Unit 1 · Angle Measures & Unit Circle
⚡ RAD × (180/π) = DEG · DEG × (π/180) = RAD
Convert 135° to radians. Express your answer as a fraction of π.
Quick Example
60° → 60 × (π/180) = π/3 rad
✗ Let's see why.
135 × (π/180) = 3π/4. A common slip is simplifying 135/180 wrong — the GCD is 45, giving 3/4, so the answer is 3π/4. Watch out for 2π/3 (= 120°) and 5π/6 (= 150°).
⚡ (cos θ, sin θ) = point on unit circle · x=cos, y=sin
What are the exact coordinates of the point on the unit circle corresponding to θ = 5π/6?
Reference Angle Strategy
5π/6 is in Quadrant II. Reference angle = π − 5π/6 = π/6. In Q II, cos is negative, sin is positive.
✗ Key detail missed!
At 5π/6: reference angle is π/6 → base values (√3/2, 1/2). In Quadrant II, cos is negative → (−√3/2, 1/2). Option A forgets the negative sign. Option D uses π/3 values.
⚡ s = rθ (θ must be in RADIANS)
A circle has radius 8 cm. Find the arc length subtended by a central angle of 45°.
Watch Out!
Always convert degrees → radians before using s = rθ. 45° = π/4 rad.
✗ Don't forget to convert!
45° = π/4 rad. s = rθ = 8 × π/4 = 2π cm. The trap answer 360 comes from using degrees directly in the formula — never do that!
⚡ Coterminal = add or subtract 360° (or 2π) as many times as needed
Which angle is coterminal with −50°?
✗ Add 360° to a negative angle!
−50° + 360° = 310°. A coterminal angle shares the same terminal side. 50° is just the reference angle (not coterminal). 230° = 180° + 50° (a completely different angle).
Unit 2 · Right Triangle Trig
⚡ SOH: sin=Opp/Hyp · CAH: cos=Adj/Hyp · TOA: tan=Opp/Adj
In a right triangle, if sin θ = 3/5, what is cos θ? (Assume θ is acute.)
Pythagorean Trick
If sin θ = 3/5, then opp = 3, hyp = 5 → adj = √(25 − 9) = √16 = 4. So cos θ = adj/hyp.
✗ Use the Pythagorean theorem on the sides!
adj = √(5²−3²) = 4. cos θ = adj/hyp = 4/5. The answer 3/4 is actually tan θ. 5/3 and 5/4 are the reciprocal trig functions (csc and sec).
⚡ csc=1/sin · sec=1/cos · cot=1/tan (CO-functions are RECIPROCALS of each other)
If cos θ = −2/3 and sin θ > 0, find sec θ.
✗ sec = 1/cos — keep the sign!
sec θ = 1/cos θ = 1/(−2/3) = −3/2. The sign does NOT change — many students flip it. Since cos is negative, sec must also be negative.
⚡ ASTC rule — All Students Take Calculus (QI:all+ · QII:sin+ · QIII:tan+ · QIV:cos+)
If tan θ > 0 and cos θ < 0, in which quadrant does θ lie?
✗ Use the ASTC mnemonic!
cos < 0 eliminates Q I and Q IV. tan > 0 means sin and cos have the same sign — both negative in Q III. In Q II, sin > 0 and cos < 0, so tan < 0 there.
Unit 3 · Graphs of Trig Functions
⚡ y=A sin(Bx+C)+D · |A|=amplitude · Period=2π/|B| · Phase=−C/B
For y = −3 sin(2x − π), identify the amplitude and period.
Common Mistake
Amplitude is |A| = |−3| = 3, NOT −3. The negative sign means reflection, not smaller amplitude.
✗ Two traps in one question!
Amplitude = |−3| = 3 (absolute value!). Period = 2π/|B| = 2π/2 = π. The period triples here because B=2 compresses the wave. Never write amplitude as negative.
⚡ Phase shift = −C/B · Positive shift = RIGHT · Negative shift = LEFT
What is the phase shift of y = cos(3x + π/2)?
✗ Don't confuse C with the shift!
Phase shift = −C/B = −(π/2)/3 = −π/6, meaning π/6 to the left. The most common error is reading C = π/2 as the shift without dividing by B.
⚡ sin & cos range: [−1, 1] · tan range: all reals · tan undefined at π/2 + nπ
Which function has a range of all real numbers and vertical asymptotes?
✗ Only tan has asymptotes!
tan x has range (−∞, ∞) and vertical asymptotes at x = π/2 + nπ because it equals sin/cos, which is undefined when cos = 0. sin x and cos x are always bounded between −1 and 1.
Unit 4 · Trigonometric Identities
⚡ sin²θ + cos²θ = 1 (memorize this FIRST — all others derive from it)
If sin θ = 5/13 and θ is in Quadrant II, find cos θ using a Pythagorean identity.
✗ Don't drop the negative sign!
cos²θ = 1 − sin²θ = 1 − 25/169 = 144/169 → cos θ = ±12/13. Since θ is in Quadrant II, cos < 0 → cos θ = −12/13. Option A is the Q I answer (wrong quadrant).
⚡ 1+tan²θ = sec²θ · 1+cot²θ = csc²θ (derived from sin²+cos²=1, divide both sides)
Simplify: (1 − sin²θ) / cos θ
✗ Replace using the Pythagorean identity!
1 − sin²θ = cos²θ. So the expression becomes cos²θ / cos θ = cos θ. Step 1: recognize the substitution. Step 2: simplify by canceling.
⚡ cos(−θ) = cos θ (EVEN) · sin(−θ) = −sin θ (ODD) · tan(−θ) = −tan θ (ODD)
Which expression equals sin(−θ) · cos(−θ)?
✗ Apply even/odd rules separately!
sin(−θ) = −sin θ (odd). cos(−θ) = +cos θ (even). Product: (−sin θ)(cos θ) = −sin θ · cos θ. Choice A forgets the negative. Choice B incorrectly treats cos as odd.
⚡ sin(2θ) = 2 sinθ cosθ · cos(2θ) = cos²θ − sin²θ = 1−2sin²θ = 2cos²θ−1
If sin θ = 1/3, find the exact value of sin(2θ). (Assume θ is in Q I.)
✗ Find cos θ first, then apply the formula!
cos θ = √(1 − 1/9) = √(8/9) = 2√2/3. Then sin(2θ) = 2·(1/3)·(2√2/3) = 4√2/9. The trap of A comes from incorrectly squaring 2sin instead of using 2 sin cos.
⚡ sin(A±B) = sinA cosB ± cosA sinB · cos(A±B) = cosA cosB ∓ sinA sinB (cos FLIPS sign!)
Use the angle addition formula to find the exact value of sin(75°).
Decompose 75°
75° = 45° + 30° — both are special angles you know exactly.
✗ Apply sin(A+B) carefully!
sin(45°+30°) = sin45·cos30 + cos45·sin30 = (√2/2)(√3/2) + (√2/2)(1/2) = √6/4 + √2/4 = (√6+√2)/4. Option B is cos(75°) = sin(15°). Option D is sin(60°).
Unit 5 · Inverse Trig & Equations
⚡ arcsin range: [−π/2, π/2] · arccos range: [0, π] · arctan range: (−π/2, π/2)
Evaluate: arcsin(−√2/2). Give the answer in radians.
Restricted Domain Rule
arcsin only outputs values in [−π/2, π/2]. So the answer must be in that range — it cannot be 3π/4 (which is Q II).
✗ Stay within the range of arcsin!
sin(π/4) = √2/2, so sin(−π/4) = −√2/2. Since arcsin outputs in [−π/2, π/2], the answer is −π/4. The tricky choice 3π/4 also satisfies sin = −√2/2 but is outside arcsin's restricted range.
⚡ sin x = k has 2 solutions per period · find reference angle, then check all quadrants
Solve for x in [0, 2π): 2 sin x − √3 = 0
✗ Two solutions for sin = positive!
2 sin x = √3 → sin x = √3/2. Reference angle: x = π/3. sin is positive in Q I and Q II → x = π/3 and π − π/3 = 2π/3. Option B confuses sin(π/6) = 1/2 ≠ √3/2.
⚡ General solution for sin x = k: x = arcsin(k) + 2nπ OR x = π − arcsin(k) + 2nπ
What is the general solution for cos x = 0?
✗ cos = 0 at π/2 AND 3π/2!
cos x = 0 at x = π/2, 3π/2, 5π/2, … These are equally spaced by π, so the pattern is x = π/2 + nπ. Option C only captures half the solutions (misses 3π/2 in [0, 2π)). Option A gives sin x = 0 solutions.
⚡ STRATEGY: convert all terms to ONE trig function using identities, then factor
Solve in [0, 2π): 2cos²x + cos x − 1 = 0
✗ Factor like a quadratic in cos x!
Let u = cos x: 2u² + u − 1 = (2u − 1)(u + 1) = 0 → cos x = 1/2 or cos x = −1.
cos x = 1/2 → x = π/3, 5π/3. cos x = −1 → x = π. All three: π/3, π, 5π/3.
⚡ Law of Cosines: c² = a² + b² − 2ab cosC · Use when you have SAS or SSS
In triangle ABC, a = 7, b = 10, and angle C = 60°. Find side c using the Law of Cosines.
SAS Tip
You have two sides and the included angle → Law of Cosines. cos(60°) = 1/2.
✗ Plug carefully into c² = a² + b² − 2ab cosC!
c² = 49 + 100 − 2(7)(10)(1/2) = 149 − 70 = 79 → c = √79 ≈ 8.89. Option C (√249) comes from forgetting the 2ab·cosC term. Option D (9) is a rounding error.