Trigonometry is nominally 15% of the CLEP Precalculus exam, but the fact sheet warns that trig permeates most major topics — in practice roughly 30–40% of questions touch it. Identity and exact-value work lands almost entirely in Section 2 (no calculator), so everything here must run by hand.
From x² + y² = 1 on the unit circle:
sin²θ + cos²θ = 1
Dividing through by cos²θ or by sin²θ produces the two siblings:
tan²θ + 1 = sec²θ 1 + cot²θ = csc²θ
Primary use: one trig value + a quadrant → every trig value. Given sin θ = 3/5 with θ in Quadrant II: cos²θ = 1 − 9/25 = 16/25, so cos θ = ±4/5 — and Quadrant II makes it −4/5. Then tan θ = sin/cos = −3/4, and the reciprocals follow. The identity supplies the magnitude; the quadrant supplies the sign. Skipping the sign step is the single most-planted distractor in this topic.
Quadrant sign chart (ASTC, counterclockwise from QI): All positive, Sine only, Tangent only, Cosine only.
sin(A ± B) = sin A cos B ± cos A sin B
cos(A ± B) = cos A cos B ∓ sin A sin B ← cosine flips the sign
Memory anchor: sine's expansion mixes the functions and keeps the sign; cosine's matches them and flips it. Unsure of a sign? Test against cos(π/2) = cos(π/4 + π/4) = 1/2 − 1/2 = 0 ✓ — a ten-second audit.
These formulas turn non-special angles into arithmetic on special ones: π/12 = π/3 − π/4, so
cos(π/12) = cos(π/3 − π/4) = cos(π/3)cos(π/4) + sin(π/3)sin(π/4) = (√2 + √6)/4 = (√6 + √2)/4 ≈ 0.966
Sanity check the size: π/12 = 15°, cosine near 1 ✓. The twelfths of π (π/12, 5π/12, 7π/12) all decompose into combinations of π/6, π/4, π/3.
Setting A = B = θ gives the double-angle formulas:
sin(2θ) = 2 sin θ cos θ
cos(2θ) = cos²θ − sin²θ = 2cos²θ − 1 = 1 − 2sin²θ
Cosine's three versions are the point: choose the one whose variable matches the rest of your equation. Solving cos(2θ) + sin θ = 0? Use 1 − 2sin²θ and the equation becomes a quadratic in sin θ alone. And beware the most tempting wrong identity in trigonometry: sin(2θ) ≠ 2 sin θ. Test θ = π/4: sin(π/2) = 1, but 2 sin(π/4) ≈ 1.41.
For sin θ = k (or cos, tan):
Example: sin θ = 1/2 on [0, 2π) → {π/6, 5π/6}. General solution (no interval): θ = π/6 + 2kπ or θ = 5π/6 + 2kπ, k any integer. For tan θ = 1: θ = π/4 + kπ — one family, because tangent's period is π, not 2π.
Equations needing algebra first. Treat the trig function as a variable and factor: 2cos²θ + cos θ − 1 = 0 factors as (2cos θ − 1)(cos θ + 1) = 0, giving cos θ = 1/2 (θ = π/3, 5π/3) or cos θ = −1 (θ = π). Never divide both sides by a trig factor that could be zero — dividing by cos θ silently discards the cos θ = 0 solutions. Factor instead.
Impossible equations. |sin θ| ≤ 1 and |cos θ| ≤ 1 force |sec θ| ≥ 1 and |csc θ| ≥ 1. An equation like csc θ = 1/2 demands sin θ = 2 — no solution. "No solution" is a legitimate answer the exam plants among ordinary-looking choices.
For sin(bθ) = k, substitute u = bθ — and stretch the window by the factor b. Solving sin(2x) = 1/2 on [−π, π]: u = 2x runs over [−2π, 2π], twice the original window, so collect sin u = 1/2 over two full periods: u = π/6, 5π/6, π/6 − 2π, 5π/6 − 2π — four values, hence four solutions in x. The official CLEP samples include exactly this stem shape: "How many solutions does the equation … have on the interval [−π, π]?" Count with the rule: sin(bθ) = k with |k| < 1 has 2 solutions per 2π of u-window (1 per π for tangent). Forgetting to stretch the window — and reporting half the count — is the classic error.
For any triangle with sides a, b, c opposite angles A, B, C:
Law of Sines: a/sin A = b/sin B = c/sin C
Law of Cosines: c² = a² + b² − 2ab·cos C
Use the Law of Sines when you have an angle–opposite-side pair; use the Law of Cosines for two sides with the included angle (or all three sides, solving for an angle). Watch the sign: the −2ab·cos C term subtracts when C is acute. At C = 90°, cos C = 0 and the formula collapses to the Pythagorean theorem — a built-in sanity check.
Example: sides 5 and 7 with included angle 60° → c² = 25 + 49 − 2(5)(7)(1/2) = 74 − 35 = 39, so c = √39.
Word problems reduce to SOH-CAH-TOA on a sketched triangle:
For "which expression is equivalent" items: convert to sines and cosines, hunt for Pythagorean substitutions (1 − cos²θ = sin²θ is the most common), then factor and cancel. Example:
sin²θ/(1 − cos θ) = (1 − cos²θ)/(1 − cos θ) = (1 − cos θ)(1 + cos θ)/(1 − cos θ) = 1 + cos θ
valid wherever cos θ ≠ 1. Difference of squares does as much work in trig as it does in algebra.
Q1. D. cos²θ = 1 − sin²θ = 1 − 9/25 = 16/25, so cos θ = ±4/5; Quadrant II has negative cosine, giving −4/5. A) finds the correct magnitude but ignores the quadrant — sign analysis is half the problem. B) treats 3-4-5 as tan θ and inverts it, confusing ratio roles. C) negates sine itself instead of computing cosine. E) inverts the fraction, writing hypotenuse over adjacent (that's sec θ, and with the wrong magnitude convention). Fix: identity for the magnitude, quadrant for the sign — two separate steps, every time.
Q2. B. sin(2θ) = 2 sin θ cos θ = 2(3/5)(4/5) = 24/25. A) drops the factor of 2, computing sin θ cos θ only. C) computes 2 sin θ = 6/5 — the "double the angle, double the sine" fantasy; sine is not linear. D) is cos(2θ) = 16/25 − 9/25 = 7/25, the wrong double-angle formula. E) adds 3/5 + 3/5 with a denominator slip. Fix: sin(2θ) = 2 sin θ cos θ — three factors, and the cos θ is not optional.
Q3. B. π/12 = π/3 − π/4, and cos(A − B) = cos A cos B + sin A sin B = (1/2)(√2/2) + (√3/2)(√2/2) = (√2 + √6)/4 = (√6 + √2)/4 ≈ 0.966, sensible for 15° where cosine is near 1. A) is sin(π/12) — the difference version with a minus sign — and ≈ 0.259, far too small for cos 15°. C) is negative, impossible for a first-quadrant cosine. D) is cos(π/6), misreading the angle. E) drops the √6 term entirely. Fix: decompose twelfths of π into π/6, π/4, π/3 pieces, run the formula, then size-check the decimal against the angle's location.
Q4. C. Substitute u = 2x: as x runs over [−π, π], u runs over [−2π, 2π] — two full periods. sin u = 1/2 at u = π/6 and 5π/6, plus the copies π/6 − 2π and 5π/6 − 2π, all inside [−2π, 2π]: four u-values, and each gives exactly one x. Four solutions (numerically x ≈ −2.880, −1.833, 0.262, 1.309). A) takes only the single arcsin output. B) counts the base period only, forgetting the doubled window — the classic multiple-angle error. D) over-extends by counting a third period that isn't in the window. E) doubles the correct answer, stretching the window twice. Fix: for sin(bx) = k, the u-window is b times the x-window — count 2 solutions per 2π of u-window, then convert back.
Q5. D. Factor: (2cos θ − 1)(cos θ + 1) = 0, so cos θ = 1/2 → θ = π/3, 5π/3, or cos θ = −1 → θ = π. Set: {π/3, π, 5π/3}. A) uses sine's partner rule (π − θ₁) on a cosine equation and drops the cos θ = −1 root. B) solves cos θ = 1/2 correctly but loses the cos θ = −1 factor entirely. C) places the cos θ = 1/2 solutions in the wrong quadrants (QII/QIII instead of QI/QIV). E) solves as though the factored value were 1/2 for sine, mixing up which function the reference angle belongs to. Fix: factor fully, solve each factor with the master routine, and use cosine's own symmetry (θ and 2π − θ) — not sine's.
Q6. C. tan(π/4) = 1, and tangent repeats every π, so one family θ = π/4 + kπ captures everything. A) uses period 2π and misses half the solutions (e.g., 5π/4). B) uses period π/2, inventing solutions where tan θ = −1. D) starts from the wrong reference angle — tan(3π/4) = −1. E) starts at tangent's asymptote, where it is undefined. Fix: tangent's period is π — general solutions for tan need only one family, spaced π apart.
Q7. D. The equation already contains sin θ, so 1 − 2sin²θ converts everything to sines: 1 − 2sin²θ + sin θ = 0, a quadratic in sin θ alone. A) is a true identity but leaves both sin and cos in the equation — no single-variable quadratic. B) is also true but converts to cosines while the equation's other term stays in sine. C) is the sin(2θ) formula, not cos(2θ) — wrong identity altogether. E) has the sign flipped on the 2sin²θ term and is not an identity at all (test θ = 0: cos 0 = 1 but 1 + 0 = 1 works; test θ = π/2: cos π = −1 but 1 + 2 = 3 fails). Fix: all three cos(2θ) forms are true — pick the one whose variable matches what else is in the equation.
Q8. A. Law of Cosines: c² = 5² + 7² − 2(5)(7)cos 60° = 25 + 49 − 70(1/2) = 74 − 35 = 39, so c = √39. B) adds the cosine term instead of subtracting — the sign error the formula's "−2ab cos C" exists to test. C) drops the cosine term entirely, defaulting to the Pythagorean theorem as if the angle were 90°. D) forgets the final square root, reporting c². E) uses cos 60° = 1 instead of 1/2, getting 74 − 70 = 4 → 2. Fix: Law of Cosines is Pythagoras with a correction term — subtract 2ab cos C, and take the square root at the end.
Q9. C. Law of Sines: b/sin B = a/sin A, so b = 8·sin 45°/sin 30° = 8(√2/2)/(1/2) = 4√2 · 2 = 8√2 ≈ 11.3. A) computes 8 sin 45° = 4√2 but never divides by sin 30° — half the setup. B) assumes sides opposite different angles are equal. D) replaces sin 45° with 1, treating B as a right angle. E) uses sin 60° instead of sin 45°, swapping the special angles. Fix: Law of Sines is a proportion — multiply by the new angle's sine AND divide by the paired angle's sine, then simplify.
Q10. E. csc θ = 1/2 requires sin θ = 2, impossible since |sin θ| ≤ 1; the reciprocal functions satisfy |csc θ| ≥ 1 always. A) is solvable — sine reaches every value in [−1, 1], including −0.99. B) is solvable — tangent's range is all reals, so 100 is fine. C) is a standard special-angle equation (θ = π/6, 11π/6). D) is solvable — |−3| ≥ 1 is within secant's range (cos θ = −1/3). Fix: sec and csc never take values strictly between −1 and 1 — a proper-fraction target for either means "no solution."
Q11. A. The tower height is opposite the 35° angle; the 40 m ground distance is adjacent. tan 35° = opposite/adjacent = h/40, so h = 40 tan 35° ≈ 28 m. B) uses sine, treating the 40 m as the hypotenuse — but the ground distance is a leg. C) inverts the ratio, solving as if 40 were the height. D) uses cosine, which relates adjacent to hypotenuse — neither unknown here. E) inverts the entire equation, dividing the tangent by the distance. Fix: sketch and label first — angle of elevation sits between the horizontal (adjacent) and the line of sight; the vertical target is opposite, so height = distance × tan(angle).
Q12. E. sin²θ = 1 − cos²θ = (1 − cos θ)(1 + cos θ); dividing by (1 − cos θ) leaves 1 + cos θ. A) cancels to the wrong surviving factor of the difference of squares. B) "cancels" a sin against sin² as though the denominator were sin θ. C) is the negative of the correct answer, from mis-ordering the difference-of-squares factors. D) converts to tangent by dividing by cosine — an operation nowhere in this simplification. Fix: 1 − cos²θ factors as (1 − cos θ)(1 + cos θ) — Pythagorean substitution plus difference of squares handles most "equivalent expression" items.
Q1. D. cos²θ = 1 − sin²θ = 1 − 9/25 = 16/25, so cos θ = ±4/5; Quadrant II has negative cosine, giving −4/5. A) finds the correct magnitude but ignores the quadrant — sign analysis is half the problem. B) treats 3-4-5 as tan θ and inverts it, confusing ratio roles. C) negates sine itself instead of computing cosine. E) inverts the fraction, writing hypotenuse over adjacent (that's sec θ, and with the wrong magnitude convention). Fix: identity for the magnitude, quadrant for the sign — two separate steps, every time.
Q2. B. sin(2θ) = 2 sin θ cos θ = 2(3/5)(4/5) = 24/25. A) drops the factor of 2, computing sin θ cos θ only. C) computes 2 sin θ = 6/5 — the "double the angle, double the sine" fantasy; sine is not linear. D) is cos(2θ) = 16/25 − 9/25 = 7/25, the wrong double-angle formula. E) adds 3/5 + 3/5 with a denominator slip. Fix: sin(2θ) = 2 sin θ cos θ — three factors, and the cos θ is not optional.
Q3. B. π/12 = π/3 − π/4, and cos(A − B) = cos A cos B + sin A sin B = (1/2)(√2/2) + (√3/2)(√2/2) = (√2 + √6)/4 = (√6 + √2)/4 ≈ 0.966, sensible for 15° where cosine is near 1. A) is sin(π/12) — the difference version with a minus sign — and ≈ 0.259, far too small for cos 15°. C) is negative, impossible for a first-quadrant cosine. D) is cos(π/6), misreading the angle. E) drops the √6 term entirely. Fix: decompose twelfths of π into π/6, π/4, π/3 pieces, run the formula, then size-check the decimal against the angle's location.
Q4. C. Substitute u = 2x: as x runs over [−π, π], u runs over [−2π, 2π] — two full periods. sin u = 1/2 at u = π/6 and 5π/6, plus the copies π/6 − 2π and 5π/6 − 2π, all inside [−2π, 2π]: four u-values, and each gives exactly one x. Four solutions (numerically x ≈ −2.880, −1.833, 0.262, 1.309). A) takes only the single arcsin output. B) counts the base period only, forgetting the doubled window — the classic multiple-angle error. D) over-extends by counting a third period that isn't in the window. E) doubles the correct answer, stretching the window twice. Fix: for sin(bx) = k, the u-window is b times the x-window — count 2 solutions per 2π of u-window, then convert back.
Q5. D. Factor: (2cos θ − 1)(cos θ + 1) = 0, so cos θ = 1/2 → θ = π/3, 5π/3, or cos θ = −1 → θ = π. Set: {π/3, π, 5π/3}. A) uses sine's partner rule (π − θ₁) on a cosine equation and drops the cos θ = −1 root. B) solves cos θ = 1/2 correctly but loses the cos θ = −1 factor entirely. C) places the cos θ = 1/2 solutions in the wrong quadrants (QII/QIII instead of QI/QIV). E) solves as though the factored value were 1/2 for sine, mixing up which function the reference angle belongs to. Fix: factor fully, solve each factor with the master routine, and use cosine's own symmetry (θ and 2π − θ) — not sine's.
Q6. C. tan(π/4) = 1, and tangent repeats every π, so one family θ = π/4 + kπ captures everything. A) uses period 2π and misses half the solutions (e.g., 5π/4). B) uses period π/2, inventing solutions where tan θ = −1. D) starts from the wrong reference angle — tan(3π/4) = −1. E) starts at tangent's asymptote, where it is undefined. Fix: tangent's period is π — general solutions for tan need only one family, spaced π apart.
Q7. D. The equation already contains sin θ, so 1 − 2sin²θ converts everything to sines: 1 − 2sin²θ + sin θ = 0, a quadratic in sin θ alone. A) is a true identity but leaves both sin and cos in the equation — no single-variable quadratic. B) is also true but converts to cosines while the equation's other term stays in sine. C) is the sin(2θ) formula, not cos(2θ) — wrong identity altogether. E) has the sign flipped on the 2sin²θ term and is not an identity at all (test θ = 0: cos 0 = 1 but 1 + 0 = 1 works; test θ = π/2: cos π = −1 but 1 + 2 = 3 fails). Fix: all three cos(2θ) forms are true — pick the one whose variable matches what else is in the equation.
Q8. A. Law of Cosines: c² = 5² + 7² − 2(5)(7)cos 60° = 25 + 49 − 70(1/2) = 74 − 35 = 39, so c = √39. B) adds the cosine term instead of subtracting — the sign error the formula's "−2ab cos C" exists to test. C) drops the cosine term entirely, defaulting to the Pythagorean theorem as if the angle were 90°. D) forgets the final square root, reporting c². E) uses cos 60° = 1 instead of 1/2, getting 74 − 70 = 4 → 2. Fix: Law of Cosines is Pythagoras with a correction term — subtract 2ab cos C, and take the square root at the end.
Q9. C. Law of Sines: b/sin B = a/sin A, so b = 8·sin 45°/sin 30° = 8(√2/2)/(1/2) = 4√2 · 2 = 8√2 ≈ 11.3. A) computes 8 sin 45° = 4√2 but never divides by sin 30° — half the setup. B) assumes sides opposite different angles are equal. D) replaces sin 45° with 1, treating B as a right angle. E) uses sin 60° instead of sin 45°, swapping the special angles. Fix: Law of Sines is a proportion — multiply by the new angle's sine AND divide by the paired angle's sine, then simplify.
Q10. E. csc θ = 1/2 requires sin θ = 2, impossible since |sin θ| ≤ 1; the reciprocal functions satisfy |csc θ| ≥ 1 always. A) is solvable — sine reaches every value in [−1, 1], including −0.99. B) is solvable — tangent's range is all reals, so 100 is fine. C) is a standard special-angle equation (θ = π/6, 11π/6). D) is solvable — |−3| ≥ 1 is within secant's range (cos θ = −1/3). Fix: sec and csc never take values strictly between −1 and 1 — a proper-fraction target for either means "no solution."
Q11. A. The tower height is opposite the 35° angle; the 40 m ground distance is adjacent. tan 35° = opposite/adjacent = h/40, so h = 40 tan 35° ≈ 28 m. B) uses sine, treating the 40 m as the hypotenuse — but the ground distance is a leg. C) inverts the ratio, solving as if 40 were the height. D) uses cosine, which relates adjacent to hypotenuse — neither unknown here. E) inverts the entire equation, dividing the tangent by the distance. Fix: sketch and label first — angle of elevation sits between the horizontal (adjacent) and the line of sight; the vertical target is opposite, so height = distance × tan(angle).
Q12. E. sin²θ = 1 − cos²θ = (1 − cos θ)(1 + cos θ); dividing by (1 − cos θ) leaves 1 + cos θ. A) cancels to the wrong surviving factor of the difference of squares. B) "cancels" a sin against sin² as though the denominator were sin θ. C) is the negative of the correct answer, from mis-ordering the difference-of-squares factors. D) converts to tangent by dividing by cosine — an operation nowhere in this simplification. Fix: 1 − cos²θ factors as (1 − cos θ)(1 + cos θ) — Pythagorean substitution plus difference of squares handles most "equivalent expression" items.