This lesson feeds the exam's largest category, Representations of Functions (30%), as well as Trigonometry (15% nominal — more in practice). "Which of the following could be an equation of the function graphed in the xy-plane above?" is among the most common CLEP stems, and sinusoids are its favorite subject. Graph anatomy and inverse-trig values are Section 2 (no-calculator) skills; in Section 1 the online calculator will graph any candidate equation, which turns matching items into elimination exercises — if you know what features to compare.
Let θ increase steadily and plot each unit-circle coordinate against θ:
[GRAPH: Two aligned panels over one period [0, 2π]. Top: y = sin θ with zeros at 0, π, 2π; maximum (π/2, 1); minimum (3π/2, −1). Bottom: y = cos θ with maximum (0, 1); zeros at π/2, 3π/2; minimum (π, −1). Dashed vertical gridlines at multiples of π/2 connect the panels; the midline y = 0 is dashed on both.]
Same wave, offset start: cos θ = sin(θ + π/2). Every stretch or shift of either curve is called sinusoidal.
| constant | feature | how to read it |
|---|---|---|
| |a| | amplitude | vertical distance from midline to peak; a < 0 reflects across the midline |
| b | period T = 2π/|b| | equivalently b = 2π/T; bigger b = faster repetition |
| c | phase shift | horizontal translation: (x − c) shifts right by c |
| d | midline y = d | vertical translation; range is [d − |a|, d + |a|] |
From a graph: midline = (max + min)/2, amplitude = (max − min)/2 (half the swing, not the whole swing), period = horizontal length of one full cycle (for example, peak to peak).
Factor the inside first. sin(2x − π/3) = sin(2(x − π/6)): the phase shift is π/6, not π/3. Reading the shift before factoring out b is the single most reliable distractor generator on these items.
Work features, not vibes — in this order:
Example: a sinusoid with maximum 7 at x = 0, minimum −1, period π has midline (7 + (−1))/2 = 3, amplitude (7 − (−1))/2 = 4, b = 2π/π = 2, and starts at its maximum: y = 4 cos(2x) + 3. Verify two points before committing: f(0) = 4 + 3 = 7 ✓ and f(π/2) = −4 + 3 = −1 ✓.
tan θ = sin θ/cos θ is the slope of the terminal side, and its graph is nothing like a wave:
[GRAPH: y = tan θ over (−3π/2, 3π/2): three identical increasing branches separated by dashed vertical asymptotes at θ = ±π/2 and ±3π/2; zeros marked at −π, 0, π; each branch rises from −∞ to +∞.]
sec x = 1/cos x csc x = 1/sin x cot x = cos x/sin x
Note the crossover in the names: sec pairs with cos, csc pairs with sin. Values come free from the originals: sec(π/3) = 1/(1/2) = 2; csc(π/6) = 2; cot(π/4) = 1.
| undefined where | vertical asymptotes at | range | period | |
|---|---|---|---|---|
| sec x | cos x = 0 | x = π/2 + kπ | (−∞, −1] ∪ [1, ∞) | 2π |
| csc x | sin x = 0 | x = kπ | (−∞, −1] ∪ [1, ∞) | 2π |
| cot x | sin x = 0 | x = kπ | all real numbers | π |
Graph intuition: sec's graph is a chain of U-shapes sitting on the peaks and hanging from the troughs of cos — where cos is small, sec is huge. Since |cos x| ≤ 1 and |sin x| ≤ 1, their reciprocals have magnitude ≥ 1: the band (−1, 1) is unreachable for sec and csc. An equation like sec x = 1/3 therefore has no solution — a planted answer choice on the exam.
[GRAPH: y = cos x (light curve) with y = sec x (bold) on [0, 2π]: an upward U of sec touching cos's maximum at (0, 1) and (2π, 1), a downward U hanging below cos's minimum at (π, −1), dashed vertical asymptotes at π/2 and 3π/2, and the horizontal band −1 < y < 1 shaded and labeled "sec never enters."]
"sin θ = 1/2 — what is θ?" has infinitely many answers (π/6, 5π/6, 13π/6, …), but a function must return exactly one. So each inverse answers from a fixed, agreed slice of the circle:
| inverse | domain | range | reads as |
|---|---|---|---|
| arcsin x (sin⁻¹) | [−1, 1] | [−π/2, π/2] | right half of the circle (QIV and QI) |
| arccos x (cos⁻¹) | [−1, 1] | [0, π] | top half (QI and QII) |
| arctan x (tan⁻¹) | all reals | (−π/2, π/2) | open — the asymptote angles are excluded |
Evaluation protocol: "arccos(−1/2) = the angle in [0, π] whose cosine is −1/2" → 2π/3. Negative inputs split by function:
That asymmetry is heavily tested: arccos(−1/2) is 2π/3, never −π/3.
Compositions. sin(arcsin x) = x always (for x in [−1, 1]). But arcsin(sin θ) = θ only when θ is already in [−π/2, π/2]. Otherwise evaluate inside-out: arcsin(sin(5π/6)) = arcsin(1/2) = π/6, not 5π/6. For mixed compositions like sin(arccos(3/5)), draw the right triangle: cos θ = 3/5 with θ in [0, π] means adjacent 3, hypotenuse 5, opposite 4, and θ is in QI (cosine positive), so sin(arccos(3/5)) = 4/5.
1. A — Correct: amplitude = |a| = |−4| = 4; the negative sign encodes a reflection across the midline, not a negative amplitude, and the +1 moves the midline without affecting the swing. B) reports a itself — amplitude is a distance and is never negative. C) adds the vertical shift to the amplitude (mixing range-maximum with amplitude). D) reports the midline value d. E) gives the full max-to-min swing (8), which is twice the amplitude. Fix: amplitude = |coefficient on the trig function| — take the absolute value and ignore the vertical shift entirely.
2. E — Correct: T = 2π/|b| = 2π/3. A) reports b itself. B) inverts to b/... producing π/3 — dividing π (instead of 2π) by b, the tangent-formula contamination. C) multiplies 2π by 3 instead of dividing. D) forgets that b ≠ 1 changes the period at all. Fix: for sine and cosine, period = 2π divided by |b| — larger b means the wave repeats faster, so the period must come out smaller than 2π when b > 1.
3. C — Correct: the midline is y = d, and here d = −5; the coefficient 2 stretches the wave but does not move its center. A) reports the amplitude as if it located the midline. B) computes −5 + 2 = −3, the graph's maximum, not its midline. D) assumes the default midline y = 0, ignoring the vertical shift. E) computes −5 − 2 = −7, the graph's minimum. Fix: midline = the constant added outside the trig function; equivalently (max + min)/2 — never the max or min themselves.
4. D — Correct: factor the argument first: sin(2x − π/3) = sin(2(x − π/6)), so the shift is right π/6. A) reads π/3 as a leftward shift, wrong on both distance and direction. B) reads the unfactored π/3 as the shift — the classic error this item exists to catch. C) gets the factored distance but flips the direction: (x − c) with c > 0 moves right. E) treats the inside constant as a vertical translation. Fix: force the form b(x − c) before reading any phase shift — divide the inside constant by b, and remember (x − c) shifts right.
5. E — Correct: midline = (7 + (−1))/2 = 3, amplitude = (7 − (−1))/2 = 4, period π → b = 2, and the curve starts at a maximum → +cos: y = 4 cos(2x) + 3. Check: f(0) = 7 ✓, f(π/2) = −1 ✓. A) is a sine starter: f(0) = 3, the midline — but the graph starts at its peak. B) swaps amplitude and midline: its maximum is 7 (looks right at x = 0!) but its minimum is 1, not −1. C) has period 2π, not π — the graph completes a full cycle by x = π. D) flips the midline's sign: maximum 1, minimum −7. Fix: extract midline, amplitude, period, and starting behavior from the graph in that order, then verify the candidate equation at two labeled points — one point is not enough to separate these distractors.
6. B — Correct: tangent's own period is π, so y = tan(2x) has period π/|b| = π/2. A) forgets that b = 2 compresses the graph. C) applies the sinusoid formula 2π/b to tangent. D) multiplies instead of divides. E) reports b itself. Fix: tangent's period formula has π on top, not 2π — memorize the pair: sin/cos → 2π/|b|, tan → π/|b|.
7. D — Correct: arccos returns the angle in [0, π] whose cosine is −1/2; cosine is negative in QII, and the reference angle for the value 1/2 is π/3, so arccos(−1/2) = π − π/3 = 2π/3. A) applies the arcsin-style rule arccos(−x) = −arccos(x), which is false — arccos never outputs a negative angle. B) drops the negative sign from the input entirely. C) negates the correct answer, again leaving [0, π]. E) uses reference angle π/6 instead of π/3 (that would be arccos(−√3/2)). Fix: negative input to arccos → obtuse output via π − arccos(x); check that any arccos answer lands in [0, π] before moving on.
8. D — Correct: evaluate inside-out: sin(5π/6) = 1/2, then arcsin(1/2) = π/6 — the angle in [−π/2, π/2] with sine 1/2. A) cancels arcsin(sin θ) = θ blindly; 5π/6 is outside arcsin's range, so the composition cannot return it. B) attaches a spurious negative — the inside value 1/2 is positive, so the output must be positive. C) grabs the wrong reference angle for the value 1/2. E) is coterminal-style reasoning (adding π) that has no role here. Fix: arcsin(sin θ) = θ only when θ ∈ [−π/2, π/2]; otherwise compute the inner value numerically first, then ask which in-range angle produces it.
9. B — Correct: arcsin returns angles from the right half of the unit circle: [−π/2, π/2], endpoints included since arcsin(±1) = ±π/2. A) is arccos's range — the top half. C) wrongly opens the interval; that open interval belongs to arctan (whose asymptote angles are unattainable). D) is arcsin's domain, not its range — the inputs are sine values in [−1, 1]. E) is a general angle-measuring window, not any inverse's range. Fix: memorize the range triple — arcsin: [−π/2, π/2], arccos: [0, π], arctan: (−π/2, π/2) — and note only arctan's is open.
10. A — Correct: let θ = arccos(3/5), so cos θ = 3/5 with θ ∈ [0, π]; since 3/5 > 0, θ is in QI. Right triangle: adjacent 3, hypotenuse 5, opposite √(25 − 9) = 4, so sin θ = 4/5 (positive in QI). B) returns the input 3/5 as if sin(arccos x) = x. C) applies a wrong sign — θ is in QI where sine is positive. D) inverts the ratio (hypotenuse over opposite), which exceeds 1 and can never be a sine. E) mixes the two legs, computing adjacent/opposite = 3/4 — that ratio is cot θ, not sin θ. Fix: for sin(arccos(a/c)) draw the triangle — adjacent a, hypotenuse c, opposite √(c² − a²) — and let the inverse's range set the sign.
11. E — Correct: sec x = 1/3 requires cos x = 3, impossible since |cos x| ≤ 1; equivalently, sec's range excludes the open band (−1, 1), and 1/3 sits inside it. A) sin x = −0.4 is solvable — sine reaches every value in [−1, 1]. B) tan x = 100 is solvable — tangent's range is all real numbers, no matter how large the target. C) cos x = 0.99 is solvable — within [−1, 1]. D) csc x = 3 is solvable — it means sin x = 1/3, and |3| ≥ 1 is in cosecant's range. Fix: convert reciprocal-function equations to their parent (sec x = k means cos x = 1/k) and test the parent's range — sec and csc can never equal a proper fraction, while tangent can equal anything.
12. C — Correct: cosecant is the reciprocal of sine: csc x = 1/sin x. A) is secant — the names cross over (co-secant pairs with sine, secant with co-sine), the exact confusion this pattern of distractors probes. B) is cotangent expressed as a reciprocal (where both are defined). D) is tan x. E) is cot x in ratio form. Fix: the "co-" prefix crosses between the pairs: sec = 1/cos and csc = 1/sin — each reciprocal function pairs with the co-named parent, not the same-named one.
1. A — Correct: amplitude = |a| = |−4| = 4; the negative sign encodes a reflection across the midline, not a negative amplitude, and the +1 moves the midline without affecting the swing. B) reports a itself — amplitude is a distance and is never negative. C) adds the vertical shift to the amplitude (mixing range-maximum with amplitude). D) reports the midline value d. E) gives the full max-to-min swing (8), which is twice the amplitude. Fix: amplitude = |coefficient on the trig function| — take the absolute value and ignore the vertical shift entirely.
2. E — Correct: T = 2π/|b| = 2π/3. A) reports b itself. B) inverts to b/... producing π/3 — dividing π (instead of 2π) by b, the tangent-formula contamination. C) multiplies 2π by 3 instead of dividing. D) forgets that b ≠ 1 changes the period at all. Fix: for sine and cosine, period = 2π divided by |b| — larger b means the wave repeats faster, so the period must come out smaller than 2π when b > 1.
3. C — Correct: the midline is y = d, and here d = −5; the coefficient 2 stretches the wave but does not move its center. A) reports the amplitude as if it located the midline. B) computes −5 + 2 = −3, the graph's maximum, not its midline. D) assumes the default midline y = 0, ignoring the vertical shift. E) computes −5 − 2 = −7, the graph's minimum. Fix: midline = the constant added outside the trig function; equivalently (max + min)/2 — never the max or min themselves.
4. D — Correct: factor the argument first: sin(2x − π/3) = sin(2(x − π/6)), so the shift is right π/6. A) reads π/3 as a leftward shift, wrong on both distance and direction. B) reads the unfactored π/3 as the shift — the classic error this item exists to catch. C) gets the factored distance but flips the direction: (x − c) with c > 0 moves right. E) treats the inside constant as a vertical translation. Fix: force the form b(x − c) before reading any phase shift — divide the inside constant by b, and remember (x − c) shifts right.
5. E — Correct: midline = (7 + (−1))/2 = 3, amplitude = (7 − (−1))/2 = 4, period π → b = 2, and the curve starts at a maximum → +cos: y = 4 cos(2x) + 3. Check: f(0) = 7 ✓, f(π/2) = −1 ✓. A) is a sine starter: f(0) = 3, the midline — but the graph starts at its peak. B) swaps amplitude and midline: its maximum is 7 (looks right at x = 0!) but its minimum is 1, not −1. C) has period 2π, not π — the graph completes a full cycle by x = π. D) flips the midline's sign: maximum 1, minimum −7. Fix: extract midline, amplitude, period, and starting behavior from the graph in that order, then verify the candidate equation at two labeled points — one point is not enough to separate these distractors.
6. B — Correct: tangent's own period is π, so y = tan(2x) has period π/|b| = π/2. A) forgets that b = 2 compresses the graph. C) applies the sinusoid formula 2π/b to tangent. D) multiplies instead of divides. E) reports b itself. Fix: tangent's period formula has π on top, not 2π — memorize the pair: sin/cos → 2π/|b|, tan → π/|b|.
7. D — Correct: arccos returns the angle in [0, π] whose cosine is −1/2; cosine is negative in QII, and the reference angle for the value 1/2 is π/3, so arccos(−1/2) = π − π/3 = 2π/3. A) applies the arcsin-style rule arccos(−x) = −arccos(x), which is false — arccos never outputs a negative angle. B) drops the negative sign from the input entirely. C) negates the correct answer, again leaving [0, π]. E) uses reference angle π/6 instead of π/3 (that would be arccos(−√3/2)). Fix: negative input to arccos → obtuse output via π − arccos(x); check that any arccos answer lands in [0, π] before moving on.
8. D — Correct: evaluate inside-out: sin(5π/6) = 1/2, then arcsin(1/2) = π/6 — the angle in [−π/2, π/2] with sine 1/2. A) cancels arcsin(sin θ) = θ blindly; 5π/6 is outside arcsin's range, so the composition cannot return it. B) attaches a spurious negative — the inside value 1/2 is positive, so the output must be positive. C) grabs the wrong reference angle for the value 1/2. E) is coterminal-style reasoning (adding π) that has no role here. Fix: arcsin(sin θ) = θ only when θ ∈ [−π/2, π/2]; otherwise compute the inner value numerically first, then ask which in-range angle produces it.
9. B — Correct: arcsin returns angles from the right half of the unit circle: [−π/2, π/2], endpoints included since arcsin(±1) = ±π/2. A) is arccos's range — the top half. C) wrongly opens the interval; that open interval belongs to arctan (whose asymptote angles are unattainable). D) is arcsin's domain, not its range — the inputs are sine values in [−1, 1]. E) is a general angle-measuring window, not any inverse's range. Fix: memorize the range triple — arcsin: [−π/2, π/2], arccos: [0, π], arctan: (−π/2, π/2) — and note only arctan's is open.
10. A — Correct: let θ = arccos(3/5), so cos θ = 3/5 with θ ∈ [0, π]; since 3/5 > 0, θ is in QI. Right triangle: adjacent 3, hypotenuse 5, opposite √(25 − 9) = 4, so sin θ = 4/5 (positive in QI). B) returns the input 3/5 as if sin(arccos x) = x. C) applies a wrong sign — θ is in QI where sine is positive. D) inverts the ratio (hypotenuse over opposite), which exceeds 1 and can never be a sine. E) mixes the two legs, computing adjacent/opposite = 3/4 — that ratio is cot θ, not sin θ. Fix: for sin(arccos(a/c)) draw the triangle — adjacent a, hypotenuse c, opposite √(c² − a²) — and let the inverse's range set the sign.
11. E — Correct: sec x = 1/3 requires cos x = 3, impossible since |cos x| ≤ 1; equivalently, sec's range excludes the open band (−1, 1), and 1/3 sits inside it. A) sin x = −0.4 is solvable — sine reaches every value in [−1, 1]. B) tan x = 100 is solvable — tangent's range is all real numbers, no matter how large the target. C) cos x = 0.99 is solvable — within [−1, 1]. D) csc x = 3 is solvable — it means sin x = 1/3, and |3| ≥ 1 is in cosecant's range. Fix: convert reciprocal-function equations to their parent (sec x = k means cos x = 1/k) and test the parent's range — sec and csc can never equal a proper fraction, while tangent can equal anything.
12. C — Correct: cosecant is the reciprocal of sine: csc x = 1/sin x. A) is secant — the names cross over (co-secant pairs with sine, secant with co-sine), the exact confusion this pattern of distractors probes. B) is cotangent expressed as a reciprocal (where both are defined). D) is tan x. E) is cot x in ratio form. Fix: the "co-" prefix crosses between the pairs: sec = 1/cos and csc = 1/sin — each reciprocal function pairs with the co-named parent, not the same-named one.