A complete review of every core Precalculus trigonometry concept, paired one-to-one with a full 20-question timed practice exam in real multiple-choice format.
Read the concept, memorize the boxed rule, check the worked example — then test yourself on the matching question in Part Two.
Concept. An angle can be measured in degrees or in radians. One full revolution equals 360° or 2π radians, so 180° = π radians.
degrees × (π/180) = radians | radians × (180/π) = degrees
Convert 60° to radians.
Solution: 60 × π/180 = π/3
Concept. Radians describe an angle as a ratio of arc length to radius — a "pure number" tied directly to the circle, rather than an arbitrary 360-part division.
30°=π/6 45°=π/4 60°=π/3 90°=π/2 180°=π 270°=3π/2 360°=2π
Convert 3π/2 radians to degrees.
Solution: 3π/2 × 180/π = 270°
Concept. For a circle of radius r, the length of an arc cut off by a central angle θ (in radians) is found by multiplying the radius by the angle.
s = rθ — θ MUST be in radians, never degrees
r = 4, θ = π/2. Find the arc length.
Solution: s = 4 × π/2 = 2π
Concept. Two angles are coterminal if they share the same terminal side — they point in the exact same direction, differing only by full rotations.
Coterminal angle = θ ± 360°n (or θ ± 2πn in radians)
Find a positive coterminal angle for −50°.
Solution: −50° + 360° = 310°
Concept. The reference angle is the acute angle formed between the terminal side and the x-axis. Trig values of any angle match ± the trig value of its reference angle.
QI: θ QII: 180°−θ QIII: θ−180° QIV: 360°−θ
Find the reference angle of 300°.
Solution: 360° − 300° = 60°
Concept. For an acute angle in a right triangle, the trig ratios compare two of the triangle's three sides: opposite, adjacent, and hypotenuse.
sinθ=opp/hyp cosθ=adj/hyp tanθ=opp/adj
opp = 3, hyp = 5. Find sinθ and cosθ.
Solution: sinθ=3/5; adj=4 (3-4-5 triple), so cosθ=4/5
Concept. On the unit circle (radius 1) centered at the origin, the point at angle θ has coordinates (cosθ, sinθ).
30°,45°,60° sine pattern: √1/2, √2/2, √3/2 (the "1-2-3 under root, over 2" trick)
Find sin(π/6).
Solution: sin(π/6) = 1/2
Concept. The same unit-circle definition applies to cosine. Because x or y becomes negative in different quadrants, the sign of sin and cos changes by quadrant.
"All Students Take Calculus" — QI All +, QII Sine +, QIII Tangent +, QIV Cosine +
Find cos(2π/3).
Solution: QII, reference π/3, cosine negative: −1/2
Concept. Each quadrant has a unique combination of signs for sine and cosine, so the signs of two ratios pin down the quadrant exactly.
Use ASTC backward: match the + functions in your data to the quadrant that makes them positive.
cosθ > 0 and sinθ < 0. Which quadrant?
Solution: Quadrant IV
Concept. Since cosθ and sinθ are the legs of a right triangle with hypotenuse 1, they always satisfy the same relationship.
sin²θ + cos²θ = 1 → 1+tan²θ=sec²θ → 1+cot²θ=csc²θ
cosθ = 3/5, θ in Quadrant I. Find sinθ.
Solution: sinθ = √(1−9/25) = 4/5
Concept. To solve a basic trig equation, find the reference angle from the ratio, then place it in every quadrant where that function has the correct sign.
Find the reference angle first, then check ALL matching quadrants within the given interval.
Solve cosθ = −1/2 on [0, 2π).
Solution: Reference π/3; cosine negative in QII, QIII → θ = 2π/3, 4π/3
Concept. Sum identities let you find the exact trig value of angles built by adding two "nice" angles (30°, 45°, 60°, 90°).
sin(A+B)=sinA·cosB+cosA·sinB cos(A+B)=cosA·cosB−sinA·sinB
Find sin(105°) using 105° = 60°+45°.
Solution: (√3/2)(√2/2)+(1/2)(√2/2) = (√6+√2)/4
Concept. The matching family of identities for differences of two angles — note the sign flips compared to the sum identities.
sin(A−B)=sinA·cosB−cosA·sinB cos(A−B)=cosA·cosB+sinA·sinB
Find cos(15°) using 15° = 45°−30°.
Solution: (√2/2)(√3/2)+(√2/2)(1/2) = (√6+√2)/4
Concept. Setting A = B = θ in the sine sum identity produces a direct shortcut for sin(2θ).
sin(2θ) = 2·sinθ·cosθ
sinθ = 1/2, θ in QI. Find sin(2θ).
Solution: cosθ=√3/2 → sin2θ = 2(1/2)(√3/2) = √3/2
Concept. cos(2θ) has three equivalent forms — pick whichever matches the ratio you already know.
cos2θ = cos²θ−sin²θ = 2cos²θ−1 = 1−2sin²θ
sinθ = 1/2. Find cos(2θ).
Solution: 1 − 2(1/2)² = 1 − 1/2 = 1/2
Concept. In ANY triangle (not just right triangles), each side is proportional to the sine of its opposite angle. Use it when you know AAS, ASA, or SSA.
a/sinA = b/sinB = c/sinC
A=30°, B=70°, a=8. Find b.
Solution: b = 8·sin70°/sin30° ≈ 15.0
Concept. A generalized Pythagorean theorem for any triangle. Use it when you know two sides and the included angle (SAS), or all three sides (SSS).
c² = a² + b² − 2ab·cosC
a=5, b=6, C=60°. Find c.
Solution: c²=25+36−30=31 → c=√31
Concept. For y = A·sin(Bx+C)+D, the constant A stretches the graph vertically, and B controls how fast one full cycle repeats.
Amplitude = |A| Period = 2π/|B|
y = 4cos(3x). Find amplitude and period.
Solution: Amplitude = 4, Period = 2π/3
Concept. Rewrite Bx+C by factoring out B as B(x + C/B) — the horizontal shift is whatever value is being added to x inside the parentheses.
Factor out B first. Phase shift = −C/B (right if positive, left if negative).
y = sin(3x−π). Find the phase shift.
Solution: 3x−π = 3(x−π/3) → shift right π/3
Concept. Inverse trig functions return the ONE angle, from a restricted range, that produces a given ratio — even though infinitely many angles share that ratio.
arcsin: [−π/2, π/2] arccos: [0, π] arctan: (−π/2, π/2)
Evaluate arcsin(1/2).
Solution: π/6 (not 5π/6 — that lies outside arcsin's range)
Real exam-style multiple choice. Press Start to begin the timer — each question reveals its solution the instant you answer.
Convert 150° to radian measure.
Multiply degrees by π/180: 150 × π/180 = 5π/6.
Convert 7π/4 radians to degree measure.
Multiply radians by 180/π: 7π/4 × 180/π = 315°.
A circle has radius 8 cm. Find the length of the arc intercepted by a central angle of 3π/4 radians.
s = rθ = 8 × 3π/4 = 6π cm.
Which angle is coterminal with 480°, where 0° ≤ θ < 360°?
480° − 360° = 120°.
Find the reference angle for 210°.
210° lies in Quadrant III, so the reference angle is 210° − 180° = 30°.
In a right triangle, angle θ has opposite side 5 and hypotenuse 13. Find cos θ.
Adjacent = √(13²−5²) = √144 = 12, so cosθ = adjacent/hypotenuse = 12/13.
Find the exact value of sin(5π/3).
5π/3 lies in Quadrant IV with reference angle π/3. Sine is negative in QIV: −√3/2.
Find the exact value of cos(2π/3).
2π/3 lies in Quadrant II with reference angle π/3. Cosine is negative in QII: −1/2.
If sin θ < 0 and cos θ < 0, in which quadrant does θ lie?
Both sine and cosine are negative only in Quadrant III (tangent is positive there).
If sin θ = 3/5 and θ is in Quadrant II, find cos θ.
cos²θ = 1 − 9/25 = 16/25, so cosθ = ±4/5. Cosine is negative in QII: −4/5.
Solve 2sin θ − 1 = 0 for θ in [0, 2π).
sinθ = 1/2. Sine is positive in QI and QII, giving θ = π/6, 5π/6.
Find the exact value of sin 75° using a sum identity.
sin75° = sin(45°+30°) = sin45°cos30°+cos45°sin30° = (√6+√2)/4.
Find the exact value of cos 105° using a sum/difference identity.
cos105° = cos(60°+45°) = cos60°cos45°−sin60°sin45° = (√2−√6)/4.
If sin θ = 3/5 and θ is in Quadrant I, find sin(2θ).
cosθ = 4/5 (QI is positive), so sin2θ = 2(3/5)(4/5) = 24/25.
If cos θ = 1/3, find cos(2θ).
cos2θ = 2cos²θ−1 = 2(1/9)−1 = 2/9−1 = −7/9.
In △ABC, A = 40°, B = 60°, and a = 10. Find side b to the nearest tenth.
b = a·sinB/sinA = 10·sin60°/sin40° ≈ 13.5.
In △ABC, a = 7, b = 9, and C = 60°. Find side c.
c² = 7²+9²−2(7)(9)cos60° = 49+81−63 = 67, so c = √67.
Find the amplitude and period of y = 3sin(2x − π/4).
Amplitude = |3| = 3. Period = 2π/|2| = π.
Find the phase shift of y = cos(2x + π/3).
2x+π/3 = 2(x+π/6), so the graph shifts π/6 to the left.
Evaluate arccos(−√2/2).
arccos has range [0, π], and cos(3π/4) = −√2/2, so the answer is 3π/4.