Trigonometry is nominally 15% of the CLEP exam, but the fact sheet warns that trig permeates most major topics — in practice roughly 30–40% of the questions require it. This lesson builds the exact-value machinery those questions run on. Special-angle values are Section 2 (no-calculator) material: they must be producible by hand. In Section 1, the online graphing calculator computes trig values — but only if it is in the correct angle mode, a trap this lesson closes.
For an acute angle θ in a right triangle:
sin θ = opposite/hypotenuse cos θ = adjacent/hypotenuse tan θ = opposite/adjacent
(Mnemonic: SOH-CAH-TOA.) Given any two sides, the Pythagorean theorem supplies the third — the 3-4-5, 5-12-13, and 8-15-17 triples appear constantly. Given one side and one acute angle, the appropriate ratio solves for a missing side: pick the ratio that connects the side you know to the side you want.
Angles of elevation and depression. The angle of elevation is measured upward from the horizontal to a sightline; the angle of depression is measured downward from the horizontal. Both produce right triangles with the ground or a horizontal line as one leg. Standing 40 m from a tower and sighting its top at a 30° elevation: the tower's height h satisfies tan 30° = h/40, so h = 40 tan 30° = 40√3/3 ≈ 23.1 m. Choose tangent when the two sides involved are the legs (height and ground distance); sine or cosine when the hypotenuse (a sightline or ramp length) is involved.
Place an angle at the center of a circle of radius r. Its radian measure is
θ = arc length / radius = s/r
— how many radius-lengths of arc the angle cuts off. A full circle: 2πr/r = 2π radians = 360°, so:
180° = π rad degrees → radians: multiply by π/180 radians → degrees: multiply by 180/π
The special angles in both languages (memorize cold):
| deg | 0° | 30° | 45° | 60° | 90° | 120° | 135° | 150° | 180° | 270° | 360° |
|---|---|---|---|---|---|---|---|---|---|---|---|
| rad | 0 | π/6 | π/4 | π/3 | π/2 | 2π/3 | 3π/4 | 5π/6 | π | 3π/2 | 2π |
Reduce before multiplying: 150° → 150/180 = 5/6 → 5π/6, done.
Arc length, rearranged: s = rθ, valid only when θ is in radians — that is what radians are for. A central angle of 2π/3 in a circle of radius 6 subtends an arc of s = 6 · (2π/3) = 4π.
Put an angle θ in standard position: vertex at the origin, initial side along the positive x-axis, counterclockwise positive. Where the terminal side crosses the unit circle (radius 1):
that point is (cos θ, sin θ)
cos θ is the x-coordinate; sin θ is the y-coordinate. Everything else follows:
Tangent: tan θ = sin θ/cos θ = y/x = the slope of the terminal side. Undefined where cos θ = 0 (vertical ray: θ = π/2, 3π/2, …).
First-quadrant column (the rest come by reflection):
| θ | 0 | π/6 | π/4 | π/3 | π/2 |
|---|---|---|---|---|---|
| sin θ | 0 | 1/2 | √2/2 | √3/2 | 1 |
| cos θ | 1 | √3/2 | √2/2 | 1/2 | 0 |
| tan θ | 0 | √3/3 | 1 | √3 | undefined |
Pattern: sin runs √0/2, √1/2, √2/2, √3/2, √4/2; cos runs the same list backwards.
For any angle θ, the reference angle is the acute angle between the terminal side and the x-axis (never the y-axis). Protocol:
Example: 7π/6 is π/6 past π — QIII, reference angle π/6, both coordinates negative: sin(7π/6) = −1/2, cos(7π/6) = −√3/2, tan(7π/6) = (−1/2)/(−√3/2) = √3/3 (positive: tangent is positive in QIII).
The same machinery runs in reverse: given a unit-circle point such as (−√2/2, √2/2), the signs (−, +) place the angle in QII and the values identify reference angle π/4, so θ = π − π/4 = 3π/4.
[GRAPH: Unit circle with the special points marked. QI points labeled with coordinates: (√3/2, 1/2) at π/6, (√2/2, √2/2) at π/4, (1/2, √3/2) at π/3. The angle 7π/6 drawn with its terminal side in QIII, reference angle π/6 shaded between the terminal side and the negative x-axis, terminal point (−√3/2, −1/2) labeled. Axis points (1, 0), (0, 1), (−1, 0), (0, −1) at angles 0, π/2, π, 3π/2.]
On Section 1, the built-in calculator evaluates trig functions in whichever angle mode is set. In degree mode, sin(π/6) returns ≈ 0.0091 — the sine of about half a degree — instead of 0.5. The output looks computed, not broken, which is exactly why it is dangerous. Check the mode before the first trig keystroke; radian mode is the default assumption whenever angles are written with π.
1. B — Correct: 150 · (π/180) = 150π/180 = 5π/6 (reduce 150/180 = 5/6 first). A) is 135° converted. C) is 120° converted. D) uses the wrong anchor, converting with 90° = π instead of 180° = π: 150π/90 = 5π/3. E) inverts the conversion, treating the factor as 180/π and reducing 180/150 = 6/5. Fix: degrees → radians is always × π/180; reduce the numeric fraction before attaching π.
2. C — Correct: opposite = 8, hypotenuse = 17, so the adjacent leg is √(17² − 8²) = √(289 − 64) = √225 = 15, and tan θ = opposite/adjacent = 8/15. A) reports sin θ (opposite over hypotenuse) — the ratio the problem hands you, not the one it asks for. B) reports cos θ. D) inverts tangent (adjacent over opposite = cot θ). E) puts the hypotenuse in a tangent ratio, which never contains the hypotenuse. Fix: when tangent is requested and only one leg is given, run the Pythagorean theorem first — tangent is leg/leg, and the hypotenuse never appears in it.
3. C — Correct: from the special-angle table, sin(π/6) = 1/2. A) is cos(π/6) — the classic coordinate swap (cos is the x-coordinate, listed first; sin is the y). B) is the value at π/4, where sine and cosine coincide. D) attaches a spurious negative — π/6 is in QI where everything is positive. E) is tan(π/6). Fix: anchor the sine pattern √0/2, √1/2, √2/2, √3/2, √4/2 across 0, π/6, π/4, π/3, π/2 — sin(π/6) = √1/2 = 1/2.
4. A — Correct: 3π/4 is in QII with reference angle π/4; cosine is negative in QII, so cos(3π/4) = −√2/2. B) takes the reference value but forgets the QII sign — the single most common special-angle error. C) uses the wrong reference value (that magnitude belongs to π/3-family angles). D) is wrong in both value-family and sign logic. E) uses the π/6-family magnitude with the right sign. Fix: quadrant and sign first, table value second — sketch the terminal side, mark the quadrant's (±, ±), then fill in the reference-angle magnitude.
5. E — Correct: π < θ < 3π/2 places θ in QIII, where sine is negative; from cos²θ + sin²θ = 1, sin²θ = 1 − 3/4 = 1/4, so sin θ = −1/2. (Equivalently: reference angle π/6, QIII signs.) A) is positive — impossible in QIII — and carries the cosine magnitude. B) has the right magnitude but the wrong sign, ignoring the quadrant restriction. C) reports cos θ itself (and even that sign copy is only accidental). D) is a tangent-family value. Fix: Pythagorean identity gives the magnitude; the stated quadrant alone decides the sign — resolve the ± with the interval before writing the answer.
6. A — Correct: sin θ < 0 means the terminal point is below the x-axis (QIII or QIV); cos θ > 0 means it is right of the y-axis (QI or QIV); the intersection is QIV. B) QI requires both positive. C) QII has the signs exactly reversed (sin > 0, cos < 0). D) QIII requires both negative. E) the two conditions together pin down a unique quadrant, so it can be determined. Fix: treat each sign condition as a half-plane — sin's sign picks above/below the x-axis, cos's sign picks right/left of the y-axis — and intersect them.
7. B — Correct: s = rθ = 6 · (2π/3) = 4π cm. A) uses θ = π/3, dropping the factor of 2 in the angle. C) uses the diameter 12 instead of the radius: 12 · (2π/3) = 8π. D) computes the full circumference 2πr = 12π, ignoring that the angle is only a fraction of a turn. E) doubles the angle (2 · 2π/3) instead of multiplying by the radius — treating the coefficient in 2π/3 as the thing to scale rather than applying s = rθ. Fix: arc length = radius × angle-in-radians, s = rθ — confirm the angle is in radians and that you multiplied by the radius, not the diameter.
8. D — Correct: tan 30° = height/40, so height = 40 tan 30° = 40(√3/3) = 40√3/3 ≈ 23.1 m. A) computes 40 sin 30° = 20, wrongly treating the 40 m ground distance as a hypotenuse. B) multiplies 40 by sin 60° ≈ cos 30°'s partner value, swapping which trig value equals √3/2 versus √3/3. C) computes 40/sin 30° = 80, inverting the ratio. E) uses tan 60° = √3 — the angle measured from the tower instead of from the ground. Fix: label the triangle before choosing a ratio: the known ground distance is the leg adjacent to the elevation angle, the unknown height is opposite, so height = (adjacent) × tan(angle).
9. E — Correct: the point (−√2/2, √2/2) has x < 0 and y > 0 — QII — and the magnitudes √2/2 mark reference angle π/4, so θ = π − π/4 = 3π/4. A) gives only the reference angle, ignoring the signs that force QII. B) is the QIII angle 5π/4, where both coordinates would be negative. C) is the axis point (0, −1), which has entirely different coordinates. D) is the QIV angle, where the y-coordinate would be negative. Fix: read the signs first to lock the quadrant, then match magnitudes to a reference angle — signs choose the quadrant, magnitudes choose the family.
10. A — Correct: cosine is an even function, so cos(−π/3) = cos(π/3) = 1/2. B) applies the odd-function rule (correct for sine, wrong for cosine) and negates. C) is cos(π/6) — a wrong-row table grab. D) combines the wrong row with the wrong sign rule. E) is a tangent-family value. Fix: negatives inside trig functions: cosine ignores the sign (even), sine and tangent pull it out front (odd) — decide which rule applies before touching the table.
11. E — Correct: 11π/6 is in QIV (just short of 2π) with reference angle π/6: cos = +√3/2, sin = −1/2, tan = (−1/2)/(√3/2) = −√3/3 — both claims in E check out. A) fails because sine is negative in QIV. B) misidentifies the reference angle as π/3; the distance from 11π/6 to 2π is π/6. C) gives tangent the wrong magnitude and the wrong sign (tan is negative in QIV). D) assigns sine the cosine's magnitude. Fix: for any "which statement is true" angle item, compute the full triple (quadrant, reference angle, signed values) once at the start — then test each claim against your computed values rather than re-deriving per option.
12. D — Correct: in degree mode the calculator reads the input π/6 ≈ 0.5236 as degrees; sin(0.5236°) ≈ 0.0091 — a plausible-looking but wrong output. A) cos(π/6) ≈ 0.866, which is not what the student sought and doesn't explain the tiny output. B) rounding cannot turn 0.5 into 0.0091. C) sine's domain is all real numbers — nothing is out of domain. E) entering 6/π ≈ 1.91 in radian mode gives sin ≈ 0.94, not 0.0091, so that error doesn't produce this symptom. Fix: an absurdly small sine for a familiar angle is the degree-mode signature — when a trig output looks wrong, check the angle mode before checking your algebra.
1. B — Correct: 150 · (π/180) = 150π/180 = 5π/6 (reduce 150/180 = 5/6 first). A) is 135° converted. C) is 120° converted. D) uses the wrong anchor, converting with 90° = π instead of 180° = π: 150π/90 = 5π/3. E) inverts the conversion, treating the factor as 180/π and reducing 180/150 = 6/5. Fix: degrees → radians is always × π/180; reduce the numeric fraction before attaching π.
2. C — Correct: opposite = 8, hypotenuse = 17, so the adjacent leg is √(17² − 8²) = √(289 − 64) = √225 = 15, and tan θ = opposite/adjacent = 8/15. A) reports sin θ (opposite over hypotenuse) — the ratio the problem hands you, not the one it asks for. B) reports cos θ. D) inverts tangent (adjacent over opposite = cot θ). E) puts the hypotenuse in a tangent ratio, which never contains the hypotenuse. Fix: when tangent is requested and only one leg is given, run the Pythagorean theorem first — tangent is leg/leg, and the hypotenuse never appears in it.
3. C — Correct: from the special-angle table, sin(π/6) = 1/2. A) is cos(π/6) — the classic coordinate swap (cos is the x-coordinate, listed first; sin is the y). B) is the value at π/4, where sine and cosine coincide. D) attaches a spurious negative — π/6 is in QI where everything is positive. E) is tan(π/6). Fix: anchor the sine pattern √0/2, √1/2, √2/2, √3/2, √4/2 across 0, π/6, π/4, π/3, π/2 — sin(π/6) = √1/2 = 1/2.
4. A — Correct: 3π/4 is in QII with reference angle π/4; cosine is negative in QII, so cos(3π/4) = −√2/2. B) takes the reference value but forgets the QII sign — the single most common special-angle error. C) uses the wrong reference value (that magnitude belongs to π/3-family angles). D) is wrong in both value-family and sign logic. E) uses the π/6-family magnitude with the right sign. Fix: quadrant and sign first, table value second — sketch the terminal side, mark the quadrant's (±, ±), then fill in the reference-angle magnitude.
5. E — Correct: π < θ < 3π/2 places θ in QIII, where sine is negative; from cos²θ + sin²θ = 1, sin²θ = 1 − 3/4 = 1/4, so sin θ = −1/2. (Equivalently: reference angle π/6, QIII signs.) A) is positive — impossible in QIII — and carries the cosine magnitude. B) has the right magnitude but the wrong sign, ignoring the quadrant restriction. C) reports cos θ itself (and even that sign copy is only accidental). D) is a tangent-family value. Fix: Pythagorean identity gives the magnitude; the stated quadrant alone decides the sign — resolve the ± with the interval before writing the answer.
6. A — Correct: sin θ < 0 means the terminal point is below the x-axis (QIII or QIV); cos θ > 0 means it is right of the y-axis (QI or QIV); the intersection is QIV. B) QI requires both positive. C) QII has the signs exactly reversed (sin > 0, cos < 0). D) QIII requires both negative. E) the two conditions together pin down a unique quadrant, so it can be determined. Fix: treat each sign condition as a half-plane — sin's sign picks above/below the x-axis, cos's sign picks right/left of the y-axis — and intersect them.
7. B — Correct: s = rθ = 6 · (2π/3) = 4π cm. A) uses θ = π/3, dropping the factor of 2 in the angle. C) uses the diameter 12 instead of the radius: 12 · (2π/3) = 8π. D) computes the full circumference 2πr = 12π, ignoring that the angle is only a fraction of a turn. E) doubles the angle (2 · 2π/3) instead of multiplying by the radius — treating the coefficient in 2π/3 as the thing to scale rather than applying s = rθ. Fix: arc length = radius × angle-in-radians, s = rθ — confirm the angle is in radians and that you multiplied by the radius, not the diameter.
8. D — Correct: tan 30° = height/40, so height = 40 tan 30° = 40(√3/3) = 40√3/3 ≈ 23.1 m. A) computes 40 sin 30° = 20, wrongly treating the 40 m ground distance as a hypotenuse. B) multiplies 40 by sin 60° ≈ cos 30°'s partner value, swapping which trig value equals √3/2 versus √3/3. C) computes 40/sin 30° = 80, inverting the ratio. E) uses tan 60° = √3 — the angle measured from the tower instead of from the ground. Fix: label the triangle before choosing a ratio: the known ground distance is the leg adjacent to the elevation angle, the unknown height is opposite, so height = (adjacent) × tan(angle).
9. E — Correct: the point (−√2/2, √2/2) has x < 0 and y > 0 — QII — and the magnitudes √2/2 mark reference angle π/4, so θ = π − π/4 = 3π/4. A) gives only the reference angle, ignoring the signs that force QII. B) is the QIII angle 5π/4, where both coordinates would be negative. C) is the axis point (0, −1), which has entirely different coordinates. D) is the QIV angle, where the y-coordinate would be negative. Fix: read the signs first to lock the quadrant, then match magnitudes to a reference angle — signs choose the quadrant, magnitudes choose the family.
10. A — Correct: cosine is an even function, so cos(−π/3) = cos(π/3) = 1/2. B) applies the odd-function rule (correct for sine, wrong for cosine) and negates. C) is cos(π/6) — a wrong-row table grab. D) combines the wrong row with the wrong sign rule. E) is a tangent-family value. Fix: negatives inside trig functions: cosine ignores the sign (even), sine and tangent pull it out front (odd) — decide which rule applies before touching the table.
11. E — Correct: 11π/6 is in QIV (just short of 2π) with reference angle π/6: cos = +√3/2, sin = −1/2, tan = (−1/2)/(√3/2) = −√3/3 — both claims in E check out. A) fails because sine is negative in QIV. B) misidentifies the reference angle as π/3; the distance from 11π/6 to 2π is π/6. C) gives tangent the wrong magnitude and the wrong sign (tan is negative in QIV). D) assigns sine the cosine's magnitude. Fix: for any "which statement is true" angle item, compute the full triple (quadrant, reference angle, signed values) once at the start — then test each claim against your computed values rather than re-deriving per option.
12. D — Correct: in degree mode the calculator reads the input π/6 ≈ 0.5236 as degrees; sin(0.5236°) ≈ 0.0091 — a plausible-looking but wrong output. A) cos(π/6) ≈ 0.866, which is not what the student sought and doesn't explain the tiny output. B) rounding cannot turn 0.5 into 0.0091. C) sine's domain is all real numbers — nothing is out of domain. E) entering 6/π ≈ 1.91 in radian mode gives sin ≈ 0.94, not 0.0091, so that error doesn't produce this symptom. Fix: an absurdly small sine for a familiar angle is the degree-mode signature — when a trig output looks wrong, check the angle mode before checking your algebra.