AP PRECALCULUS — UNIT 3A · TRIGONOMETRIC & POLAR FUNCTIONS
3.3B — Sine and Cosine Function Values
Notes — Finding Values Without the Full Table
💡 Learning Objectives (3.3.A Part 2)
By the end of this lesson you will be able to:
- Use the Pythagorean identity to find one trig value given the other
- Solve trig equations on a restricted interval using reference angles
- Reason about which angles share the same sine or cosine value
- Translate between (cos θ, sin θ) coordinates and a given angle
1. The Pythagorean Identity as a Tool
You already know cos²(θ) + sin²(θ) = 1. Turn this into a problem-solving tool: if you know ONE of sin(θ) or cos(θ), plus the quadrant, you can find the other.
📘 Example — Finding a partner trig value
Suppose sin(θ) = 3/5 and θ is in Quadrant II. Find cos(θ).
- Use identity: cos²(θ) = 1 − sin²(θ) = 1 − 9/25 = 16/25
- Take square root: cos(θ) = ±4/5
- Apply quadrant rule: in Quadrant II, cosine is negative, so cos(θ) = −4/5
2. Equations of the Form sin(θ) = c
On one full revolution, the equation sin(θ) = c (where |c| ≤ 1) has either ONE or TWO solutions:
- If c = 1: θ = π/2 only
- If c = −1: θ = 3π/2 only
- If −1 < c < 1 and c ≠ 0: there are exactly two solutions in [0, 2π)
- If c = 0: θ = 0 and θ = π in [0, 2π)
📘 Example — Solve sin(θ) = √3/2 on [0, 2π)
- Where is sine positive? Quadrants I and II
- Reference angle with sin = √3/2 is π/3
- Quadrant I solution: θ = π/3
- Quadrant II solution: θ = π − π/3 = 2π/3
- Solutions: θ ∈ {π/3, 2π/3}
3. Equations of the Form cos(θ) = c
Same structure, different quadrants:
📘 Example — Solve cos(θ) = −1/2 on [0, 2π)
- Where is cosine negative? Quadrants II and III
- Reference angle with cos = 1/2 is π/3
- Quadrant II solution: θ = π − π/3 = 2π/3
- Quadrant III solution: θ = π + π/3 = 4π/3
- Solutions: θ ∈ {2π/3, 4π/3}
4. All Solutions Over the Whole Real Line
Because sine and cosine are periodic with period 2π, if θ₀ is a solution then so is θ₀ + 2πk for every integer k. Write general solutions this way:
📘 Example — General solutions
Solve cos(θ) = 1/2 over all real θ.
- Quadrant I: θ = π/3 + 2πk
- Quadrant IV: θ = −π/3 + 2πk (equivalent to 5π/3 + 2πk)
- Combined: θ = ±π/3 + 2πk, k ∈ ℤ
5. From a Point to an Angle
Sometimes you are given the (x, y) coordinates of a point on the unit circle and asked for the angle. Use both coordinates together to pin down the angle uniquely (or to the right family of coterminal values):
📘 Example — Point to angle
Which angle in [0, 2π) has unit-circle coordinates (−√2/2, −√2/2)?
- Both coordinates negative ⇒ Quadrant III
- Reference angle with |cos| = |sin| = √2/2 is π/4
- So θ = π + π/4 = 5π/4
6. From a Non-Unit Circle
If your point is on a circle of radius r (not 1), convert before computing. If (x, y) is on a circle of radius r centered at the origin, then:
cos(θ) = x/r, sin(θ) = y/r
📘 Example — Circle of radius 5
Point (3, −4) lies on a circle of radius 5 (check: 3² + 4² = 25).
- cos(θ) = 3/5
- sin(θ) = −4/5
- Quadrant IV (positive x, negative y), so θ is between 3π/2 and 2π
7. Useful Identities Beyond Pythagoras
Two symmetries are worth knowing for quick mental calculation:
- Supplement: sin(π − θ) = sin(θ) and cos(π − θ) = −cos(θ)
- Complement: sin(π/2 − θ) = cos(θ) and cos(π/2 − θ) = sin(θ)
The complement identity says sine and cosine swap under the substitution θ → π/2 − θ. That is why they are called ‘co-functions’ of each other.
8. Summary
- Use cos²(θ) + sin²(θ) = 1 and the quadrant to pin down one trig value from the other
- Equations sin(θ) = c or cos(θ) = c typically have two solutions per rotation — one per quadrant where the sign matches
- Add ‘+ 2πk, k ∈ ℤ’ to extend solutions to all real angles
- For a point on a circle of radius r, divide coordinates by r to get trig values
- Supplement and complement identities produce quick related-angle values