Simplifying Expressions with Inverse Trig Functions

5 min read

Reference: Stewart §6.6

Quick Reference: Common Simplifications

Expression	Result	Valid for
$\cos(\arcsin x)$	$\sqrt{1-x^2}$	$-1 \le x \le 1$
$\tan(\arcsin x)$	$\frac{x}{\sqrt{1-x^2}}$	$-1 < x < 1$
$\sin(\arccos x)$	$\sqrt{1-x^2}$	$-1 \le x \le 1$
$\sin(\arctan x)$	$\frac{x}{\sqrt{1+x^2}}$	all real $x$
$\cos(\arctan x)$	$\frac{1}{\sqrt{1+x^2}}$	all real $x$

Key Insight: The range of the inverse function determines the sign of the result.

Skip to Practice Problems

From Angles to Ratios and Back

What is $\tan(\arcsin(x))$? This asks: “If $\theta$ is the angle whose sine is $x$, what is the tangent of $\theta$?”

You could try using identities like $\tan\theta = \sin\theta/\cos\theta$, but there’s a more intuitive approach: draw a right triangle. This “triangle method” converts inverse trig problems into simple geometry.

The same triangle method derives the formula for $\frac{d}{dx}\arcsin(x)$ and drives integration by trigonometric substitution.

📋 Before You Start: Check Your Prerequisites

Inverse trig ranges: What is the range of $\arcsin(x)$? What is the range of $\arctan(x)$?
- If unsure: Review Inverse Trig Definitions
Pythagorean theorem: In a right triangle with legs $a$ and $b$, the hypotenuse is...?
- Answer: $\sqrt{a^2 + b^2}$
SOH-CAH-TOA: Given an angle $\theta$ in a right triangle, what is $\tan\theta$ in terms of the sides?
- Answer: opposite/adjacent

Prerequisite Map

Quick Reference

Property	Value
Concept	Inverse Functions
Difficulty	Intermediate
Time	~20 minutes

Key Concepts

The Triangle Method

Goal: Simplify $\text{trig}(\text{inverse-trig}(x))$.

Strategy:

Let $\theta = \text{inverse-trig}(x)$
Draw a right triangle with angle $\theta$
Label two sides based on the inverse trig definition
Use the Pythagorean theorem to find the third side
Read off the desired trig ratio

Example: $\tan(\arcsin(x))$

Let $\theta = \arcsin(x)$, so $\sin\theta = x = \frac{x}{1}$.

                    /|
                   / |
                  /  |
                1/   | x    ← opposite = x
                /    |
               / θ   |
              /______|
                 ?

By Pythagorean theorem: adjacent = √(1 - x²)

Therefore: $$\tan(\arcsin(x)) = \frac{\text{opposite}}{\text{adjacent}} = \frac{x}{\sqrt{1-x^2}}$$

The Key Insight: Range Determines Quadrant

The range of the inverse trig function tells you which quadrant $\theta$ is in, which determines the signs:

Function	Range	Quadrant	Signs
$\arcsin$	$[-\pi/2, \pi/2]$	I or IV	$\cos\theta \ge 0$
$\arccos$	$[0, \pi]$	I or II	$\sin\theta \ge 0$
$\arctan$	$(-\pi/2, \pi/2)$	I or IV	$\cos\theta > 0$

Why this matters: When you use $\sqrt{\cos^2\theta} = \vert \cos\theta\vert $, you need to know if you can drop the absolute value.

Reference Triangles for Common Cases

For $\theta = \arcsin(x)$: (where $\sin\theta = x/1$)

        hyp = 1
       /|
      / |
     /  | opp = x
    /θ  |
   /____|
   adj = √(1-x²)

For $\theta = \arctan(x)$: (where $\tan\theta = x/1$)

        hyp = √(1+x²)
       /|
      / |
     /  | opp = x
    /θ  |
   /____|
   adj = 1

For $\theta = \arccos(x)$: (where $\cos\theta = x/1$)

        hyp = 1
       /|
      / |
     /  | opp = √(1-x²)
    /θ  |
   /____|
   adj = x

Common Simplifications Table

Expression	Result	Valid for
$\cos(\arcsin x)$	$\sqrt{1-x^2}$	$-1 \le x \le 1$
$\tan(\arcsin x)$	$\frac{x}{\sqrt{1-x^2}}$	$-1 < x < 1$
$\sin(\arccos x)$	$\sqrt{1-x^2}$	$-1 \le x \le 1$
$\tan(\arccos x)$	$\frac{\sqrt{1-x^2}}{x}$	$-1 \le x \le 1$, $x \ne 0$
$\sin(\arctan x)$	$\frac{x}{\sqrt{1+x^2}}$	all real $x$
$\cos(\arctan x)$	$\frac{1}{\sqrt{1+x^2}}$	all real $x$

Alternative Method: Using Identities

Sometimes identities are faster, especially when you already know one trig value.

Example: Find $\cos(\arctan x)$.

Let $\theta = \arctan x$. Then: $$\sec^2\theta = 1 + \tan^2\theta = 1 + x^2$$ $$\sec\theta = \sqrt{1+x^2} \quad \text{(positive because } \cos\theta > 0 \text{ for } \theta \in (-\pi/2, \pi/2)\text{)}$$ $$\cos\theta = \frac{1}{\sqrt{1+x^2}}$$

Common Pitfalls

Mistake	Why It’s Wrong	Correct Approach
Writing $\cos(\arcsin x) = \pm\sqrt{1-x^2}$	The range of $\arcsin$ determines the sign	Since $\arcsin x \in [-\pi/2, \pi/2]$, cosine is non-negative, so $\cos(\arcsin x) = \sqrt{1-x^2}$ (positive root only)
Forgetting domain restrictions	$\tan(\arcsin x) = \frac{x}{\sqrt{1-x^2}}$ is undefined at $x = \pm 1$	Check that your expression is defined; here $x \ne \pm 1$
Drawing triangle with wrong labels	Must match the inverse trig definition exactly	For $\arcsin(x)$: opposite = $x$, hypotenuse = $1$. For $\arctan(x)$: opposite = $x$, adjacent = $1$
Using wrong Pythagorean identity	$\sin^2 + \cos^2 = 1$ vs. $1 + \tan^2 = \sec^2$	Pick the identity that matches what you know and what you want

Practice Problems

Level 1 Basic Triangle Setup

Simplify $\sin(\arctan(3))$ (exact value, no calculator).

Thought Process

Let $\theta = \arctan(3)$. Draw a right triangle where $\tan\theta = 3 = \frac{3}{1}$.

Opposite = 3
Adjacent = 1
Hypotenuse = $\sqrt{1^2 + 3^2} = \sqrt{10}$

Now read off sine = opposite/hypotenuse.

Show Answer

Let $\theta = \arctan(3)$. Then $\tan\theta = 3$.

Triangle: opposite = 3, adjacent = 1, hypotenuse = $\sqrt{10}$.

$$\sin(\arctan(3)) = \frac{3}{\sqrt{10}} = \frac{3\sqrt{10}}{10}$$

Level 2 Simplification with Variable

Simplify $\sec(\arctan(x))$ as an algebraic expression in $x$.

Thought Process

Let $\theta = \arctan(x)$, so $\tan\theta = x/1$.

Triangle: opposite = $x$, adjacent = $1$, hypotenuse = $\sqrt{1+x^2}$.

$\sec\theta = \frac{\text{hyp}}{\text{adj}} = \frac{\sqrt{1+x^2}}{1}$.

Check: Since $\theta \in (-\pi/2, \pi/2)$, we have $\cos\theta > 0$, so $\sec\theta > 0$. ✓

Show Answer

$$\sec(\arctan(x)) = \sqrt{1+x^2}$$

Level 3 Double Angle with Inverse Trig

Simplify $\cos(2\arctan(x))$ for all real $x$.

Thought Process

Let $\theta = \arctan(x)$. We want $\cos(2\theta)$.

Use the double angle formula: $\cos(2\theta) = \cos^2\theta - \sin^2\theta$.

From the reference triangle for $\theta = \arctan(x)$:

opposite = $x$, adjacent = $1$, hypotenuse = $\sqrt{1+x^2}$

So $\cos\theta = \frac{1}{\sqrt{1+x^2}}$ and $\sin\theta = \frac{x}{\sqrt{1+x^2}}$.

Show Answer

Let $\theta = \arctan(x)$, so $\tan\theta = x$.

From the reference triangle: opposite = $x$, adjacent = $1$, hypotenuse = $\sqrt{1+x^2}$.

Thus: $\cos\theta = \frac{1}{\sqrt{1+x^2}}$ and $\sin\theta = \frac{x}{\sqrt{1+x^2}}$

Using $\cos(2\theta) = \cos^2\theta - \sin^2\theta$:

$$\cos(2\arctan(x)) = \frac{1}{1+x^2} - \frac{x^2}{1+x^2} = \frac{1-x^2}{1+x^2}$$

Note: This result is valid for all real $x$, since $\arctan$ has domain $\mathbb{R}$.

Level 4 Sum of Inverse Functions

Simplify $\sin(\arctan(1) + \arctan(2))$.

Thought Process

Let $\alpha = \arctan(1)$ and $\beta = \arctan(2)$.

Use the sine addition formula: $\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta$.

For $\alpha = \arctan(1)$: triangle has opp=1, adj=1, hyp=$\sqrt{2}$.

$\sin\alpha = 1/\sqrt{2}$, $\cos\alpha = 1/\sqrt{2}$

For $\beta = \arctan(2)$: triangle has opp=2, adj=1, hyp=$\sqrt{5}$.

$\sin\beta = 2/\sqrt{5}$, $\cos\beta = 1/\sqrt{5}$

The key is building both triangles and reading off all four trig values needed for the addition formula.

Show Answer

Let $\alpha = \arctan(1)$ and $\beta = \arctan(2)$.

Building the reference triangles:

For $\alpha = \arctan(1)$: opposite = 1, adjacent = 1, hypotenuse = $\sqrt{1^2+1^2} = \sqrt{2}$

$\sin\alpha = \frac{1}{\sqrt{2}}$, $\cos\alpha = \frac{1}{\sqrt{2}}$

For $\beta = \arctan(2)$: opposite = 2, adjacent = 1, hypotenuse = $\sqrt{1^2+2^2} = \sqrt{5}$

$\sin\beta = \frac{2}{\sqrt{5}}$, $\cos\beta = \frac{1}{\sqrt{5}}$

Applying the addition formula:

$$\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta$$

$$= \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{5}} + \frac{1}{\sqrt{2}} \cdot \frac{2}{\sqrt{5}}$$

$$= \frac{1}{\sqrt{10}} + \frac{2}{\sqrt{10}} = \frac{3}{\sqrt{10}} = \frac{3\sqrt{10}}{10}$$

Level 5 Proving a Simplification Formula

Prove that $\cos(\sin^{-1}(x)) = \sqrt{1-x^2}$ for $x \in [-1, 1]$ using both:

(a) The triangle method (b) The Pythagorean identity

Thought Process

For part (a): Draw the standard triangle for $\theta = \arcsin(x)$, where $\sin\theta = x$.

For part (b): Start with $\sin^2\theta + \cos^2\theta = 1$ and solve for $\cos\theta$. The key is justifying why we take the positive root (use the range of arcsine).

Show Answer

(a) Triangle Method:

Let $\theta = \sin^{-1}(x)$, so $\sin\theta = x = \frac{x}{1}$.

Draw a right triangle with angle $\theta$:

Opposite = $x$
Hypotenuse = $1$
Adjacent = $\sqrt{1-x^2}$ (by Pythagorean theorem)

Therefore: $\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \sqrt{1-x^2}$

(b) Pythagorean Identity:

Let $\theta = \sin^{-1}(x)$. Then $\sin\theta = x$.

By the Pythagorean identity: $$\cos^2\theta = 1 - \sin^2\theta = 1 - x^2$$ $$\cos\theta = \pm\sqrt{1-x^2}$$

Since $\theta = \sin^{-1}(x) \in [-\pi/2, \pi/2]$, we know $\cos\theta \ge 0$.

Therefore: $\cos\theta = \sqrt{1-x^2}$ (taking the positive root).

$\square$

Conceptual Check (CCI-Style)

Question: A student claims that $\tan(\arccos(x)) = \frac{\sqrt{1-x^2}}{x}$ is always valid. What’s wrong with this statement?

(A) The formula is completely incorrect (B) The formula fails when $x = 0$ (C) The formula fails when $x < 0$ (D) Both (B) and (C)

Answer

(B) The formula fails when $x = 0$ because division by zero is undefined.

Note: The formula IS valid for $x < 0$. Even though $\arccos(x)$ is in the second quadrant when $x < 0$, we have $\cos\theta = x$ (which is negative) and $\sin\theta = \sqrt{1-x^2}$ (positive), so $\tan\theta = \sqrt{1-x^2}/x$ is negative, which is correct for angles in $(π/2, π)$.

Common Misconceptions

Common misconception

$\arcsin(\sin(\theta)) = \theta$ for all values of $\theta$.

This is the concept-image-conflicts-definition error. The cancellation $\arcsin(\sin(\theta)) = \theta$ holds only when $\theta$ is in the range of $\arcsin$, namely $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. Outside that interval, $\arcsin$ returns the unique value in that range with the same sine. For $\theta = \frac{2\pi}{3}$, $\sin\!\left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2}$, so $\arcsin\!\left(\sin\!\left(\frac{2\pi}{3}\right)\right) = \arcsin\!\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3} \neq \frac{2\pi}{3}$.

Mastery Checklist

I can draw a reference triangle for any inverse trig expression
I can use the Pythagorean theorem to find the missing side
I understand how the range of the inverse function determines sign choices
I can simplify expressions like $\sin(\arctan(x))$ or $\cos(\arcsin(x))$
I can handle double-angle formulas combined with inverse trig functions

Mental Model

The Triangle as a Translator: Think of the reference triangle as a “translation device” between the algebraic world (where you have $x$) and the trigonometric world (where you have $\sin$, $\cos$, $\tan$). The inverse trig function tells you the angle; the triangle tells you everything else about that angle.

Connections

Looking back:

The inverse trig definitions give us the ranges we need to determine signs
Pythagorean identities from MATH161 are the algebraic version of this triangle method

Looking ahead:

The derivative formulas for inverse trig are derived using exactly this technique
In Chapter 7, trig substitution reverses this process: given $\sqrt{1-x^2}$, let $x = \sin\theta$

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Inverse Trig Definitions	Skills Index	Derivatives of Inverse Trig

Last updated: 2026-01-22