Inverse Trig Functions Worksheet for Unit 5C, featuring exercises on domains, graphs, and values of inverse trigonometric functions.

Problem Analysis:

The worksheet focuses on inverse trigonometric functions, their domains, ranges, and evaluations. Let's solve the problems step by step.

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1) Give the restricted domain of the original function that we use to make the inverse into a function.

#### a) \( y = \tan x \)
- The tangent function is periodic with period \( \pi \).
- To make \( \tan^{-1} x \) (arctan) a function, we restrict the domain of \( \tan x \) to one period where it is one-to-one.
- The standard restricted domain is: \( \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) \).

#### b) \( y = \sin x \)
- The sine function is periodic with period \( 2\pi \).
- To make \( \sin^{-1} x \) (arcsin) a function, we restrict the domain of \( \sin x \) to one interval where it is one-to-one.
- The standard restricted domain is: \( \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \).

#### c) \( y = \cos x \)
- The cosine function is periodic with period \( 2\pi \).
- To make \( \cos^{-1} x \) (arccos) a function, we restrict the domain of \( \cos x \) to one interval where it is one-to-one.
- The standard restricted domain is: \( [0, \pi] \).

Answer:
\[
\boxed{
\text{a) } \left( -\frac{\pi}{2}, \frac{\pi}{2} \right), \quad
\text{b) } \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right], \quad
\text{c) } [0, \pi]
}
\]

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2) Graph each with labels.

#### a) \( y = \sin^{-1} x \)
- Domain: \( [-1, 1] \)
- Range: \( \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \)
- Key points: \( (-1, -\frac{\pi}{2}) \), \( (0, 0) \), \( (1, \frac{\pi}{2}) \)

#### b) \( y = \tan^{-1} x \)
- Domain: \( (-\infty, \infty) \)
- Range: \( \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) \)
- Key points: As \( x \to -\infty \), \( y \to -\frac{\pi}{2} \); as \( x \to \infty \), \( y \to \frac{\pi}{2} \); \( (0, 0) \)

#### c) \( y = \arccos x \)
- Domain: \( [-1, 1] \)
- Range: \( [0, \pi] \)
- Key points: \( (-1, \pi) \), \( (0, \frac{\pi}{2}) \), \( (1, 0) \)

Graphs:
- These graphs can be sketched using the above information, but since this is a text-based response, I'll describe them instead of drawing them.

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3) Give the domain and range for each in interval notation.

#### a) \( y = \cos^{-1} x \)
- Domain: \( [-1, 1] \)
- Range: \( [0, \pi] \)

#### b) \( y = \arcsin x \)
- Domain: \( [-1, 1] \)
- Range: \( \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \)

#### c) \( y = \arctan x \)
- Domain: \( (-\infty, \infty) \)
- Range: \( \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) \)

Answer:
\[
\boxed{
\text{a) Domain: } [-1, 1], \text{ Range: } [0, \pi], \quad
\text{b) Domain: } [-1, 1], \text{ Range: } \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right], \quad
\text{c) Domain: } (-\infty, \infty), \text{ Range: } \left( -\frac{\pi}{2}, \frac{\pi}{2} \right)
}
\]

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4) Evaluate the given expressions.

#### a) \( \arcsin \left( \frac{1}{2} \right) \)
- \( \arcsin \left( \frac{1}{2} \right) \) is the angle \( \theta \) such that \( \sin \theta = \frac{1}{2} \) and \( \theta \in \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \).
- \( \theta = \frac{\pi}{6} \).

#### b) \( \arcsin \left( -\frac{1}{2} \right) \)
- \( \arcsin \left( -\frac{1}{2} \right) \) is the angle \( \theta \) such that \( \sin \theta = -\frac{1}{2} \) and \( \theta \in \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \).
- \( \theta = -\frac{\pi}{6} \).

#### c) \( \arcsin (0) \)
- \( \arcsin (0) \) is the angle \( \theta \) such that \( \sin \theta = 0 \) and \( \theta \in \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \).
- \( \theta = 0 \).

#### d) \( \arcsin (-1) \)
- \( \arcsin (-1) \) is the angle \( \theta \) such that \( \sin \theta = -1 \) and \( \theta \in \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \).
- \( \theta = -\frac{\pi}{2} \).

#### e) \( \arccos \left( \frac{1}{2} \right) \)
- \( \arccos \left( \frac{1}{2} \right) \) is the angle \( \theta \) such that \( \cos \theta = \frac{1}{2} \) and \( \theta \in [0, \pi] \).
- \( \theta = \frac{\pi}{3} \).

#### f) \( \arccos \left( -\frac{1}{2} \right) \)
- \( \arccos \left( -\frac{1}{2} \right) \) is the angle \( \theta \) such that \( \cos \theta = -\frac{1}{2} \) and \( \theta \in [0, \pi] \).
- \( \theta = \frac{2\pi}{3} \).

#### g) \( \arccos (0) \)
- \( \arccos (0) \) is the angle \( \theta \) such that \( \cos \theta = 0 \) and \( \theta \in [0, \pi] \).
- \( \theta = \frac{\pi}{2} \).

#### h) \( \arccos (1) \)
- \( \arccos (1) \) is the angle \( \theta \) such that \( \cos \theta = 1 \) and \( \theta \in [0, \pi] \).
- \( \theta = 0 \).

#### i) \( \arctan (1) \)
- \( \arctan (1) \) is the angle \( \theta \) such that \( \tan \theta = 1 \) and \( \theta \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) \).
- \( \theta = \frac{\pi}{4} \).

#### j) \( \arctan (-1) \)
- \( \arctan (-1) \) is the angle \( \theta \) such that \( \tan \theta = -1 \) and \( \theta \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) \).
- \( \theta = -\frac{\pi}{4} \).

#### k) \( \arctan (0) \)
- \( \arctan (0) \) is the angle \( \theta \) such that \( \tan \theta = 0 \) and \( \theta \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) \).
- \( \theta = 0 \).

#### l) \( \arctan (\sqrt{3}) \)
- \( \arctan (\sqrt{3}) \) is the angle \( \theta \) such that \( \tan \theta = \sqrt{3} \) and \( \theta \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) \).
- \( \theta = \frac{\pi}{3} \).

#### m) \( \arctan \left( -\sqrt{3} \right) \)
- \( \arctan \left( -\sqrt{3} \right) \) is the angle \( \theta \) such that \( \tan \theta = -\sqrt{3} \) and \( \theta \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) \).
- \( \theta = -\frac{\pi}{3} \).

#### n) \( \arctan (\infty) \)
- \( \arctan (\infty) \) approaches \( \frac{\pi}{2} \) as \( x \to \infty \).
- \( \theta = \frac{\pi}{2} \).

#### o) \( \arctan (-\infty) \)
- \( \arctan (-\infty) \) approaches \( -\frac{\pi}{2} \) as \( x \to -\infty \).
- \( \theta = -\frac{\pi}{2} \).

Answer:
\[
\boxed{
\text{a) } \frac{\pi}{6}, \quad
\text{b) } -\frac{\pi}{6}, \quad
\text{c) } 0, \quad
\text{d) } -\frac{\pi}{2}, \quad
\text{e) } \frac{\pi}{3}, \quad
\text{f) } \frac{2\pi}{3}, \quad
\text{g) } \frac{\pi}{2}, \quad
\text{h) } 0, \quad
\text{i) } \frac{\pi}{4}, \quad
\text{j) } -\frac{\pi}{4}, \quad
\text{k) } 0, \quad
\text{l) } \frac{\pi}{3}, \quad
\text{m) } -\frac{\pi}{3}, \quad
\text{n) } \frac{\pi}{2}, \quad
\text{o) } -\frac{\pi}{2}
}
\]

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5) Evaluate the given composite expressions.

#### a) \( \sin \left( \cos^{-1} \left( \frac{1}{2} \right) \right) \)
- Let \( \theta = \cos^{-1} \left( \frac{1}{2} \right) \). Then \( \cos \theta = \frac{1}{2} \) and \( \theta \in [0, \pi] \).
- \( \theta = \frac{\pi}{3} \).
- \( \sin \left( \frac{\pi}{3} \right) = \frac{\sqrt{3}}{2} \).

#### b) \( \arccos \left( \sin \left( \frac{3\pi}{2} \right) \right) \)
- \( \sin \left( \frac{3\pi}{2} \right) = -1 \).
- \( \arccos (-1) = \pi \).

#### c) \( \tan^{-1} \left( \tan \left( \frac{3\pi}{4} \right) \right) \)
- \( \tan \left( \frac{3\pi}{4} \right) = -1 \).
- \( \tan^{-1} (-1) = -\frac{\pi}{4} \).

#### d) \( \arccos \left( \cos \left( \frac{5\pi}{6} \right) \right) \)
- \( \cos \left( \frac{5\pi}{6} \right) = -\frac{\sqrt{3}}{2} \).
- \( \arccos \left( -\frac{\sqrt{3}}{2} \right) = \frac{5\pi}{6} \).

#### e) \( \arcsin \left( \sin \left( \frac{5\pi}{6} \right) \right) \)
- \( \sin \left( \frac{5\pi}{6} \right) = \frac{1}{2} \).
- \( \arcsin \left( \frac{1}{2} \right) = \frac{\pi}{6} \).

#### f) \( \cos \left( \arctan \left( -\sqrt{3} \right) \right) \)
- Let \( \theta = \arctan \left( -\sqrt{3} \right) \). Then \( \tan \theta = -\sqrt{3} \) and \( \theta \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) \).
- \( \theta = -\frac{\pi}{3} \).
- \( \cos \left( -\frac{\pi}{3} \right) = \frac{1}{2} \).

#### g) \( \tan \left( \cos^{-1} \left( -\frac{\sqrt{2}}{2} \right) \right) \)
- Let \( \theta = \cos^{-1} \left( -\frac{\sqrt{2}}{2} \right) \). Then \( \cos \theta = -\frac{\sqrt{2}}{2} \) and \( \theta \in [0, \pi] \).
- \( \theta = \frac{3\pi}{4} \).
- \( \tan \left( \frac{3\pi}{4} \right) = -1 \).

#### h) \( \sin \left( \sin^{-1} \left( -\frac{1}{2} \right) \right) \)
- Let \( \theta = \sin^{-1} \left( -\frac{1}{2} \right) \). Then \( \sin \theta = -\frac{1}{2} \) and \( \theta \in \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \).
- \( \theta = -\frac{\pi}{6} \).
- \( \sin \left( -\frac{\pi}{6} \right) = -\frac{1}{2} \).

#### i) \( \sin \left( \sin^{-1} (5) \right) \)
- Since \( \sin^{-1} (x) \) is only defined for \( x \in [-1, 1] \), \( \sin^{-1} (5) \) is undefined.
- Answer: Undefined.

#### j) \( \cos \left( \cos^{-1} \left( \frac{\sqrt{3}}{2} \right) \right) \)
- Let \( \theta = \cos^{-1} \left( \frac{\sqrt{3}}{2} \right) \). Then \( \cos \theta = \frac{\sqrt{3}}{2} \) and \( \theta \in [0, \pi] \).
- \( \theta = \frac{\pi}{6} \).
- \( \cos \left( \frac{\pi}{6} \right) = \frac{\sqrt{3}}{2} \).

#### k) \( \cos^{-1} \left( \cos \left( \frac{2\pi}{3} \right) \right) \)
- \( \cos \left( \frac{2\pi}{3} \right) = -\frac{1}{2} \).
- \( \cos^{-1} \left( -\frac{1}{2} \right) = \frac{2\pi}{3} \).

#### l) \( \sin^{-1} \left( \sin \left( \frac{2\pi}{3} \right) \right) \)
- \( \sin \left( \frac{2\pi}{3} \right) = \frac{\sqrt{3}}{2} \).
- \( \sin^{-1} \left( \frac{\sqrt{3}}{2} \right) = \frac{\pi}{3} \).

#### m) \( \cos^{-1} \left( \cos \left( \frac{11\pi}{6} \right) \right) \)
- \( \cos \left( \frac{11\pi}{6} \right) = \cos \left( 2\pi - \frac{\pi}{6} \right) = \cos \left( -\frac{\pi}{6} \right) = \cos \left( \frac{\pi}{6} \right) = \frac{\sqrt{3}}{2} \).
- \( \cos^{-1} \left( \frac{\sqrt{3}}{2} \right) = \frac{\pi}{6} \).

#### n) \( \sin^{-1} \left( \sin \left( \frac{11\pi}{6} \right) \right) \)
- \( \sin \left( \frac{11\pi}{6} \right) = \sin \left( 2\pi - \frac{\pi}{6} \right) = -\sin \left( \frac{\pi}{6} \right) = -\frac{1}{2} \).
- \( \sin^{-1} \left( -\frac{1}{2} \right) = -\frac{\pi}{6} \).

Answer:
\[
\boxed{
\text{a) } \frac{\sqrt{3}}{2}, \quad
\text{b) } \pi, \quad
\text{c) } -\frac{\pi}{4}, \quad
\text{d) } \frac{5\pi}{6}, \quad
\text{e) } \frac{\pi}{6}, \quad
\text{f) } \frac{1}{2}, \quad
\text{g) } -1, \quad
\text{h) } -\frac{1}{2}, \quad
\text{i) } \text{Undefined}, \quad
\text{j) } \frac{\sqrt{3}}{2}, \quad
\text{k) } \frac{2\pi}{3}, \quad
\text{l) } \frac{\pi}{3}, \quad
\text{m) } \frac{\pi}{6}, \quad
\text{n) } -\frac{\pi}{6}
}
\]

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Final Answer:

\[
\boxed{
\text{1) } \left( -\frac{\pi}{2}, \frac{\pi}{2} \right), \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right], [0, \pi], \quad
\text{2) } \text{(Graphs described)}, \quad
\text{3) } [-1, 1], [0, \pi], [-1, 1], \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right], (-\infty, \infty), \left( -\frac{\pi}{2}, \frac{\pi}{2} \right), \quad
\text{4) } \frac{\pi}{6}, -\frac{\pi}{6}, 0, -\frac{\pi}{2}, \frac{\pi}{3}, \frac{2\pi}{3}, \frac{\pi}{2}, 0, \frac{\pi}{4}, -\frac{\pi}{4}, 0, \frac{\pi}{3}, -\frac{\pi}{3}, \frac{\pi}{2}, -\frac{\pi}{2}, \quad
\text{5) } \frac{\sqrt{3}}{2}, \pi, -\frac{\pi}{4}, \frac{5\pi}{6}, \frac{\pi}{6}, \frac{1}{2}, -1, -\frac{1}{2}, \text{Undefined}, \frac{\sqrt{3}}{2}, \frac{2\pi}{3}, \frac{\pi}{3}, \frac{\pi}{6}, -\frac{\pi}{6}
}
\]