Cogito
AP Precalculus · Chapter 6 · Lesson 3
Inverse Trigonometric Functions
Going back from a ratio to an angle.
12 problems · about 22 minutes · AP Precalculus 3.11, 3.12
What this lesson teaches
The student applies inverse trigonometric functions and finds all solutions to trigonometric equations.
- An inverse trigonometric function needs a restricted range to exist.
- A calculator returns one solution; the others must be constructed.
- For sine use θ and π − θ; for cosine use θ and −θ.
Warm Up
Straightforward practice. Get the method working first.
5 problemssin θ = 0.5 gives 30°. What is the other solution below 360°, in degrees?
Answer 150
Why 150°.
Why does a calculator return only one solution?
Answer The inverse has a restricted range, so it must pick one.
Why The restricted range.
sin θ = 1. What is θ in degrees?
Answer 90
Why The peak.
cos θ = 1. What is θ in degrees?
Answer 0
Why The point (1, 0).
cos θ = −1. What is θ in degrees?
Answer 180
Why The point (−1, 0).
Build It Up
The same ideas with more to keep track of.
3 problemssin θ = 0.866 gives 60°. The other solution below 360°?
Answer 120
Why 180 − 60.
sin θ = 2. How many solutions?
Answer 0
Why Sine never exceeds 1.
sin θ = 0.5 with θ between 0° and 720°. How many solutions?
Answer 4
Why Two per turn, over two turns.
Stretch Yourself
Mixed problems. Work out what kind of question it is before you start.
4 problemsMatch each inverse function with its range.
Answer arcsin → Negative 90 to 90 degrees, ends included; arccos → Zero to 180 degrees; arctan → Negative 90 to 90 degrees, ends excluded
Why Only arccos covers the upper half of the circle.
sin θ = 0.643 gives 40°. The other solution below 360°?
Answer 140
Why 180 − 40.
The Second Solution: sin θ = 0.5 gives 30°. What is the other solution below 360°, in degrees?
Answer 150 degrees
Why 150°.
The Range: How many degrees does the range of arccos span?
Answer 180
Why 180.