Inverse Trigonometric Functions

🔄 Inverse Trigonometric Functions — Principal Values & Restricted Domains

Part 1 of 7

Trig functions are not one-to-one, so to define inverses we must restrict the domain to an interval where the function passes the horizontal line test.

The Three Main Inverse Functions

Key Idea

$\boxed{\text{Inverse trig functions output ANGLES, not ratios}}$

$\arcsin\!\left(\frac{1}{2}\right) = \frac{\pi}{6}$ means "the angle (in the principal range) whose sine is $\frac{1}{2}$".

Notation warning: $\sin^{-1}x$ means $\arcsin x$, NOT $\frac{1}{\sin x}$ (that's $\csc x$).

📝 Worked Examples

Example 1: Evaluate $\arcsin\!\left(\frac{\sqrt{3}}{2}\right)$

Ask: "What angle $\theta \in [-\frac{\pi}{2}, \frac{\pi}{2}]$ has $\sin\theta = \frac{\sqrt{3}}{2}$?"

$\theta = \frac{\pi}{3}$

Example 2: Evaluate $\arccos(-1)$

Ask: "What angle $\theta \in [0, \pi]$ has $\cos\theta = -1$?"

$\theta = \pi$

Example 3: Evaluate $\arctan(-1)$

Ask: "What angle $\theta \in (-\frac{\pi}{2}, \frac{\pi}{2})$ has $\tan\theta = -1$?"

$\theta = -\frac{\pi}{4}$

Example 4: Why $\arcsin\!\left(\sin \frac{4\pi}{3}\right) \neq \frac{4\pi}{3}$

$\sin \frac{4\pi}{3} = -\frac{\sqrt{3}}{2}$. The principal value of $\arcsin(-\frac{\sqrt{3}}{2})$ is $-\frac{\pi}{3}$, not $\frac{4\pi}{3}$, because $\arcsin$ outputs must be in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.

🔍 Why We Restrict the Domain

Without Restriction: Infinitely Many Answers

$\sin\theta = \frac{1}{2}$ has solutions $\theta = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{13\pi}{6}, \frac{17\pi}{6}, \ldots$ and also $-\frac{7\pi}{6}, -\frac{11\pi}{6}, \ldots$

A function can only return one output. So we pick the interval where each trig function is one-to-one:

Quick Reference: Special Angle Outputs

📏 Radians only. The AP exam measures every angle in radians, so inverse trig answers are written as $\frac{\pi}{6}$, not $30°$.

Concept Check 🎯

Inverse Trig Evaluation 🧮

Write each answer as a multiple of $\pi$ and enter the coefficient (type 1/6 for $\frac{\pi}{6}$).

1) $\arcsin\!\left(\frac{\sqrt{2}}{2}\right) = \,?\cdot\pi$ (e.g., $\arcsin(\frac{\sqrt{3}}{2}) = \frac{\pi}{3}$, coefficient 1/3)

2) $\arccos(0) = \,?\cdot\pi$ (e.g., $\arccos(-1) = \pi$, coefficient 1)

3) $\arctan(\sqrt{3}) = \,?\cdot\pi$ (e.g., $\arctan(1) = \frac{\pi}{4}$, coefficient 1/4)

Domain & Range Matching 🔽

Exit Quiz ✅

📈 Graphs of Inverse Trig Functions

Part 2 of 7

Each inverse trig graph is the reflection of the restricted trig graph across the line $y = x$.

Arcsin Graph: $y = \arcsin x$

Arccos Graph: $y = \arccos x$

Arctan Graph: $y = \arctan x$

🔑 Key Graphing Relationships

Reflection Property

If $(a, b)$ is on $y = \sin x$ (restricted), then $(b, a)$ is on $y = \arcsin x$.

For example: $(\frac{\pi}{6}, \frac{1}{2})$ on $\sin$ → $(\frac{1}{2}, \frac{\pi}{6})$ on $\arcsin$

Complementary Identity

$\boxed{\arcsin x + \arccos x = \frac{\pi}{2} \quad \text{for all } x \in [-1,1]}$

This means the arcsin and arccos graphs are "complementary" — at any $x$-value, their outputs sum to $\frac{\pi}{2}$.

Symmetry

📝 Worked Examples

Example 1: Key Points on the Arcsin Graph

Plot: $(-1, -\frac{\pi}{2}),\; (-\frac{\sqrt{3}}{2}, -\frac{\pi}{3}),\; (-\frac{1}{2}, -\frac{\pi}{6}),\; (0, 0),\; (\frac{1}{2}, \frac{\pi}{6}),\; (\frac{\sqrt{3}}{2}, \frac{\pi}{3}),\; (1, \frac{\pi}{2})$

Connect with a smooth increasing curve from $(-1, -\frac{\pi}{2})$ to $(1, \frac{\pi}{2})$.

Example 2: Using the Complementary Identity

Find $\arccos(\frac{1}{2})$ given that $\arcsin(\frac{1}{2}) = \frac{\pi}{6}$:

$\arccos\!\left(\frac{1}{2}\right) = \frac{\pi}{2} - \arcsin\!\left(\frac{1}{2}\right) = \frac{\pi}{2} - \frac{\pi}{6} = \frac{\pi}{3}$

Example 3: Negative Input Symmetry

$\arctan(-\sqrt{3}) = -\arctan(\sqrt{3}) = -\frac{\pi}{3}$

We used the odd-function property: $\arctan(-x) = -\arctan x$.

Graph Features 🎯

Using Graphing Properties 🧮

Write each answer as a multiple of $\pi$ and enter the coefficient (type -1/4 for $-\frac{\pi}{4}$).

1) If $\arcsin(\frac{\sqrt{3}}{2}) = \frac{\pi}{3}$, then $\arccos(\frac{\sqrt{3}}{2}) = \,?\cdot\pi$ (e.g., $\arcsin(\frac{1}{2}) = \frac{\pi}{6}$ gives $\arccos(\frac{1}{2}) = \frac{\pi}{2} - \frac{\pi}{6} = \frac{\pi}{3}$)

2) $\arctan(-1) = \,?\cdot\pi$ (e.g., $\arctan(-\sqrt{3}) = -\frac{\pi}{3}$ by the odd-function property)

3) $\arcsin(-\frac{1}{2}) = \,?\cdot\pi$ (e.g., $\arcsin(-\frac{\sqrt{2}}{2}) = -\frac{\pi}{4}$ by the odd-function property)

Graph Identification 🔽

Exit Quiz ✅

🎯 Evaluating Inverse Trig — Exact Values

Part 3 of 7

Evaluating inverse trig functions means finding exact angle values from the unit circle. The key is memorizing outputs for special inputs.

Complete Special-Value Table

Arctan Special Values

🧠 Evaluation Strategy

Step-by-Step Process

$\boxed{\text{1. Identify the function → 2. Recall its range → 3. Find the angle in that range}}$

Example 1: $\arcsin\!\left(-\frac{\sqrt{2}}{2}\right)$

1. Function: $\arcsin$ → range is $[-\frac{\pi}{2}, \frac{\pi}{2}]$
2. Need $\theta$ with $\sin\theta = -\frac{\sqrt{2}}{2}$ and $\theta \in [-\frac{\pi}{2}, \frac{\pi}{2}]$
3. $\theta = -\frac{\pi}{4}$ ✓

Example 2: $\arccos\!\left(-\frac{1}{2}\right)$

1. Function: $\arccos$ → range is $[0, \pi]$
2. Need $\theta$ with $\cos\theta = -\frac{1}{2}$ and $\theta \in [0, \pi]$
3. $\theta = \frac{2\pi}{3}$ ✓

Example 3: Undefined Inputs

$\arcsin(2)$ is undefined — there's no angle whose sine equals $2$ (sine only outputs $[-1, 1]$).

$\arccos(-3)$ is also undefined for the same reason.

$\arctan(100)$ IS defined — tangent can take any real value, so arctan accepts all reals.

📝 More Practice

Example 4: Answers Are in Radians

$\arcsin(\frac{1}{2}) = \frac{\pi}{6}$. If you first think "$30°$", convert with $30 \cdot \frac{\pi}{180} = \frac{\pi}{6}$ before you answer: the AP exam (and a calculator in radian mode) gives every angle in radians.

Example 5: Tricky Negative Values

$\arccos(-\frac{\sqrt{3}}{2})$:

• We know $\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}$
• For the negative input, the angle must be in Quadrant II: $\pi - \frac{\pi}{6} = \frac{5\pi}{6}$
• $\arccos(-\frac{\sqrt{3}}{2}) = \frac{5\pi}{6}$

Pattern for Negative Inputs

Exact Value Quiz 🎯

Compute Exact Values 🧮

Write each answer as a multiple of $\pi$ and enter the coefficient (type 2/3 for $\frac{2\pi}{3}$).

1) $\arcsin(-1) = \,?\cdot\pi$ (e.g., $\arcsin(1) = \frac{\pi}{2}$ since $\sin \frac{\pi}{2} = 1$)

2) $\arccos(-\frac{1}{2}) = \,?\cdot\pi$ (e.g., $\arccos(\frac{1}{2}) = \frac{\pi}{3}$ since $\cos \frac{\pi}{3} = \frac{1}{2}$)

3) $\arctan(-\sqrt{3}) = \,?\cdot\pi$ (e.g., $\arctan(\sqrt{3}) = \frac{\pi}{3}$ since $\tan \frac{\pi}{3} = \sqrt{3}$)

Quick Evaluation 🔽

Exit Quiz ✅

🔗 Compositions of Trig & Inverse Trig

Part 4 of 7

One of the most important skills is simplifying compositions like $\sin(\arccos x)$ or $\cos(\arctan x)$.

Two Types of Compositions

Type 1: Trig(InverseTrig) — e.g., $\sin(\arccos \frac{3}{5})$

Strategy: Draw a right triangle from the inverse trig value.

Type 2: InverseTrig(Trig) — e.g., $\arcsin(\sin \frac{5\pi}{6})$

Strategy: Check if the angle is in the principal range. If not, find the equivalent angle.

Type 1 — The Right Triangle Method

$\boxed{\text{Let } \theta = \arccos x \text{, draw triangle, find desired ratio}}$

If $\theta = \arccos\!\left(\frac{3}{5}\right)$, then $\cos\theta = \frac{3}{5}$.

Draw a right triangle: adjacent = $3$, hypotenuse = $5$, so opposite = $\sqrt{25 - 9} = 4$.

$\sin(\arccos \tfrac{3}{5}) = \sin\theta = \frac{4}{5}$

📝 Worked Examples

Example 1: $\tan(\arcsin \frac{5}{13})$

Let $\theta = \arcsin \frac{5}{13}$, so $\sin\theta = \frac{5}{13}$.

Right triangle: opposite = $5$, hypotenuse = $13$, adjacent = $\sqrt{169 - 25} = 12$

$\tan(\arcsin \tfrac{5}{13}) = \frac{5}{12}$

Example 2: $\cos(\arctan 2)$

Let $\theta = \arctan 2$, so $\tan\theta = \frac{2}{1}$.

Right triangle: opposite = $2$, adjacent = $1$, hypotenuse = $\sqrt{4 + 1} = \sqrt{5}$

$\cos(\arctan 2) = \frac{1}{\sqrt{5}} = \frac{\sqrt{5}}{5}$

Example 3: General Formula $\sin(\arccos x)$

Let $\theta = \arccos x$. Then $\cos\theta = x = \frac{x}{1}$.

Adjacent = $x$, hypotenuse = $1$, opposite = $\sqrt{1 - x^2}$

$\boxed{\sin(\arccos x) = \sqrt{1 - x^2}}$

🔄 Type 2: InverseTrig(Trig)

The Cancellation Rules

These only work when the angle is in the principal range:

Example 4: $\arcsin(\sin \frac{7\pi}{6})$

$\frac{7\pi}{6}$ is NOT in $[-\frac{\pi}{2}, \frac{\pi}{2}]$, so we can't just cancel.

$\sin \frac{7\pi}{6} = -\frac{1}{2}$

$\arcsin(-\frac{1}{2}) = -\frac{\pi}{6}$

Example 5: $\arccos(\cos(-\frac{\pi}{3}))$

$-\frac{\pi}{3}$ is NOT in $[0, \pi]$, so we can't cancel.

$\cos(-\frac{\pi}{3}) = \frac{1}{2}$

$\arccos(\frac{1}{2}) = \frac{\pi}{3}$

Composition Quiz 🎯

Evaluate Compositions 🧮

Write answers as simplified fractions or integers.

1) $\tan(\arccos \frac{3}{5})$ = ? (e.g., $\sin(\arccos \frac{4}{5}) = \frac{3}{5}$ using a 3-4-5 triangle)

2) $\arcsin(\sin \frac{7\pi}{6}) = \,?\cdot\pi$. Enter the coefficient of $\pi$ (e.g., $\arcsin(\sin \frac{5\pi}{6}) = \frac{\pi}{6}$, coefficient 1/6, since $\sin \frac{5\pi}{6} = \frac{1}{2}$)

3) $\cos(\arctan \frac{3}{4})$ = ? Write as a decimal like 0.8 (e.g., $\cos(\arctan 1) = \frac{\sqrt{2}}{2} \approx 0.707$)

True or False 🔽

Exit Quiz ✅

📐 Inverse Trig with Right Triangles

Part 5 of 7

Inverse trig functions let us find angles in right triangles when we know the sides. As everywhere in AP Precalculus, the angles come out in radians (calculator in radian mode). A right angle is $\frac{\pi}{2} \approx 1.571$, so the two acute angles of a right triangle add to $\frac{\pi}{2}$.

The Setup

Given a right triangle with known side lengths, find an angle $\theta$:

$\boxed{\theta = \arctan\!\left(\frac{\text{opposite}}{\text{adjacent}}\right) = \arcsin\!\left(\frac{\text{opposite}}{\text{hypotenuse}}\right) = \arccos\!\left(\frac{\text{adjacent}}{\text{hypotenuse}}\right)}$

Example: Triangle with sides 5, 12, 13

For the angle opposite the side of length 5:

$\theta = \arcsin\!\left(\frac{5}{13}\right) = \arctan\!\left(\frac{5}{12}\right) \approx 0.395$

For the angle opposite the side of length 12:

$\alpha = \arcsin\!\left(\frac{12}{13}\right) = \arctan\!\left(\frac{12}{5}\right) \approx 1.176$

Verify: $0.395 + 1.176 = 1.571 \approx \frac{\pi}{2}$ ✓

🔢 Algebraic Expressions with Inverse Trig

Writing Trig Ratios as Algebraic Expressions

Problem: Write $\sin(\arctan \frac{x}{3})$ as an algebraic expression in $x$.

Solution:

1. Let $\theta = \arctan \frac{x}{3}$, so $\tan\theta = \frac{x}{3}$
2. Right triangle: opposite = $x$, adjacent = $3$
3. Hypotenuse = $\sqrt{x^2 + 9}$
4. $\sin\theta = \frac{x}{\sqrt{x^2 + 9}}$

$\boxed{\sin\!\left(\arctan \frac{x}{3}\right) = \frac{x}{\sqrt{x^2 + 9}}}$

Common General Formulas

✏️ Solving for Missing Angles

Example 1: Ladder Problem

A 20-foot ladder leans against a wall with its base 8 feet from the wall. Find the angle with the ground.

Adjacent = $8$, hypotenuse = $20$.

$\theta = \arccos\!\left(\frac{8}{20}\right) = \arccos(0.4) \approx 1.159$

Example 2: Finding Both Acute Angles

In a right triangle with legs $a = 7$ and $b = 24$:

$\alpha = \arctan\!\left(\frac{7}{24}\right) \approx 0.284$ $\beta = \arctan\!\left(\frac{24}{7}\right) \approx 1.287$

Check: $0.284 + 1.287 = 1.571 \approx \frac{\pi}{2}$ ✓ (The acute angles in a right triangle sum to $\frac{\pi}{2}$.)

Example 3: Using a Known Hypotenuse

Right triangle with opposite $= 6$, hypotenuse $= 10$.

$\theta = \arcsin\!\left(\frac{6}{10}\right) = \arcsin(0.6) \approx 0.644$

This is a 3-4-5 triangle scaled by 2 (sides 6, 8, 10), so the same angle is also $\arctan\!\left(\frac{6}{8}\right) \approx 0.644$.

Triangle & Algebra Quiz 🎯

Solving Triangles 🧮

Give each angle in radians, rounded to two decimal places.

1) Right triangle: opposite = 3, adjacent = 4. Find angle $\theta$. (e.g., if opposite = 5, adjacent = 12, then $\theta = \arctan(\frac{5}{12}) \approx 0.39$)

2) Right triangle: opposite = 7, hypotenuse = 25. Find angle $\theta$. (e.g., if opp = 5, hyp = 13, then $\theta = \arcsin(\frac{5}{13}) \approx 0.39$)

3) Right triangle: adjacent = 9, hypotenuse = 15. Find angle $\theta$. (e.g., if adj = 12, hyp = 13, then $\theta = \arccos(\frac{12}{13}) \approx 0.39$)

Choose the Right Expression 🔽

Exit Quiz ✅

🌍 Applications of Inverse Trig

Part 6 of 7

Inverse trig functions appear everywhere in real-world problems — navigation, physics, engineering, and more.

Angle of Elevation & Depression

$\boxed{\text{Angle of Elevation/Depression} = \arctan\!\left(\frac{\text{vertical distance}}{\text{horizontal distance}}\right)}$

Example 1: Angle of Elevation

A 6-foot person looks up at the top of a 50-foot building from 80 feet away. What is the angle of elevation?

Vertical distance = $50 - 6 = 44$ ft, horizontal distance = $80$ ft.

$\theta = \arctan\!\left(\frac{44}{80}\right) = \arctan(0.55) \approx 0.503 \text{ radians}$

Example 2: Angle of Depression

A drone at 200 feet altitude spots a target 500 feet away horizontally. The angle of depression is:

$\theta = \arctan\!\left(\frac{200}{500}\right) = \arctan(0.4) \approx 0.381 \text{ radians}$

🎡 Inverse Trig in Periodic Models

A rider's height on a Ferris wheel is $h(t) = 10\sin\!\left(\frac{\pi}{20}t\right) + 12$ meters, $t$ in seconds, for one 40-second turn ($0 \le t \le 40$).

Example 3: A special value

When is the rider first $17$ m high?

$10\sin\!\left(\frac{\pi}{20}t\right) + 12 = 17 \implies \sin\!\left(\frac{\pi}{20}t\right) = 0.5$

$\frac{\pi}{20}t = \arcsin(0.5) = \frac{\pi}{6} \implies t = \frac{20}{6} = \frac{10}{3} \approx 3.33 \text{ s}$

Example 4: A calculator value, and the second solution

When is the rider $15$ m high?

$\sin\!\left(\frac{\pi}{20}t\right) = 0.3 \implies \frac{\pi}{20}t = \arcsin(0.3) \approx 0.3047 \implies t \approx 1.94 \text{ s}$

$\arcsin$ only returns the angle in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. Sine is also $0.3$ at $\pi - 0.3047$, which gives the second time in the turn:

$\frac{\pi}{20}t = \pi - 0.3047 \implies t \approx 20 - 1.94 = 18.06 \text{ s}$

⚠️ An inverse trig function gives one angle. In a context, check whether the domain allows other solutions, and adjust the calculator's answer to find them.

🔧 Solving Inverse Trig Equations

Example 6: Solve $2\arcsin(x) = \frac{\pi}{3}$

$\arcsin(x) = \frac{\pi}{6}$ $x = \sin\!\left(\frac{\pi}{6}\right) = \frac{1}{2}$

Example 7: Solve $\arctan(2x - 1) = \frac{\pi}{4}$

$2x - 1 = \tan\!\left(\frac{\pi}{4}\right) = 1$ $2x = 2 \implies x = 1$

Key Strategy for Solving

$\boxed{\text{Isolate the inverse trig function, then apply the corresponding trig function to both sides}}$

If $\arcsin(\text{expr}) = \theta$, then $\text{expr} = \sin\theta$.

If $\arccos(\text{expr}) = \theta$, then $\text{expr} = \cos\theta$.

If $\arctan(\text{expr}) = \theta$, then $\text{expr} = \tan\theta$.

Applications Quiz 🎯

Solve Equations 🧮

1) Solve $\arcsin(x) = \frac{\pi}{6}$. What is $x$? Write as a decimal. (e.g., If $\arccos(x) = \frac{\pi}{3}$, then $x = \cos(\frac{\pi}{3}) = 0.5$)

2) Solve $\arctan(x) = \frac{\pi}{4}$. What is $x$? (e.g., If $\arctan(x) = 0$, then $x = \tan(0) = 0$)

3) A tree casts a 40-foot shadow when the sun's angle of elevation is $0.9$ radians. Tree height = $40\tan(0.9) \approx$ ? feet. Round to nearest integer, calculator in radian mode. (e.g., $40\tan(\frac{\pi}{4}) = 40$ since $\tan \frac{\pi}{4} = 1$)

Application Matching 🔽

Exit Quiz ✅

🏆 Inverse Trig — Full Synthesis

Part 7 of 7

This part brings together everything from Parts 1–6: domains & ranges, graphs, exact values, compositions, triangle problems, and applications.

Master Summary

Key Identities

$\boxed{\arcsin x + \arccos x = \frac{\pi}{2}}$

$\arcsin(-x) = -\arcsin x, \quad \arctan(-x) = -\arctan x$

$\arccos(-x) = \pi - \arccos x$

📝 Mixed Review Problems

Problem 1: Exact Value

$\arccos\!\left(-\frac{\sqrt{2}}{2}\right) = \frac{3\pi}{4}$ because $\cos\frac{3\pi}{4} = -\frac{\sqrt{2}}{2}$ and $\frac{3\pi}{4} \in [0, \pi]$.

Problem 2: Composition

$\sin(\arctan \frac{3}{4})$: Triangle with opp = 3, adj = 4, hyp = 5. Answer: $\frac{3}{5}$.

Problem 3: InverseTrig(Trig)

$\arccos(\cos \frac{7\pi}{4})$: $\cos\frac{7\pi}{4} = \frac{\sqrt{2}}{2}$. $\arccos(\frac{\sqrt{2}}{2}) = \frac{\pi}{4}$.

Problem 4: Equation

Solve $\arcsin(2x - 1) = -\frac{\pi}{6}$: $2x - 1 = \sin(-\frac{\pi}{6}) = -\frac{1}{2}$ $2x = \frac{1}{2} \implies x = \frac{1}{4}$

Problem 5: Application

Lighthouse 150 ft tall, boat 400 ft away. Angle of depression: $\arctan\!\left(\frac{150}{400}\right) = \arctan(0.375) \approx 0.359 \text{ radians}$

⚠️ Common Mistakes to Avoid

Comprehensive Quiz 🎯

Mixed Skill Check 🧮

1) $\tan(\arcsin \frac{8}{17})$ = ? Write as a fraction. (e.g., $\tan(\arcsin \frac{3}{5}) = \frac{3}{4}$ using a 3-4-5 triangle)

2) Solve $\arccos(x) = \frac{2\pi}{3}$. What is $x$? Write as a decimal. (e.g., $\arccos(x) = \frac{\pi}{3}$ gives $x = \cos\frac{\pi}{3} = 0.5$)

3) $\arcsin(\frac{\sqrt{3}}{2}) + \arccos(\frac{\sqrt{3}}{2}) = \,?\cdot\pi$. Enter the coefficient of $\pi$. (e.g., $\arcsin(\frac{1}{2}) + \arccos(\frac{1}{2}) = \frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{2}$, coefficient 1/2)

Final Review 🔽

Final Exit Quiz ✅