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Lesson 4: Inverse Trig Functions

Estimated time: 35-40 minutes

Learning Objectives

Why We Need Restricted Domains

Trig functions are not one-to-one (they fail the horizontal line test). To define an inverse, we restrict the domain so each output corresponds to exactly one input.

The Three Main Inverse Functions

y = sin¹(x) = arcsin(x)

Domain: [−1, 1]  |  Range: [−π/2, π/2]

Returns the angle in [−π/2, π/2] whose sine is x.

y = cos¹(x) = arccos(x)

Domain: [−1, 1]  |  Range: [0, π]

Returns the angle in [0, π] whose cosine is x.

y = tan¹(x) = arctan(x)

Domain: (−∞, ∞)  |  Range: (−π/2, π/2)

Returns the angle in (−π/2, π/2) whose tangent is x.

Evaluating Inverse Functions

Example 1: arcsin(1/2)

What angle in [−π/2, π/2] has sine = 1/2? Answer: π/6.

Example 2: arccos(−√2/2)

What angle in [0, π] has cosine = −√2/2? Answer: 3π/4.

Example 3: arctan(1)

What angle in (−π/2, π/2) has tangent = 1? Answer: π/4.

Example 4: arcsin(−1)

What angle in [−π/2, π/2] has sine = −1? Answer: −π/2.

Compositions

sin(arcsin x) = x for x in [−1, 1]

arcsin(sin x) = x only when x is in [−π/2, π/2]

Example 5: Composition

Find cos(arcsin(3/5)).

Let θ = arcsin(3/5), so sinθ = 3/5 and θ in QI. cosθ = 4/5. Answer: 4/5.

Example 6: arcsin(sin(5π/6))

sin(5π/6) = 1/2. arcsin(1/2) = π/6. Not 5π/6 because 5π/6 is outside [−π/2, π/2]. Answer: π/6.

Check Your Understanding

1. Find arccos(0).

π/2 (the angle in [0,π] whose cosine is 0).

2. Find arctan(−√3).

−π/3 (in (−π/2, π/2)).

3. Find sin(arccos(5/13)).

Let θ=arccos(5/13). sinθ=12/13 (QI). Answer: 12/13.

Key Takeaways

Module 6 Complete!