Inverse Trig Functions

Section 6.3 — arcsin, arccos, arctan: working backwards from a ratio to an angle

MAT 172 — Week 3
Regular trig asks: given an angle, what's the ratio? Inverse trig asks: given the ratio, what's the angle? It's the same relationship — just run backwards. The ladder is on the wall; you know how high — what's the angle?
Inverse trig functions undo trig functions. sin(30°) = 0.5, so arcsin(0.5) = 30°. They answer the question: "what angle produces this ratio?"
Forward vs inverse — the same relationship, two directions Forward trig sin(θ) = x Input: angle θ Output: ratio x sin(30°) = 0.5 sin(45°) = √2/2 sin(60°) = √3/2 angle → ratio inverse forward Inverse trig arcsin(x) = θ Input: ratio x Output: angle θ arcsin(0.5) = 30° arcsin(√2/2) = 45° arcsin(√3/2) = 60° ratio → angle Two notations arcsin(x) = sin⁻¹(x) arccos(x) = cos⁻¹(x) arctan(x) = tan⁻¹(x) same thing!
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sin⁻¹(x) does NOT mean 1/sin(x)
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This is the most common notation confusion. sin⁻¹(x) is the inverse function, not the reciprocal. The reciprocal of sin is csc (cosecant): csc(x) = 1/sin(x). But sin⁻¹(x) = arcsin(x) — the angle whose sine is x. If you want to avoid the confusion, use arcsin, arccos, arctan notation — it's unambiguous. Your calculator may show sin⁻¹ as a button, which means the inverse, not 1/sin.
✗ sin⁻¹(x) ≠ 1/sin(x)    ✓ sin⁻¹(x) = arcsin(x)
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The ladder — inverse trig in real life
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The unit circle ladder: sin = how high, cos = how wide. Forward trig gives you the height from the angle. Inverse trig gives you the angle from the height. A 15-foot ladder leans against a wall with its base 6 feet out. What's the angle of elevation? tan(θ) = opposite/adjacent... but we want θ. So: θ = arctan(opposite/adjacent). That's inverse trig. You know the ratio, you want the angle. Same ladder, different question.
Why domains are restricted: sin(30°) = sin(150°) = 0.5. If arcsin had to undo both, it couldn't pick one answer. So we restrict inverse trig functions to specific intervals where they're one-to-one.
arcsin(x)
Domain: [−1, 1]
Range: [−π/2, π/2]
(−90° to 90°)
QI and QIV only
arccos(x)
Domain: [−1, 1]
Range: [0, π]
(0° to 180°)
QI and QII only
arctan(x)
Domain: (−∞, ∞)
Range: (−π/2, π/2)
(−90° to 90°)
QI and QIV only
Restricted ranges on the unit circle arcsin — right half QI ✓ QIV ✓ QII ✗ QIII ✗ −π/2 to π/2 · output always in this range arccos — top half QI ✓ QII ✓ QIII ✗ QIV ✗ 0 to π · output always in this range arctan — QI & QIV QI ✓ QIV ✓ QII ✗ QIII ✗ −π/2 to π/2 (open) · domain: all reals
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Why these specific intervals?
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For a function to have an inverse, it must be one-to-one — each output corresponds to exactly one input. Regular sine repeats every 360°, so we'd have infinite choices for the inverse. The restricted intervals are the conventional choices that include all possible output values while being one-to-one.

arcsin uses [−π/2, π/2] because that's where sine is increasing and covers all values from −1 to 1. arccos uses [0, π] because that's where cosine is decreasing through all values from 1 to −1. arctan uses (−π/2, π/2) open because tangent approaches but never reaches ±90°.
To evaluate inverse trig: ask "what angle in the restricted range has this trig value?" Use your special angle table — 30°, 45°, 60° — and check the sign using the quadrant restrictions.
Worked examples — step by step arcsin(−√3/2) = ? Step 1: Ignore the sign. sin(?) = √3/2 → 60° (π/3) Step 2: arcsin range is [−π/2, π/2]. Step 3: Negative sin → QIV → angle is −60° = −π/3 (= −60°) arccos(−1/2) = ? Step 1: Ignore the sign. cos(?) = 1/2 → 60° (π/3) Step 2: arccos range is [0, π]. Step 3: Negative cos → QII → 180° − 60° = 120° = 2π/3 (= 120°) arctan(−1) = ? Step 1: Ignore the sign. tan(?) = 1 → 45° (π/4) Step 2: arctan range is (−π/2, π/2). Step 3: Negative tan → QIV → angle is −45° = −π/4 (= −45°) sin(arccos(3/5)) = ? Step 1: Let θ = arccos(3/5) → cos θ = 3/5 Step 2: Draw the triangle. hyp=5, adj=3, opp=? Step 3: opp = √(25−9) = 4 → sin θ = 4/5 = 4/5
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Composition with trig — draw the triangle
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When you see sin(arctan(x)) or cos(arcsin(x)) — draw a right triangle. Let the inverse trig expression define the angle, figure out the triangle sides from the given ratio, use Pythagorean theorem for the missing side, then read off what the outer trig function wants. This is the most reliable method and the one that makes the most sense geometrically. The unit circle is just a triangle where the hypotenuse is always 1.
💡 Always check the sign based on which quadrant the inner inverse function puts you in. arcsin and arctan land in QI or QIV. arccos lands in QI or QII.
Interactive inverse trig calculator. Slide a value, pick a function, see the angle — and watch it on the unit circle.
0.50
arcsin(x)
π/6
In degrees
30°
Decimal (rad)
0.524

Showing arcsin(0.5) = 30° on the unit circle

Quick reference

Ranges to memorize
arcsin: [−π/2, π/2]
arccos: [0, π]
arctan: (−π/2, π/2)
Special values:
arcsin(0) = 0
arcsin(1/2) = π/6
arcsin(√2/2) = π/4
arcsin(√3/2) = π/3
arcsin(1) = π/2

Key vocabulary

Inverse function
Undoes the original function. If f(a) = b, then f⁻¹(b) = a. arcsin undoes sin.
Principal value
The one output in the restricted range. arcsin always gives a value in [−π/2, π/2] even though sin has infinitely many angles with that value.
One-to-one
Each output maps to exactly one input. Required for a function to have an inverse. Regular sin is NOT one-to-one — hence the restriction.
Composition
sin(arcsin(x)) = x (for x in [−1,1]). arcsin(sin(x)) = x only if x is in [−π/2, π/2].

Exam reminders

arcsin range: [−π/2, π/2]. Negative outputs are in QIV.
arccos range: [0, π]. Negative cos values give answers between π/2 and π.
arctan range: (−π/2, π/2) open — it never reaches ±π/2.
sin⁻¹(x) ≠ 1/sin(x). It means arcsin(x).
For compositions: draw the right triangle. Let the inverse define the angle, find the missing side, read off the answer.