A circle has radius r = 5 cm. What is the arc length subtended by a central angle of 2π/3 radians?
Formula: s = rθ — θ must be in radians!
📐 Explanation
s = rθ = 5 × (2π/3) = 10π/3
Common mistake: confusing s = rθ with the area formula A = ½r²θ. Arc length is LINEAR — it's just radius times the angle.
Unit 2 — The Unit Circle
03 / 20
KEY (cos θ, sin θ)
Unit Circle Coordinates
What is sin(5π/6)?
5π/6 is in Quadrant II. Reference angle = π/6. Sin is positive in QII.
Reference Angle Trick
If θ is in QII: reference angle = π − θ · sin(5π/6) = sin(π − 5π/6) = sin(π/6)
📐 Explanation
5π/6 lives in Quadrant II. Its reference angle is π − 5π/6 = π/6.
sin(π/6) = 1/2, and since sin is positive in QII → sin(5π/6) = 1/2.
04 / 20
KEY ASTC (All Students Take Calculus)
Signs by Quadrant
If cos θ < 0 and sin θ < 0, which quadrant is θ in?
Use ASTC: All–Sin–Tan–Cos are positive in QI, QII, QIII, QIV respectively.
📐 Explanation
ASTC rule: in QIII, both sin and cos are negative. Only tan is positive.
Both cos < 0 AND sin < 0 → answer is Quadrant III.
05 / 20
KEY cos(7π/6) → QIII → negative
Exact Values
Which of the following equals cos(7π/6)? (Be careful about the quadrant!)
📐 Explanation
7π/6 = π + π/6 → Quadrant III. Reference angle = π/6.
cos(π/6) = √3/2. In QIII, cosine is negative → cos(7π/6) = −√3/2.
⚠ Trap: choosing −1/2 (that's sin, not cos!).
Unit 3 — Trig Functions & Their Graphs
06 / 20
KEY Period = 2π/|B|
Period of Sine Function
What is the period of f(x) = sin(3x)?
The B-value stretches/compresses the period. Period = 2π divided by |B|.
Quick Example
sin(x): period = 2π · sin(2x): period = π · sin(½x): period = 4π
📐 Explanation
B = 3. Period = 2π/|3| = 2π/3.
Bigger B → shorter period (graph is compressed horizontally). Smaller B → longer period.
07 / 20
KEY Amplitude = |A|
Amplitude
What is the amplitude of g(x) = −4 cos(x) + 1?
Amplitude is always positive. The −4 flips the graph, but amplitude = |A|.
📐 Explanation
Amplitude = |A| = |−4| = 4. The negative sign reflects the graph but does NOT change amplitude.
The +1 is a vertical shift — it moves the midline up, not the amplitude.
08 / 20
KEY Phase shift = −C/B
Phase Shift
For h(x) = sin(2x − π), what is the phase shift?
Write as sin(B(x − C/B)). Factor out B first if needed.
📐 Explanation
sin(2x − π) = sin(2(x − π/2)). Phase shift = π/2 to the right.
⚠ Trap: using −π directly without dividing by B=2. Always factor B out first!
Unit 4 — Right Triangle Trigonometry
09 / 20
KEY SOH-CAH-TOA
Basic Ratios
In a right triangle, the side opposite angle θ = 3 and the hypotenuse = 5. What is cos θ?
Find the adjacent side first using the Pythagorean theorem: a² + b² = c².
3-4-5 Family
3² + 4² = 5² · So adjacent = 4 · Then CAH: cos = adj/hyp
📐 Explanation
Adjacent = √(5² − 3²) = √(25−9) = √16 = 4.
cos θ = adjacent/hypotenuse = 4/5.
3/5 is sin θ, 3/4 is tan θ — don't mix them up!
10 / 20
KEY tan = sin/cos = opp/adj
Reciprocal Identities
If sin θ = 5/13 and θ is in QI, what is csc θ?
csc is the reciprocal of sin. Simple flip!
📐 Explanation
csc θ = 1/sin θ = 1/(5/13) = 13/5.
Reciprocal pairs: sin↔csc, cos↔sec, tan↔cot. Just flip the fraction!
12/13 is cos θ (from 5-12-13 triangle). 13/12 is sec θ.
Unit 5 — Pythagorean Identities
11 / 20
KEY sin²+cos²=1 → always!
Core Identity
If cos θ = 3/5, what is sin²θ?
Use: sin²θ + cos²θ = 1 → sin²θ = 1 − cos²θ
📐 Explanation
sin²θ = 1 − cos²θ = 1 − (3/5)² = 1 − 9/25 = 16/25.
Note: sin²θ is (sinθ)², not sin(θ²). This is the most-used identity in all of trig!
12 / 20
KEY 1+tan²=sec², 1+cot²=csc²
Derived Identities
Simplify: 1 + tan²θ
📐 Explanation
Divide sin²θ + cos²θ = 1 by cos²θ:
→ tan²θ + 1 = sec²θ → so 1 + tan²θ = sec²θ.
Memory: tan & sec are a pair. sin & csc are a pair. cos & sec are a pair.
Unit 6 — Sum & Difference Formulas
13 / 20
KEY cos(A−B) = cosA·cosB + sinA·sinB
Cosine Difference Formula
Find the exact value of cos(15°).
Split 15° = 45° − 30°, then apply the cosine difference formula.
Formula Recall
cos(A − B) = cos A cos B + sin A sin B
📐 Explanation
cos(45°−30°) = cos45°cos30° + sin45°sin30°
= (√2/2)(√3/2) + (√2/2)(1/2)
= √6/4 + √2/4 = (√6 + √2)/4
⚠ Trap: using subtraction instead of addition (that would be sin formula!).
14 / 20
KEY sin(A+B) = sinA·cosB + cosA·sinB
Sine Sum Formula
Expand sin(x + π/3) using the sum formula.
📐 Explanation
sin(A + B) = sinA cosB + cosA sinB
→ sin(x + π/3) = sin x · cos(π/3) + cos x · sin(π/3)
= sin x · (1/2) + cos x · (√3/2)
⚠ sin(x + π/3) ≠ sin x + sin(π/3). Trig doesn't distribute like that!
Unit 7 — Double & Half Angle Formulas
15 / 20
KEY sin(2θ) = 2 sin θ cos θ
Double Angle — Sine
If sin θ = 1/2 and θ is in QI, what is sin(2θ)?
Find cos θ first. Then apply: sin(2θ) = 2 sin θ cos θ.
📐 Explanation
sin θ = 1/2 → cos θ = √3/2 (QI, so positive).
sin(2θ) = 2 · (1/2) · (√3/2) = √3/2.
Alternatively: sin θ = 1/2 → θ = π/6, so 2θ = π/3, and sin(π/3) = √3/2. ✓
16 / 20
KEY cos(2θ) has THREE forms — pick wisely!
Double Angle — Cosine
Which expression is NOT equal to cos(2θ)?
📐 Explanation
Options A, B, C are all valid forms of cos(2θ).
Option D — 2sinθ cosθ = sin(2θ), NOT cos(2θ). This is the double angle for SINE.
All three cos(2θ) forms come from substituting sin²+cos²=1 into cos²θ−sin²θ.
Unit 8 — Inverse Trig & Equations
17 / 20
KEY arcsin range: [−π/2, π/2]
Inverse Sine Range
What is arcsin(−1/2)?
arcsin outputs angles only in [−π/2, π/2]. Look for the QI or QIV angle.
📐 Explanation
arcsin outputs values in [−π/2, π/2] only.
sin(−π/6) = −1/2 and −π/6 ∈ [−π/2, π/2] → arcsin(−1/2) = −π/6.
7π/6 also has sin = −1/2, but it's outside the range of arcsin. ⚠
18 / 20
KEY 2 solutions per period for sin/cos!
Solving Trig Equations
Solve for x in [0, 2π): 2sinx = √2
Isolate sin x. Then find ALL angles in [0, 2π) where sin equals that value.
📐 Explanation
2sin x = √2 → sin x = √2/2.
sin is positive in QI and QII → x = π/4 and x = π − π/4 = 3π/4.
⚠ 7π/4 gives sin = −√2/2 (QIV). Always check the sign of the original value!
19 / 20
KEY Factor → Zero Product Property
Factoring Trig Equations
Find all solutions in [0, 2π) for: 2sin²x − sinx − 1 = 0
Let u = sin x. Factor like a quadratic: 2u² − u − 1 = 0.
Factoring Hint
2u² − u − 1 = (2u + 1)(u − 1) = 0 → u = −1/2 or u = 1
📐 Explanation
(2sinx + 1)(sinx − 1) = 0
sin x = −1/2 → x = 7π/6, 11π/6 (QIII, QIV)
sin x = 1 → x = π/2
All solutions: x = π/2, 7π/6, 11π/6
20 / 20
KEY Verify both sides ≠ simplify one side only
Verifying Identities
Which step correctly begins verifying tan²θ + 1 = sec²θ from the left side?
Replace tan with sin/cos, then simplify. Only work on ONE side at a time!
📐 Explanation
Option A is the correct verification:
tan²θ + 1 = (sin²θ/cos²θ) + (cos²θ/cos²θ) = (sin²θ + cos²θ)/cos²θ = 1/cos²θ = sec²θ ✓
Option B "moves terms across" — that's solving, not verifying.
Option C: √(tan²+1) ≠ tanθ+1 (can't distribute square roots like that!).
Option D: circular reasoning — never substitute the thing you're trying to prove.
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Quiz Complete!
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You've completed all 20 core trigonometry problems.