) = \frac{3\pi}{4}
\]

### e) \( \cos^{-1} 0 \)

We are looking for the angle \( \theta \) such that \( \cos \theta = 0 \). This occurs at \( \frac{\pi}{2} \) (90 degrees), so:

\[
\cos^{-1}(0) = \frac{\pi}{2}
\]

### f) \( \sin^{-1} \left( \sin \frac{7\pi}{6} \right) \)

We know that \( \sin \left( \frac{7\pi}{6} \right) = -\frac{1}{2} \). The inverse sine function will return the angle within \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \) that gives \( -\frac{1}{2} \). The corresponding angle is \( -\frac{\pi}{6} \), so:

\[
\sin^{-1} \left( \sin \frac{7\pi}{6} \right) = -\frac{\pi}{6}
\]

### Final Answers:
a) \( \frac{\pi}{2} \)  
b) \( \frac{7\pi}{6} \)  
c) \( \frac{\pi}{4} \)  
d) \( \frac{3\pi}{4} \)  
e) \( \frac{\pi}{2} \)  
f) \( -\frac{\pi}{6} \)

Would you like further details on any of these calculations?

---

Here are 5 related questions you might want to explore:
1. What is \( \sin^{-1}(-1) \)?
2. How do you find \( \cos^{-1}(-1) \)?
3. Can you explain why the domain of \( \sin^{-1}(x) \) is restricted to \( [-1, 1] \)?
4. What is the range of \( \cos^{-1}(x) \)?
5. How do inverse trigonometric functions differ from regular trigonometric functions?

**Tip:** The range of inverse sine (arcsin) is always between \( -\frac{\pi}{2} \) and \( \frac{\pi}{2} \), so keep that in mind when calculating values!

Find the exact value of each expression, if it is defined.
a) sin^−1 1
b) cos^−1(cos 7π/6)
c) sin^−1 √2/2
d) cos^−1 −√2/2
e) cos^−1 0
f) sin^−1(sin 7π/6)

Solve problems involving inverse trigonometric functions like sin^−1(1) and cos^−1(cos 7π/6). Learn how to determine exact angles using inverse sine and cosine functions, suitable for Grades 10-12 students.

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