Navigation: Wiki Home > Skills > Simplifying Expressions with Inverse Trig Functions
| 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.
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.
This skill is essential for derivatives (you'll need it to derive the formula for $\frac{d}{dx}\arcsin(x)$) and for integration by trigonometric substitution.
| Property | Value |
|---|---|
| Concept | Inverse Functions |
| Week | Week 2 |
| Difficulty | Intermediate |
| Time | ~20 minutes |
Goal: Simplify $\text{trig}(\text{inverse-trig}(x))$.
Strategy:
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 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.
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 = 1For $\theta = \arccos(x)$: (where $\cos\theta = x/1$)
hyp = 1
/|
/ |
/ | opp = √(1-x²)
/θ |
/____|
adj = x| 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 $x$ |
Sometimes identities are faster, especially when you already know one trig value.
Example: Find $\cos(\arctan x)$.
Let $\theta = \arctan x$. Then: $$\sec^2\theta = 1 + \tan^2\theta = 1 + x^2$$ $$\sec\theta = \sqrt{1+x^2} \quad \text{(positive because } \cos\theta > 0 \text{ for } \theta \in (-\pi/2, \pi/2)\text{)}$$ $$\cos\theta = \frac{1}{\sqrt{1+x^2}}$$
| 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 |
Simplify $\sin(\arctan(3))$ (exact value, no calculator).
Simplify $\sec(\arctan(x))$ as an algebraic expression in $x$.
Simplify $\cos(2\arctan(x))$ for all real $x$.
Simplify $\sin(\arctan(1) + \arctan(2))$.
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
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)
(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 $(π/2, π)$.
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.
Looking back:
Looking ahead:
| Previous | Up | Next |
|---|---|---|
| Inverse Trig Definitions | Skills Index | Derivatives of Inverse Trig |
Last updated: 2026-01-22