Unit 1

Angles & Radian Measure

📐 Core Concepts

Angle Measurement: Angles are measured in degrees (°) or radians (rad).

Radian Definition: 1 radian = the angle subtended by an arc equal in length to the radius.

Key Equivalences:

360° = 2π rad  →  180° = π rad

Arc Length:

s = r · θ   (θ must be in radians)

Sector Area:

A = ½ r² θ   (θ in radians)

Coterminal Angles: Angles sharing the same terminal side. Add or subtract multiples of 360° (or 2π).

Reference Angle: The acute positive angle between the terminal side and the x-axis.

⭐ Must Memorize — Degree ↔ Radian
30° = π/6  |  45° = π/4  |  60° = π/3  |  90° = π/2 120° = 2π/3  |  135° = 3π/4  |  150° = 5π/6  |  180° = π 210° = 7π/6  |  225° = 5π/4  |  240° = 4π/3  |  270° = 3π/2 300° = 5π/3  |  315° = 7π/4  |  330° = 11π/6  |  360° = 2π To convert: deg × (π/180) = rad  |  rad × (180/π) = deg
✏️ Worked Example

Convert 210° to radians. Find a positive coterminal angle.

1210° × (π/180) = 210π/180 = 7π/6 rad
2Positive coterminal: 7π/6 + 2π = 7π/6 + 12π/6 = 19π/6
Answer: 7π/6 rad  |  positive coterminal = 19π/6
Q01
Radian Conversion
Convert 315° to radians. Express your answer as an exact fraction of π.

Also state one negative coterminal angle (in radians).

Work Space
Q02
Arc Length & Sector Area
A circle has radius 8 cm. A central angle of 5π/6 radians subtends an arc.

(a) Find the arc length.  (b) Find the area of the sector. Leave answers exact.

Work Space
Q03
Reference Angle
Find the reference angle for each: (a) 7π/4   (b) 5π/3   (c) 240°

Express radian answers as exact fractions of π.

Work Space
Q04
Coterminal & Quadrant
Angle θ = -13π/6.  (a) Find the smallest positive coterminal angle.  (b) In which quadrant does the terminal side lie?
Work Space
Unit 2

The Unit Circle & Trigonometric Functions

📐 Core Concepts

Unit Circle: Radius = 1, centered at origin. Any point = (cos θ, sin θ).

Six Trig Functions:

sin θ = y/r  |  cos θ = x/r  |  tan θ = y/x csc θ = r/y  |  sec θ = r/x  |  cot θ = x/y   (r = 1 on unit circle)

Signs by Quadrant — ASTC Rule: All positive (Q1), Sin (Q2), Tan (Q3), Cos (Q4).

Pythagorean Identity:

sin²θ + cos²θ = 1

Even/Odd: cos(−θ) = cos θ (even)  |  sin(−θ) = −sin θ (odd)  |  tan(−θ) = −tan θ (odd)

⭐ Must Memorize — Unit Circle Key Values
θ = 0:    (1, 0)         sin 0 = 0,     cos 0 = 1 θ = π/6:  (√3/2, 1/2)   sin π/6 = 1/2,   cos π/6 = √3/2 θ = π/4:  (√2/2, √2/2) sin π/4 = √2/2, cos π/4 = √2/2 θ = π/3:  (1/2, √3/2)   sin π/3 = √3/2, cos π/3 = 1/2 θ = π/2:  (0, 1)         sin π/2 = 1,   cos π/2 = 0

Extend to all quadrants using symmetry and ASTC signs.

✏️ Worked Example

Find all six trig values for θ = 5π/3.

15π/3 is in Q4 (between 3π/2 and 2π). Reference angle = 2π − 5π/3 = π/3
2Unit circle at π/3: cos = 1/2, sin = √3/2
3Q4 signs: cos > 0, sin < 0
sin(5π/3) = −√3/2  |  cos(5π/3) = 1/2  |  tan(5π/3) = −√3
csc(5π/3) = −2√3/3  |  sec(5π/3) = 2  |  cot(5π/3) = −√3/3
Q05
Unit Circle Values
Find exact values of all six trig functions for θ = 7π/6.

Show which quadrant and the reference angle used.

Work Space
Q06
Pythagorean Identity
Given sin θ = −3/5 and θ is in Quadrant III, find cos θ, tan θ, and sec θ.

Show all algebraic steps.

Work Space
Q07
Even/Odd & Symmetry
Without a calculator, evaluate: (a) cos(−7π/6)   (b) sin(−5π/4)   (c) tan(−π/3)

Justify each answer using even/odd properties.

Work Space
Q08
Finding Angles from Values
Find all angles θ ∈ [0, 2π) such that cos θ = −1/2.

Express answers as exact values in radians.

Work Space
Unit 3

Graphs of Trigonometric Functions

📐Core Concepts

Standard Form:

y = A sin(Bx + C) + D   or   y = A cos(Bx + C) + D

Amplitude: |A| — half the distance from midline to peak

Period: T = 2π/|B| for sin/cos  |  T = π/|B| for tan/cot

Phase Shift: −C/B (negative = left, positive = right)

Vertical Shift: D — the midline is y = D

sin x key points: 0 → max (π/2) → 0 (π) → min (3π/2) → 0 (2π)

cos x key points: max (0) → 0 (π/2) → min (π) → 0 (3π/2) → max (2π)

tan x: period = π; vertical asymptotes at x = π/2 + nπ

⭐Must Memorize — Transformation Steps
1. Identify A, B, C, D from y = A sin(Bx + C) + D 2. Amplitude = |A|  |  Period = 2π/|B|  |  Phase shift = −C/B  |  Midline y = D 3. Key x-values: start, 1/4, 1/2, 3/4, full period 4. If A < 0 → graph is reflected vertically
✏️Worked Example

State all features of y = −3 sin(2x − π) + 1

1A = −3 → Amplitude = 3, graph reflected vertically
2B = 2 → Period = 2π/2 = π
3C = −π → Phase shift = −(−π)/2 = π/2 (right)
4D = 1 → Midline y = 1; Max = 4; Min = −2
Amplitude 3 | Period π | Phase shift π/2 right | Midline y = 1 | Max 4 | Min −2
Q09
Amplitude, Period, Phase Shift
For y = 4 cos(3x + π/2) − 2, find: (a) amplitude, (b) period, (c) phase shift, (d) midline, (e) max and min values.
Work Space
Q10
Writing the Equation
A sinusoidal function has amplitude 5, period 4π, phase shift π/3 right, midline y = −1, and starts at a maximum. Write the equation in the form y = A cos(Bx + C) + D.
Work Space
Q11
Tangent Graph
For y = 2 tan(x/2 − π/4): (a) find the period, (b) find two consecutive vertical asymptotes, (c) identify the phase shift.
Work Space
Q12
Graph Analysis
A graph completes one full cycle from x = π/6 to x = 7π/6 with maximum 5 and minimum −1.

(a) Find amplitude, period, and midline.  (b) Write a sine equation for the graph.

Work Space
Unit 4

Trigonometric Identities & Equations

📐Core Concepts

Pythagorean Identities:

sin²θ + cos²θ = 1  |  1 + tan²θ = sec²θ  |  1 + cot²θ = csc²θ

Sum & Difference:

sin(A ± B) = sin A cos B ± cos A sin B cos(A ± B) = cos A cos B ∓ sin A sin B tan(A ± B) = (tan A ± tan B) / (1 ∓ tan A tan B)

Double Angle:

sin 2θ = 2 sin θ cos θ cos 2θ = cos²θ − sin²θ = 2cos²θ − 1 = 1 − 2sin²θ

Half Angle:

sin(θ/2) = ±√((1−cosθ)/2)  |  cos(θ/2) = ±√((1+cosθ)/2)
⭐Must Memorize — Identity & Equation Strategy
1. Work on ONE side only (usually the more complex side) 2. Convert everything to sin and cos if stuck 3. Look for Pythagorean substitution opportunities 4. Factor or multiply by conjugate when needed 5. For equations: isolate → unit circle → all solutions in interval
✏️Worked Example

Verify the identity: tan²θ + 1 = sec²θ

1Start with left side: tan²θ + 1 = (sin²θ / cos²θ) + 1
2= (sin²θ + cos²θ) / cos²θ
3= 1 / cos²θ   [since sin²θ + cos²θ = 1]
= sec²θ  ✓
Q13
Identity Verification
Prove the identity: (sin θ + cos θ)² = 1 + 2 sin θ cos θ

Show every algebraic step clearly.

Work Space
Q14
Sum/Difference Formulas
Find the exact value of cos(75°) using the identity cos(A − B).

Express the answer with rationalized denominator.

Work Space
Q15
Double-Angle Formula
Given sin θ = 5/13 and θ is in Quadrant II, find exact values of: (a) sin 2θ   (b) cos 2θ   (c) tan 2θ
Work Space
Q16
Solving Trig Equations
Solve for all x ∈ [0, 2π):   2 sin²x − sin x − 1 = 0

Give all solutions as exact values. Show the factoring step.

Work Space
Unit 5

Laws of Sines & Cosines / Applications

📐Core Concepts

Law of Sines:

a / sin A = b / sin B = c / sin C

Use when: AAS, ASA, or SSA (ambiguous case)

Ambiguous Case (SSA): Let h = b sin A. If a < h → no triangle; a = h → 1 right triangle; h < a < b → 2 triangles; a ≥ b → 1 triangle.

Law of Cosines:

c² = a² + b² − 2ab cos C cos C = (a² + b² − c²) / (2ab)

Use when: SAS or SSS

Area of Triangle:

Area = ½ ab sin C

Heron's Formula (SSS): s = (a+b+c)/2; Area = √(s(s−a)(s−b)(s−c))

⭐Must Memorize — Which Law to Use?
AAS or ASA  →  Law of Sines SSA          →  Law of Sines (check ambiguous case!) SAS          →  Law of Cosines (find opposite side first) SSS          →  Law of Cosines (find any angle)
✏️Worked Example

In triangle ABC: a = 7, B = 38°, C = 65°. Find b.

1A = 180° − 38° − 65° = 77°
2Law of Sines: b / sin B = a / sin A
3b = 7 sin 38° / sin 77° ≈ 7(0.6157) / 0.9744
b ≈ 4.42
Q17
Law of Sines — AAS
In triangle ABC: angle A = 40°, angle B = 75°, side a = 12. Find sides b and c to the nearest hundredth.
Work Space
Q18
Ambiguous Case (SSA)
In triangle ABC: a = 10, b = 14, A = 30°. Determine how many triangles are possible and find all valid solutions.

Show the height comparison step (h = b sin A).

Work Space
Q19
Law of Cosines — SAS
In triangle ABC: a = 8, c = 11, B = 52°. (a) Find side b. (b) Find angle A. (c) Find the area of the triangle.

Round all answers to 2 decimal places.

Work Space
Q20
Real-World Application
Two ships leave a harbor simultaneously. Ship A sails 15 km on bearing N 35° E. Ship B sails 22 km on bearing S 60° E. Find the distance between the two ships.

Draw and label a diagram. Use the Law of Cosines. Round to 2 decimal places.

Work Space

Answer Key

Full Step-by-Step Solutions — All 20 Problems

Q01
Radian Conversion

315° × (π/180) = 315π/180; GCD = 45

315° = 7π/4 rad

Negative coterminal: 7π/4 − 2π = −π/4

Negative coterminal = −π/4

Q02
Arc Length & Sector Area

r = 8, θ = 5π/6

(a) s = rθ = 8 · (5π/6) = 40π/6

(a) s = 20π/3 cm ≈ 20.94 cm

(b) A = ½r²θ = ½(64)(5π/6) = 160π/6

(b) A = 80π/3 cm² ≈ 83.78 cm²

Q03
Reference Angle

(a) 7π/4 in Q4: ref = 2π − 7π/4

(a) π/4

(b) 5π/3 in Q4: ref = 2π − 5π/3

(b) π/3

(c) 240° in Q3: ref = 240° − 180°

(c) 60°

Q04
Coterminal & Quadrant

−13π/6 + 2π = −π/6; −π/6 + 2π = 11π/6

(a) Smallest positive coterminal = 11π/6

11π/6 is between 3π/2 and 2π

(b) Quadrant IV

Q05
Unit Circle Values

7π/6 in Q3; ref angle = π/6; both negative

sin = −1/2 | cos = −√3/2 | tan = √3/3

csc = −2 | sec = −2√3/3 | cot = √3

Q06
Pythagorean Identity

sin²θ + cos²θ = 1 → 9/25 + cos²θ = 1 → cos²θ = 16/25

Q3: cos θ < 0

cos θ = −4/5 | tan θ = 3/4 | sec θ = −5/4

Q07
Even/Odd & Symmetry

(a) cos even: cos(−7π/6) = cos(7π/6)

(a) −√3/2

(b) sin odd: −sin(5π/4) = −(−√2/2)

(b) √2/2

(c) tan odd: −tan(π/3)

(c) −√3

Q08
Finding Angles

cos θ = −1/2; ref angle = π/3; neg in Q2 & Q3

θ = 2π/3 and θ = 4π/3

Q09
Amplitude, Period, Phase Shift

A = 4, B = 3, C = π/2, D = −2

(a) Amplitude = 4

(b) Period = 2π/3

(c) Phase shift = −π/6 (left)

(d) Midline y = −2

(e) Max = 2; Min = −6

Q10
Writing the Equation

A = 5; B = 2π/(4π) = 1/2; C = −π/6; D = −1

y = 5 cos(½x − π/6) − 1

Q11
Tangent Graph

B = 1/2, C = −π/4

(a) Period = π/(1/2) = 2π

(b) Asymptotes: x = 3π/2 and x = 7π/2

(c) Phase shift = π/2 right

Q12
Graph Analysis

Period = 7π/6 − π/6 = π; Amp = (5−(−1))/2 = 3; Midline y = 2

(a) Amp = 3 | Period = π | Midline y = 2

B = 2; phase shift π/6 → C = −π/3

(b) y = 3 sin(2x − π/3) + 2

Q13
Identity Verification

(sin θ + cos θ)² = sin²θ + 2 sin θ cos θ + cos²θ

= (sin²θ + cos²θ) + 2 sin θ cos θ

= 1 + 2 sin θ cos θ  ✓

Q14
Sum/Difference Formula

cos 75° = cos(45°+30°) = cos45°cos30° − sin45°sin30°

= (√2/2)(√3/2) − (√2/2)(1/2) = √6/4 − √2/4

cos 75° = (√6 − √2) / 4

Q15
Double-Angle Formula

sin θ = 5/13, Q2 → cos θ = −12/13

(a) sin 2θ = 2(5/13)(−12/13) = −120/169

(b) cos 2θ = 144/169 − 25/169 = 119/169

(c) tan 2θ = −120/119

Q16
Solving Trig Equations

Factor: (2 sin x + 1)(sin x − 1) = 0

sin x = −1/2 → x = 7π/6, 11π/6

sin x = 1 → x = π/2

x = π/2, 7π/6, 11π/6

Q17
Law of Sines — AAS

C = 65°; b = 12 sin75°/sin40°; c = 12 sin65°/sin40°

b ≈ 18.03 | c ≈ 16.92

Q18
Ambiguous Case (SSA)

h = 14 sin30° = 7; h < a=10 < b=14 → 2 triangles

Triangle 1: B≈44.43°, C≈105.57°, c≈19.27

Triangle 2: B≈135.57°, C≈14.43°, c≈4.99

Q19
Law of Cosines — SAS

b² = 64+121−2(8)(11)cos52° = 185−108.36 = 76.64

(a) b ≈ 8.75

cosA = (76.64+121−64)/(2·8.75·11) ≈ 0.6942

(b) A ≈ 46.10°

(c) Area = ½(8)(11)sin52° ≈ 34.67 sq units

Q20
Real-World Application

Angle between bearings = 180°−35°−60° = 85°

d² = 15²+22²−2(15)(22)cos85° = 709−57.53 = 651.47

Distance ≈ 25.52 km

All answers verified. Approximate decimals rounded to 2 decimal places unless stated otherwise.