AP Precalculus 3.9 Inverse Trigonometric Functions — Notes

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Notation & The Big Idea

Input and output are swapped — the output is always an angle

📌 Two Equivalent Notations

sin⁻¹(x) = arcsin(x) cos⁻¹(x) = arccos(x) tan⁻¹(x) = arctan(x)

Read aloud as "arcsine of x", "arccosine of x", "arctangent of x".
The output is always an angle — inverse trig functions undo the original trig function.

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What does an inverse trig function ask?

sin(π6) = 12 asks: "What is the sine of π6?"
arcsin(12) = π6 asks: "What angle has a sine of 12?"
The input and output are swapped. The output of arcsin/arccos/arctan is an angle.

📌 Example 1 — Write sin(π6) = 12 in equivalent arcsine notation.

1

sin(π6) = 12 means: the angle π6 has a sine of 12.

2

Swapping input and output: "12 has an arcsine of π6."

arcsin(12) = π6 or equivalently sin⁻¹(12) = π6

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Restricted Domains & Ranges

Trig functions are periodic — we must restrict the domain to make inverses work

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Why restrict the domain?

A function only has an inverse if it is one-to-one (each output has exactly one input). Trig functions repeat — many angles give the same sincos/tan value. We restrict to a specific interval so the function is one-to-one, then the inverse is well-defined.

sin⁻¹(x) = arcsin(x)

Domain: [−1, 1]

[−π2, π2]

Covers Q4 and Q1 (right half of unit circle). Includes endpoints (closed brackets).

cos⁻¹(x) = arccos(x)

Domain: [−1, 1]

[0, π]

Covers Q1 and Q2 (top half of unit circle). Includes endpoints (closed brackets).

tan⁻¹(x) = arctan(x)

Domain: (−∞, ∞)

(−π2, π2)

Covers Q4 and Q1. Open brackets — π2 and −π2 excluded (asymptotes).

Shaded region = valid output angles for each inverse trig function

Function	Restricted Domain of Original	Range of Inverse	Quadrants
sin⁻¹(x)	[−π2, π2]	[−π2, π2]	Q4 and Q1
cos⁻¹(x)	[0, π]	[0, π]	Q1 and Q2
tan⁻¹(x)	(−π2, π2)	(−π2, π2)	Q4 and Q1

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Key rule: range determines which quadrant your answer is in

arcsin always gives an angle in [−π2, π2] → Q4 or Q1. If the value is negative, the answer is in Q4 (negative angle).
arccos always gives an angle in [0, π] → Q1 or Q2. If the value is negative, the answer is in Q2 (angle between π2 and π).
arctan always gives an angle in (−π2, π2) → Q4 or Q1.

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Example 2 — Evaluating Inverse Trig Expressions

Use unit circle + range restriction to find the correct angle

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Method: find reference angle, then apply the range restriction

Step 1: Find the reference angle using the magnitude of the input.
Step 2: Determine the sign/quadrant based on the range of the inverse function and the sign of the input.
Step 3: Write the final angle (positive if in Q1/Q2 for arccos, negative if in Q4 for arcsin/arctan).

cos⁻¹(−√22)

Reference angle: cos = √22 at π4
Input is negative → angle in Q2 (range [0,π])
Q2 angle with ref π4: π − π4

= 3π4

sin⁻¹(−√32)

Reference angle: sin = √32 at π3
Input is negative → angle in Q4 (range [−π2,π2])
Q4 angle: −π3

= −π3

tan⁻¹(√3)

tan = √3 at π3 (from unit circle)
Input is positive → Q1 (range (−π2,π2))
Answer: π3

= π3

✅ Summary — Example 2 Answers

        a) cos⁻¹(−√22) = 3π4  · 
        b) sin⁻¹(−√32) = −π3  · 
        c) tan⁻¹(√3) = π3
      

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Example 3 — Reflected Points on the Unit Circle

P=(x,y) in Q1 · Q, R, S are reflections · match each inverse expression to the correct point

P=(x,y) in Q1 · Q=(−x,y) in Q2 · R=(−x,−y) in Q3 · S=(x,−y) in Q4

Point	Coordinates	How obtained	Quadrant
P	`(x, y)`	Original point — angle θ	Q1
Q	`(−x, y)`	Reflect over y-axis	Q2
R	`(−x, −y)`	Reflect over origin	Q3
S	`(x, −y)`	Reflect over x-axis	Q4

📌 Example 3 — For each expression, which labeled point does the terminal ray intersect?

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Strategy: The inverse trig function gives an angle. Match the (cos, sin) coordinates of that angle to one of P, Q, R, S. The range restriction tells you which quadrant to look in.

cos⁻¹(x)

cos = x → angle in [0,π]. Since x > 0 (P is in Q1), the angle is in Q1. The point at angle θ is P(x,y).

→ P

sin⁻¹(−y)

sin = −y → range [−π2,π2] → Q4. The angle with sin = −y is in Q4, where the point is (x, −y) = S.

→ S

cos⁻¹(−x)

cos = −x → range [0,π] → Q2 (cosine is negative). The point at angle in Q2 with cos = −x is (−x, y) = Q.

→ Q

tan⁻¹(−yx)

tan = −yx → negative → range (−π2,π2) → Q4. Point in Q4: (x, −y) = S.

→ S

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Quick Reference

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🔑 Key Facts

Output of inverse trig = ANGLE

arcsin range: [−π2, π2] → Q4, Q1

arccos range: [0, π] → Q1, Q2

arctan range: (−π2, π2) → Q4, Q1

Negative input → Q2 (cos) or Q4 (sin, tan)

⚠️ Common Mistakes

❌ arcsin(−12) = 5π6

arcsin range is [−π2, π2]. 5π6 is not in range. Correct: arcsin(−12) = −π6 (Q4, negative angle).

❌ arccos(−1) = π (that's actually correct!) but arccos(−√22) = −3π4

arccos range is [0,π] — NEVER negative. arccos(−√22) = 3π4 (Q2). Arccos cannot give negative angles.

❌ sin⁻¹(x) means 1sin(x)

The −1 exponent on a function name means INVERSE, not reciprocal. sin⁻¹(x) = arcsin(x). The reciprocal of sin is csc(x) = 1sin(x).

❌ arctan has range [−π2, π2] with closed brackets

arctan range is (−π2, π2) — OPEN brackets. ±π2 are excluded because tan is undefined there (asymptotes).

Inverse TrigonometricFunctions

What does an inverse trig function ask?

Why restrict the domain?

Key rule: range determines which quadrant your answer is in

Method: find reference angle, then apply the range restriction

🔗 Related Topics

Inverse Trigonometric
Functions