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Pre-Calculus · Trigonometry

Master Trig
from Zero

20 core questions — covering every unit that trips students up. Study, type, repeat.

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Unit 1

Angles & Radian Measure

🔁
⚡ Quick Recall
RADIAN = ARC / RADIUS  |  180° = π rad  |  To convert °→rad: multiply by π/180
1
Degree ↔ Radian

Convert 270° to radians. Express your answer as a fraction of π.

Multiply degrees by π/180 and simplify.

270 × (π / 180) = ?
Solution: 270 × π/180 = 270π/180 = 3π/2.
Think: 90° = π/2, so 270° = 3 × π/2. ✓
✅ Correct!
2
Arc Length

A circle has radius r = 5 cm. A central angle is θ = π/3 radians.
What is the arc length s?

Arc length formula: s = rθ

s = r · θ = 5 · (π/3)
Solution: s = r · θ = 5 · π/3 = 5π/3 cm.
Always keep the fraction form unless decimals are required.
✅ Correct!
Unit 2

Unit Circle & Exact Values

⭕
⚡ Quick Recall
SOHCAHTOA — Sine=Opp/Hyp · Cos=Adj/Hyp · Tan=Opp/Adj
All Students Take Calculus → A·S·T·C = signs by quadrant (I+, II sin+, III tan+, IV cos+)
3
Exact Values

What is cos(π/3)?

π/3 = 60°. Recall the 30-60-90 triangle: sides 1, √3, 2.

In a 30-60-90 triangle: cos 60° = adjacent/hypotenuse = 1/2.
Tricky: cos(π/3) = 1/2 but sin(π/3) = √3/2. Don't swap them!
✅ Correct!
4
Quadrant Signs

An angle θ is in Quadrant III. Which trig function is positive in this quadrant?

Recall: A·S·T·C — All, Sin, Tan, Cos — starting Quadrant I going counterclockwise.

A·S·T·C: Quadrant I = All positive; II = Sine; III = Tangent; IV = Cosine.
In Q III both sin and cos are negative, so tan = sin/cos = (−)/(−) = positive. ✓
✅ Correct!
5
Reference Angle

What is the reference angle for θ = 5π/6?

5π/6 lies in Quadrant II. Reference angle = π − θ.

θ′ = π − (5π/6) = ?
π − 5π/6 = 6π/6 − 5π/6 = π/6.
The reference angle is always the acute angle to the nearest x-axis.
✅ Correct!
Unit 3

Trig Functions & Their Graphs

📈
⚡ Quick Recall
y = A·sin(Bx + C) + D
Amplitude=|A| · Period=2π/|B| · Phase Shift=−C/B · Vertical Shift=D
6
Amplitude & Period

For y = 3 sin(2x), what is the period?

Period = 2π / |B|, where B is the coefficient of x.

B = 2, so Period = 2π / 2 = π.
Amplitude is |3| = 3 (a separate concept—don't mix them up!).
✅ Correct!
7
Phase Shift

What is the phase shift of y = cos(x − π/4)?

Phase shift = −C/B. Here the function is cos(Bx + C) form — find C carefully.

y = cos(x − π/4)  ⟹  C = −π/4, B = 1
Phase shift = −C/B = −(−π/4)/1 = π/4 to the right.
⚠️ (x − π/4) → shift RIGHT. (x + π/4) → shift LEFT. Easy to confuse!
✅ Correct!
8
Period of Tangent

What is the period of y = tan(3x)?

tan has period π (not 2π!), so period = π / |B|.

Period of tangent = π / B = π / 3 = π/3.
Common mistake: using 2π/B (that's for sine/cosine). For tan/cot, always divide π by B.
✅ Correct!
Unit 4

Inverse Trig Functions

🔄
⚡ Quick Recall
arcsin range: [−π/2, π/2] (Q I & IV)
arccos range: [0, π] (Q I & II)   arctan range: (−π/2, π/2)
9
arcsin

Evaluate arcsin(−1/2). Give the answer in radians.

arcsin outputs in [−π/2, π/2]. sin(π/6) = 1/2, so sin(−π/6) = −1/2.

arcsin(−1/2) = −π/6.
arcsin always returns a value in [−π/2, π/2], so NOT 7π/6 (that's outside the range).
✅ Correct!
10
Composition

Evaluate cos(arcsin(3/5)) exactly. (Assume angle in Q I.)

Let θ = arcsin(3/5). Then sin θ = 3/5. Use a right triangle to find cos θ.

sin θ = 3/5  →  opp=3, hyp=5  →  adj=?
adj = √(5²−3²) = √(25−9) = √16 = 4. So cos θ = adj/hyp = 4/5.
This "right triangle trick" is the fastest method for trig composition problems.
✅ Correct!
Unit 5

Trig Identities

🧮
⚡ Quick Recall
sin²θ + cos²θ = 1  → divide by cos² →  tan²θ + 1 = sec²θ
Reciprocals: csc=1/sin · sec=1/cos · cot=1/tan
11
Pythagorean Identity

If sin θ = 5/13 and θ is in Quadrant I, find cos θ.

Use sin²θ + cos²θ = 1.

cos²θ = 1 − (5/13)² = 1 − 25/169 = 144/169.
cos θ = √(144/169) = 12/13 (positive since Q I).
✅ Correct!
12
Simplify Using Identities

Simplify: sec²θ − 1

Recall the identity derived from the Pythagorean identity by dividing by cos²θ.

From tan²θ + 1 = sec²θ, subtract 1: sec²θ − 1 = tan²θ.
This is one of the most frequently tested simplification patterns!
✅ Correct!
13
Double Angle Formula

If sin θ = 3/5 and θ ∈ Q I, find sin(2θ).

sin(2θ) = 2 sin θ cos θ. First find cos θ.

sin(2θ) = 2 · sinθ · cosθ
cos θ = 4/5 (from Pythagorean identity, Q I).
sin(2θ) = 2 · (3/5) · (4/5) = 2 · 12/25 = 24/25. ✓
✅ Correct!
Unit 6

Solving Trig Equations

🎯
⚡ Quick Recall
Find the reference angle → apply A·S·T·C for all quadrant solutions → add + 2πk (sin/cos) or + πk (tan) for general solutions
14
Basic Trig Equation

Solve 2 sin x − 1 = 0 for x ∈ [0, 2π). List both solutions.

Isolate sin x first, then find all angles in [0, 2π) where sin has that value.

sin x = 1/2 → reference angle = π/6.
sin is positive in Q I and Q II: x = π/6 and x = π − π/6 = 5π/6.
✅ Correct!
15
Quadratic-Type Equation

Solve 2cos²x − cos x − 1 = 0 for x ∈ [0, 2π).

Factor like a quadratic: let u = cos x, solve (2u + 1)(u − 1) = 0.

2cos²x − cosx − 1 = (2cosx + 1)(cosx − 1) = 0
Case 1: cos x = 1 → x = 0.
Case 2: cos x = −1/2 → ref = π/3, Q II & Q III → x = 2π/3 and 4π/3.
Three solutions total: 0, 2π/3, 4π/3.
✅ Correct!
Unit 7

Sum, Difference & Half-Angle Formulas

➕
⚡ Quick Recall
sin(A±B) = sinA cosB ± cosA sinB
cos(A±B) = cosA cosB ∓ sinA sinB  (signs FLIP for cos!)
Half-angle: sin(θ/2) = ±√((1−cosθ)/2) · cos(θ/2) = ±√((1+cosθ)/2)
16
Sum Formula

Find the exact value of sin(75°) using sin(45° + 30°).

sin(A+B) = sin A cos B + cos A sin B. Use exact values for 45° and 30°.

sin 45°·cos 30° + cos 45°·sin 30° = (√2/2)(√3/2) + (√2/2)(1/2)
= √6/4 + √2/4 = (√6 + √2)/4. ✓
✅ Correct!
17
Half-Angle Formula

Use the half-angle formula to find cos(π/8) exactly.

π/8 = (π/4)/2. Use cos(θ/2) = +√((1 + cos θ)/2) — positive since π/8 is in Q I.

cos(π/8) = √( (1 + cos(π/4)) / 2 )
cos(π/4) = √2/2, so: √((1 + √2/2)/2) = √((2+√2)/4) = √(2+√2)/2.
Leave in exact radical form — this is the expected final answer.
✅ Correct!
Unit 8

Law of Sines & Law of Cosines

📐
⚡ Quick Recall
Law of Sines: a/sinA = b/sinB = c/sinC
Law of Cosines: c² = a² + b² − 2ab·cosC  (use when SAS or SSS)
18
Law of Sines — Missing Side

In triangle ABC: A = 30°, B = 45°, a = 6. Find side b (round to 1 decimal).

Use a/sin A = b/sin B, cross multiply and solve for b.

6/sin(30°) = b/sin(45°)
b = 6·sin(45°)/sin(30°) = 6·(√2/2)/(1/2) = 6·√2 ≈ 6 × 1.414 ≈ 8.5. ✓
✅ Correct!
19
Law of Cosines — Missing Side

Triangle: a = 5, b = 7, C = 60°. Find side c (round to 2 decimal places).

c² = a² + b² − 2ab cos C. Plug in and simplify.

c² = 5² + 7² − 2(5)(7)cos(60°)
c² = 25 + 49 − 70·(1/2) = 74 − 35 = 39.
c = √39 ≈ 6.24. ✓
✅ Correct!
20
★ Final Boss — Mixed Identities

Prove and simplify:

sin²θ / (1 − cosθ) = ?

Replace sin²θ using the Pythagorean identity, then factor and cancel.

sin²θ = 1 − cos²θ = (1 − cosθ)(1 + cosθ).
Divide: (1 − cosθ)(1 + cosθ)/(1 − cosθ) = 1 + cosθ. ✓
Key: always try factoring with Pythagorean identity first!
✅ Brilliant! 🎉