Home ›
Do My Trigonometry Homework ›
Inverse Trig Functions Explained
Inverse Trig Functions: Why arcsin Only Gives One Answer (And How to Find the Rest)
Quick Answer
Inverse trig functions take a ratio and return an angle. arcsin(x) gives the angle whose sine is x. arccos(x) gives the angle whose cosine is x. arctan(x) gives the angle whose tangent is x. They are also written sin⁻¹, cos⁻¹, and tan⁻¹.
arcsin
Output: [−90°, 90°]
arccos
Output: [0°, 180°]
arctan
Output: (−90°, 90°)
Critical: sin⁻¹(x) means arcsin, not 1/sin(x). See the notation trap section.
In This Guide
- The Core Concept: Undoing a Trig Function
- The Three Main Inverse Functions
- Why the Range Is Restricted
- The Notation Trap: sin⁻¹ ≠ 1/sin
- The Reciprocal Inverses: arccsc, arcsec, arccot
- Finding Angles in All Four Quadrants
- Which Inverse Function to Use
- Worked Examples
- Common Mistakes
- Practice Problem
- The Calculus Connection
- Platform Tips
- Frequently Asked Questions
The Core Concept: Undoing a Trig Function
The forward trig functions take an angle and return a ratio. Sin(30°) = 0.5 means: start with the angle 30°, the sine function gives you the ratio 0.5. Inverse trig functions reverse that process — they take a ratio and return the angle that produces it.
The inputs and outputs swap. Sin takes an angle, returns a ratio. Arcsin takes a ratio, returns an angle.
This is exactly what you do when finding a missing angle in a right triangle. If you know the opposite side is 5 and the hypotenuse is 10, then sin θ = 5/10 = 0.5. To find θ, you ask: “what angle has a sine of 0.5?” That’s arcsin(0.5) = 30°. The inverse function is the answer to “what angle produces this ratio?”
💡 The Three Equivalent Notations
arcsin(x), sin⁻¹(x), and asin(x) all mean the same thing — the inverse sine function. Your textbook may use any of these. On a TI-84, the button is labeled sin⁻¹ (accessed via 2nd → sin). In code and on some platforms, it’s written asin. They are identical.
The Three Main Inverse Functions
| Function | Also written | Input (domain) | Output (range) | Example |
|---|---|---|---|---|
| arcsin(x) | sin⁻¹(x), asin(x) | −1 ≤ x ≤ 1 | [−90°, 90°] | arcsin(0.5) = 30° |
| arccos(x) | cos⁻¹(x), acos(x) | −1 ≤ x ≤ 1 | [0°, 180°] | arccos(0.5) = 60° |
| arctan(x) | tan⁻¹(x), atan(x) | All real numbers | (−90°, 90°) | arctan(1) = 45° |
Three things worth noting: arctan accepts any real number as input (unlike arcsin and arccos, which only accept values from −1 to 1 since those are the bounds of sin and cos). Arctan’s output range uses parentheses instead of brackets because 90° and −90° are excluded — tangent is undefined there, so they can never be outputs. And arccos always returns a value between 0° and 180°, which means it never returns a negative angle — if you expect a negative result from arccos, something is wrong.
Exact Values Reference
These are the outputs your homework expects at standard angles. Platforms like ALEKS and Knewton Alta require exact radian form — not 0.785 for arctan(1), but π/4.
| Input x | arcsin(x) | arccos(x) | arctan(x) |
|---|---|---|---|
| −1 | −90° (−π/2) | 180° (π) | — |
| −√3/2 | −60° (−π/3) | 150° (5π/6) | — |
| −√2/2 | −45° (−π/4) | 135° (3π/4) | — |
| −1/2 | −30° (−π/6) | 120° (2π/3) | — |
| 0 | 0° | 90° (π/2) | 0° |
| 1/2 | 30° (π/6) | 60° (π/3) | — |
| √2/2 | 45° (π/4) | 45° (π/4) | — |
| √3/2 | 60° (π/3) | 30° (π/6) | — |
| 1 | 90° (π/2) | 0° | — |
| 1/√3 | — | — | 30° (π/6) |
| 1 | — | — | 45° (π/4) |
| √3 | — | — | 60° (π/3) |
Notice that arcsin and arccos of the same input always sum to 90°. arcsin(1/2) = 30° and arccos(1/2) = 60°. arcsin(√3/2) = 60° and arccos(√3/2) = 30°. This is not a coincidence.
arcsin(x) + arccos(x) = 90° for all x in [−1, 1]
If arcsin(x) gives you one acute angle in a right triangle, arccos(x) gives you the other — they must sum to 90° because all three angles in a right triangle sum to 180° and one is already 90°. This identity shows up on exams as a simplification: if you see arcsin(x) + arccos(x), the answer is always π/2 (or 90°), no calculation needed.
Why the Range Is Restricted
This is the question every textbook buries in a footnote. Here’s the actual reason.
For a function to have an inverse, it must be one-to-one — each input must produce exactly one output, and each output must come from exactly one input. Sin is not one-to-one across all angles: sin(30°) = 0.5 and sin(150°) = 0.5. If you asked “what angle has a sine of 0.5?” without any restriction, the answer would be 30°, 150°, 390°, 510°, and infinitely more. That’s not a function — it’s a chaos of answers.
The solution is to restrict the domain of sin to a range where it is one-to-one, then define arcsin as the inverse on that restricted domain. The chosen range for each function is the smallest interval that covers all possible output values exactly once:
Each highlighted arc shows the exact portion of the unit circle used as the output range. Closed dots = endpoints included. Open dots = endpoints excluded.
The practical consequence: arcsin always returns an angle in Q I or Q IV (between −90° and 90°). Arccos always returns an angle in Q I or Q II (between 0° and 180°). Arctan always returns an angle in Q I or Q IV, never exactly ±90°. If your problem needs an angle outside those ranges — for example, the Q III solution to sin θ = −0.5 — you have to find it yourself using reference angles and quadrant signs. The inverse function alone will not give it to you.
⚠️ This Is Where Most Exam Errors Happen
When a problem asks for “all solutions in [0°, 360°],” the inverse function only gives you one of them. The others come from applying symmetry rules. See Finding Angles in All Four Quadrants below.
The Notation Trap: sin⁻¹ ≠ 1/sin
This was covered on the Trig Functions Explained page but it’s worth a direct reminder here since it appears constantly when working with inverse trig.
sin⁻¹(x)
= arcsin(x)
Input: a ratio. Output: an angle.
sin⁻¹(0.5) = 30°
(sin x)⁻¹ or 1/sin(x)
= csc(x)
Input: an angle. Output: a ratio.
1/sin(30°) = 1/0.5 = 2
The −1 in sin⁻¹ is borrowed from function inverse notation (f⁻¹), not from the algebraic rule x⁻¹ = 1/x. In algebra, negative exponents mean reciprocals. In trig function notation, they mean inverse function. These are different things in the same notation — a genuine ambiguity in mathematical convention that textbooks rarely acknowledge clearly.
The Reciprocal Inverses: arccsc, arcsec, arccot
The inverses of the three reciprocal functions exist but appear rarely — mainly in advanced calculus and engineering contexts. Most pre-calculus and Calculus I courses focus exclusively on arcsin, arccos, and arctan.
| Function | Input domain | Output range | On calculator |
|---|---|---|---|
| arccsc(x) | |x| ≥ 1 | [−90°, 0) ∪ (0°, 90°] | sin⁻¹(1/x) |
| arcsec(x) | |x| ≥ 1 | [0°, 90°) ∪ (90°, 180°] | cos⁻¹(1/x) |
| arccot(x) | All real numbers | (0°, 180°) | tan⁻¹(1/x) |
No standard calculator has dedicated arccsc, arcsec, or arccot buttons. Enter them using the reciprocal relationship — arccsc(x) = arcsin(1/x), arcsec(x) = arccos(1/x), arccot(x) = arctan(1/x). Note the input constraint: arccsc and arcsec require |x| ≥ 1 because csc and sec never produce values between −1 and 1.
Finding Angles in All Four Quadrants
This is the skill that separates students who understand inverse trig from those who just memorize outputs. When a problem asks for all solutions in [0°, 360°], the inverse function gives you one answer. Finding the rest requires four steps.
Use the inverse function to find the reference angle
Apply arcsin, arccos, or arctan to get the principal value — the angle in the restricted range. This becomes your reference angle (take the absolute value if it’s negative).
Determine which quadrants have the correct sign
Use ASTC. If the ratio is positive, find the two quadrants where that function is positive. If negative, find the two where it’s negative.
Build the angles from the reference angle
Q I: θ = reference angle. Q II: θ = 180° − reference. Q III: θ = 180° + reference. Q IV: θ = 360° − reference.
Keep only the angles in the requested range
Discard any solutions outside the specified interval. If no interval is given, state the general solution using + 360°n (or + 2πn in radians).
arcsin gives you −30°. The two solutions in [0°, 360°] come from applying the reference angle in Q III and Q IV — the shaded region where sin is negative.
Find all θ in [0°, 360°] where sin θ = −0.5.
Step 1: arcsin(−0.5) = −30°. Reference angle = 30°.
Step 2: Sin is negative in Q III and Q IV.
Step 3: Q III angle = 180° + 30° = 210°. Q IV angle = 360° − 30° = 330°.
Answer: θ = 210° and θ = 330°. (arcsin gave −30°, which is outside [0°, 360°] and would have been wrong by itself.)
Which Inverse Function to Use
In a right triangle problem, the answer depends on which two sides you know relative to angle θ. This diagram makes the decision immediate — no memorization required beyond knowing that the inverse function undoes the forward function.
SOH-CAH-TOA tells you the ratio — the inverse function inverts it to find θ.
Worked Examples
Example 1 — Basic Evaluation
Find arccos(√3/2).
Ask: what angle in [0°, 180°] has a cosine of √3/2?
cos(30°) = √3/2, and 30° is in [0°, 180°]. Answer: 30°.
Example 2 — Missing Angle in a Right Triangle
A right triangle has opposite = 8, hypotenuse = 17. Find θ.
sin θ = 8/17. Apply inverse: θ = arcsin(8/17) = arcsin(0.4706).
θ ≈ 28.07°. (Use a calculator — this isn’t a standard angle.)
Example 3 — Negative Input
Find arctan(−1).
Ask: what angle in (−90°, 90°) has a tangent of −1?
tan(45°) = 1, so tan(−45°) = −1, and −45° is in (−90°, 90°).
Answer: −45° (or equivalently −π/4 radians).
Example 4 — Composition of Functions
Evaluate sin(arccos(3/5)).
Let θ = arccos(3/5), so cos θ = 3/5 and θ is in [0°, 180°].
Draw a right triangle: adjacent = 3, hypotenuse = 5, opposite = √(25−9) = 4.
sin θ = opposite/hypotenuse = 4/5. Answer: 4/5.
Common Mistakes
❌ Expecting arcsin to return all solutions
arcsin(−0.5) = −30°, not 210° or 330°. The restricted range means you always get exactly one answer. For all solutions in [0°, 360°], use the four-step process above.
❌ Inputting a value outside the domain
arcsin(2) and arccos(−3) are undefined — those values are outside [−1, 1]. If your calculator returns an error on arcsin or arccos, check that the input is between −1 and 1. If you’re getting a value outside that range, the problem setup is wrong.
❌ Wrong calculator mode
arcsin(0.5) in degree mode = 30°. arcsin(0.5) in radian mode = 0.5236 radians (≈ π/6). Both are correct — but if the problem asks for degrees and your calculator is in radian mode, 0.5236 will be marked wrong. Check your mode before every inverse trig calculation.
❌ Using arccos when arcsin is needed (or vice versa)
arccos(0.5) = 60°, not 30°. arcsin(0.5) = 30°, not 60°. These are different functions. Match the inverse function to the trig function in the original equation — if you have sin θ = 0.5, use arcsin, not arccos.
❌ Thinking arccos(x) can return a negative angle
Arccos output range is [0°, 180°] — it never returns a negative angle. If you expect arccos to return something like −60°, either the problem calls for a different approach or the expected answer is actually 120° (the supplement). Check which quadrant the problem is asking about.
Stuck on an Inverse Trig Problem?
We complete trig assignments and exams on ALEKS, WebAssign, Hawkes, Knewton Alta, and every other platform — A/B guaranteed.
Practice Problem
📝 Problem 1 — Basic Evaluation
Find the exact values of arcsin(√2/2), arccos(0), and arctan(−√3). Give answers in degrees and radians.
Click to reveal answer
arcsin(√2/2): what angle in [−90°, 90°] has sin = √2/2? That’s 45°. Answer: 45° (π/4)
arccos(0): what angle in [0°, 180°] has cos = 0? That’s 90°. Answer: 90° (π/2)
arctan(−√3): what angle in (−90°, 90°) has tan = −√3? tan(60°) = √3, so tan(−60°) = −√3. Answer: −60° (−π/3)
📝 Problem 2 — Composition
Evaluate tan(arcsin(5/13)) without a calculator.
Click to reveal answer
Let θ = arcsin(5/13). Then sin θ = 5/13, meaning opposite = 5, hypotenuse = 13.
Find adjacent using Pythagorean theorem: adjacent = √(13² − 5²) = √(169 − 25) = √144 = 12.
tan θ = opposite/adjacent = 5/12. Answer: 5/12.
📝 Problem 3 — All Quadrant Solutions
Find all θ in [0°, 360°] where cos θ = −√3/2. Give exact answers.
Click to reveal answer
Step 1: arccos(−√3/2). We know arccos(√3/2) = 30°, so arccos(−√3/2) = 180° − 30° = 150°. Reference angle = 30°.
Step 2: Cos is negative in Q II and Q III.
Step 3: Q II: θ = 180° − 30° = 150°. Q III: θ = 180° + 30° = 210°.
Answer: θ = 150° and θ = 210°.
The Calculus Connection
Inverse trig functions appear heavily in Calculus I and II, primarily through their derivatives and in integration via trig substitution. If you’re heading into calculus, knowing these formulas cold will save significant time.
Derivatives of the three main inverse functions:
| Function | Derivative | Domain |
|---|---|---|
| d/dx [arcsin(x)] | 1 / √(1 − x²) | −1 < x < 1 |
| d/dx [arccos(x)] | −1 / √(1 − x²) | −1 < x < 1 |
| d/dx [arctan(x)] | 1 / (1 + x²) | All real numbers |
The arctan derivative is the most commonly tested. The integral ∫ 1/(1 + x²) dx = arctan(x) + C appears frequently in Calc II, and recognizing the integrand pattern requires knowing the derivative cold. The arcsin derivative drives trig substitution: when you see √(1 − x²) in an integrand, the substitution x = sin θ is often the path forward.
💡 Notice
The derivatives of arcsin and arccos are negatives of each other. This makes sense: arcsin(x) + arccos(x) = 90° for all x in [−1, 1], so their rates of change must cancel out to a constant derivative of zero. Their derivatives summing to zero is the formal confirmation of that identity.
Platform Tips
MyMathLab — inverse trig answers on MyMathLab are entered using the equation editor. Look for the trig function palette and select sin⁻¹, cos⁻¹, or tan⁻¹ from the inverse section. For exact answers, use the fraction tool and the π symbol rather than typing decimals. MyMathLab sometimes displays the answer in degrees even when the internal grading expects radians — if your answer is being marked wrong and you’re confident it’s correct, try converting to the other unit.
ALEKS — inverse trig inputs usually use the sin⁻¹ notation in the expression editor. Click the function palette and look for sin⁻¹, cos⁻¹, tan⁻¹. When ALEKS asks “give your answer in degrees,” make sure your calculator is in degree mode before evaluating — ALEKS will accept the degree value directly. When it says “give your answer in radians,” enter the exact radian form if it’s a standard angle (π/6, π/4, etc.) rather than a decimal.
Hawkes Learning — Certify mode sometimes presents inverse trig as “find the angle θ in [0°, 360°] such that…” which requires the full four-step quadrant process, not just the inverse function output. Entering only the principal value will be marked wrong if the problem specifically asks for all solutions or for a solution in a non-standard quadrant.
Knewton Alta — accepts exact radian answers. arcsin(1/2) should be entered as π/6, not 0.5236. Use the fraction and π tools in the input palette. If the answer doesn’t simplify to a standard angle, enter the decimal rounded to the specified number of places.
WebAssign — composition problems like sin(arccos(x)) are common. Work through these algebraically using a reference triangle rather than trying to evaluate numerically — WebAssign expects exact simplified form and a decimal approximation will often fail the tolerance check.
Frequently Asked Questions
What is the difference between arcsin and sin⁻¹?
Nothing — they are the same function written in two different notations. arcsin(x), sin⁻¹(x), and asin(x) all refer to the inverse sine function. Your textbook or platform may use any of these. On a TI-84, the button is labeled sin⁻¹.
Why does arcsin(sin(150°)) not equal 150°?
Because arcsin’s output range is [−90°, 90°], which doesn’t include 150°. sin(150°) = 0.5, and arcsin(0.5) = 30° — the angle in [−90°, 90°] with a sine of 0.5. The composition arcsin(sin(x)) only returns x unchanged when x is already in [−90°, 90°].
Can arctan output 90° or −90°?
No — arctan’s range is the open interval (−90°, 90°), meaning those endpoints are excluded. Tangent is undefined at ±90°, so they can never be outputs of arctan. As the input to arctan gets very large (positive), the output approaches 90° but never reaches it.
How do I find a missing angle in a right triangle?
Identify which two sides you know relative to the unknown angle. If you know opposite and hypotenuse, use arcsin(opposite/hypotenuse). If you know adjacent and hypotenuse, use arccos(adjacent/hypotenuse). If you know opposite and adjacent, use arctan(opposite/adjacent). The result is the angle in the restricted range — for a right triangle problem, this is always the correct answer since all acute angles are between 0° and 90°.
Why does my calculator give a different answer than expected?
Most likely you’re in the wrong mode. arcsin(0.5) = 30° in degree mode, but = 0.5236 in radian mode. Do a quick check: type arcsin(0.5). If you get 30, you’re in degree mode. If you get 0.5236, you’re in radian mode. Switch modes accordingly based on what your problem requires.
What happens when I apply sin to arcsin, or arcsin to sin?
sin(arcsin(x)) = x for all x in [−1, 1] — applying a function and then its inverse always returns the original input. But arcsin(sin(x)) = x only when x is in [−90°, 90°]. Outside that range, arcsin(sin(x)) returns the equivalent angle within the restricted range instead.
Need Help With Your Trigonometry Course?
We complete trig assignments and exams on every platform — guaranteed A/B grades.
Related Resources
Prerequisites:
- Trig Functions Explained — all six functions, the notation trap, and exact values
- SOH-CAH-TOA Explained — right triangle setup used in finding missing angles
- Unit Circle Explained — the standard angle values arcsin and arccos draw from
What uses inverse trig next:
- Solving Trig Equations — inverse functions are the core tool for isolating θ
- Do My Trigonometry Homework