Skip to main content
course review

Honors Pre-Calculus Unit 6 Review: Periodic Functions

This unit turns the unit circle into graphs. You'll learn to sketch and transform all six trig functions, read periods and shifts from equations, find amplitudes for sine and cosine, and use inverse trig functions to get an angle back from a ratio.

Read the overview first, then work through each topic note and try its checkpoint question before checking the answer.

start unit practicereview topicsReview with a tutorFree, 24/7, real people, from UPchieve

What is Honors Pre-Calculus unit 6?

A periodic function repeats its outputs on a fixed interval called the period. Trig functions are the main examples, and this unit is about graphing them and reading their features. By the end you should be able to go from an equation to a graph, from a graph to an equation, and from a ratio to an angle.

Honors Pre-Calculus Unit 6 covers the graphs of the trig functions and their inverses. You start with sine and cosine in the form y=asin⁡(b(x−c))+dy = a\sin(b(x - c)) + d. When aa and bb are nonzero, ∣a∣|a| is the amplitude, 2π∣b∣\frac{2\pi}{|b|} is the period, cc is the phase shift, and y=dy = d is the midline. Next come tangent, cotangent, secant, and cosecant, whose graphs have vertical asymptotes wherever their denominators equal zero. The unit ends with arcsin, arccos, and arctan, which return angles from restricted ranges so each input has exactly one output.

Four numbers describe every sine and cosine graph.

In y=asin⁡(b(x−c))+dy = a\sin(b(x - c)) + d, aa sets the height, bb sets how fast the wave repeats, cc slides it sideways, and dd moves the midline. Once you can read those four values, you can sketch the graph or write a model for data that cycles, like tides or daily temperature.

The other four trig functions break wherever a denominator hits zero.

Tangent and secant have cosine in the denominator, so they have vertical asymptotes where cos⁡x=0\cos x = 0. Cotangent and cosecant have sine in the denominator, so their asymptotes fall where sin⁡x=0\sin x = 0. Knowing which denominator matters tells you where the graph can't exist.

Inverse trig functions only work on a restricted range.

Because trig functions repeat, many angles share the same sine or cosine. To make an inverse function, restrict the original function's inputs to an interval where it is one-to-one. That's why arccos⁡(−12)\arccos(-\frac{1}{2}) is 2π3\frac{2\pi}{3} and not 4π3\frac{4\pi}{3}.

All six trig graphs and their inverses come from the unit circle.

Sine and cosine are the coordinates of a point moving around the unit circle, which is why they repeat every 2π2\pi. The other four functions are ratios of those coordinates, and the inverses undo them on one carefully chosen stretch. If a graph or answer seems strange, go back to the unit circle and check what the coordinates are doing.

Honors Pre-Calculus unit 6 topics

Unit 6 review notes

6.1

Graphs of the Sine and Cosine Functions

Every sine or cosine graph is the same wave, stretched and slid around. In the form y=asin⁡(b(x−c))+dy = a\sin(b(x - c)) + d (or with cosine), when aa and bb are nonzero, the amplitude is ∣a∣|a|, the period is 2π∣b∣\frac{2\pi}{|b|}, the phase shift is cc, and the midline is y=dy = d. If aa or bb is zero, the graph is constant and has no least positive period. A negative aa flips the wave across its midline.

The parent graphs differ only in where they start. y=sin⁡xy = \sin x starts on the midline at x=0x = 0 and rises, while y=cos⁡xy = \cos x starts at its maximum. That's because cos⁡x=sin⁡(x+π2)\cos x = \sin(x + \frac{\pi}{2}), so cosine is sine shifted π2\frac{\pi}{2} to the left. Both have domain all real numbers and range [−1,1][-1, 1].

To sketch one cycle, split the period into four equal parts. Those five x-values give you the max, the min, and the midline crossings.

Worked example. Graph y=3sin⁡(2x−π2)+1y = 3\sin(2x - \frac{\pi}{2}) + 1. First factor the inside so it matches the form: 2x−π2=2(x−π4)2x - \frac{\pi}{2} = 2(x - \frac{\pi}{4}). The amplitude is 3, the period is 2π2=π\frac{2\pi}{2} = \pi, the phase shift is π4\frac{\pi}{4} to the right, and the midline is y=1y = 1. The range is [1−3,1+3]=[−2,4][1 - 3, 1 + 3] = [-2, 4]. One cycle runs from x=π4x = \frac{\pi}{4} to x=5π4x = \frac{5\pi}{4} in steps of π4\frac{\pi}{4}. The graph is on the midline at π4\frac{\pi}{4}, reaches 4 at π2\frac{\pi}{2}, returns to the midline at 3π4\frac{3\pi}{4}, hits −2-2 at π\pi, and ends on the midline at 5π4\frac{5\pi}{4}.

  • Amplitude: The distance from the midline to a peak, equal to ∣a∣|a| or to half of (max minus min).
  • Period: The horizontal length of one full cycle, found with 2π∣b∣\frac{2\pi}{|b|}.
  • Frequency: The number of cycles per unit of input, equal to 1 divided by the period.
  • Midline: The horizontal line y=dy = d halfway between the maximum and minimum values.

Try this. A Ferris wheel seat goes from a low of 2 m to a high of 30 m. What are the amplitude and midline of a model for its height? (Answer: The amplitude is 30−22=14\frac{30 - 2}{2} = 14 m and the midline is y=30+22=16y = \frac{30 + 2}{2} = 16 m.)

6.2

Graphs of the Other Trigonometric Functions

Tangent, cotangent, secant, and cosecant are all built from sine and cosine, so their graphs break wherever the denominator is zero. Those breaks are vertical asymptotes.

Since tan⁡x=sin⁡xcos⁡x\tan x = \frac{\sin x}{\cos x}, tangent has asymptotes where cos⁡x=0\cos x = 0, at x=±π2,±3π2x = \pm\frac{\pi}{2}, \pm\frac{3\pi}{2}, and so on. Between asymptotes it rises from −∞-\infty to ∞\infty and crosses zero at multiples of π\pi. Its period is π\pi, not 2π2\pi. Cotangent, cos⁡xsin⁡x\frac{\cos x}{\sin x}, has asymptotes at multiples of π\pi and falls from left to right instead of rising.

Secant and cosecant are reciprocals, sec⁡x=1cos⁡x\sec x = \frac{1}{\cos x} and csc⁡x=1sin⁡x\csc x = \frac{1}{\sin x}. To sketch one, lightly draw the cosine or sine curve, put asymptotes at its zeros, and then draw U-shaped branches that touch each peak and valley and open away from the x-axis. Their range is (−∞,−1]∪[1,∞)(-\infty, -1] \cup [1, \infty) and their period is 2π2\pi.

Transformations work as in 6.1. For nonzero aa and bb, the period of y=atan⁡(bx)y = a\tan(bx) or y=acot⁡(bx)y = a\cot(bx) is π∣b∣\frac{\pi}{|b|}. For y=asec⁡(bx)y = a\sec(bx) or y=acsc⁡(bx)y = a\csc(bx), it is 2π∣b∣\frac{2\pi}{|b|}.

Worked example. Find the period and asymptotes of y=tan⁡(2x)y = \tan(2x). The period is π2\frac{\pi}{2}. Asymptotes occur where 2x=π2+kπ2x = \frac{\pi}{2} + k\pi, so x=π4+kπ2x = \frac{\pi}{4} + \frac{k\pi}{2}. That gives x=±π4,±3π4x = \pm\frac{\pi}{4}, \pm\frac{3\pi}{4}, and so on.

  • Vertical asymptote: A line x=cx = c where the function approaches positive or negative infinity as xx approaches cc from at least one side. For the four parent functions in this section, these lines occur where the denominator is zero.
  • Reciprocal function: A function equal to 1 divided by another, such as sec⁡x=1cos⁡x\sec x = \frac{1}{\cos x}.
  • Cotangent: The ratio cos⁡xsin⁡x\frac{\cos x}{\sin x}, with period π\pi and a graph that decreases between asymptotes.

Try this. Where are the vertical asymptotes of y=csc⁡xy = \csc x, and why? (Answer: At x=kπx = k\pi, such as 0,±π,±2π0, \pm\pi, \pm2\pi, because csc⁡x=1sin⁡x\csc x = \frac{1}{\sin x} and sin⁡x=0\sin x = 0 at those points.)

FunctionPeriodVertical asymptotesRange
$\tan x$$\pi$$x = \frac{\pi}{2} + k\pi$All real numbers
$\cot x$$\pi$$x = k\pi$All real numbers
$\sec x$$2\pi$$x = \frac{\pi}{2} + k\pi$$(-\infty, -1] \cup [1, \infty)$
$\csc x$$2\pi$$x = k\pi$$(-\infty, -1] \cup [1, \infty)$
6.3

Inverse Trigonometric Functions

An inverse trig function takes a ratio and gives back an angle. arcsin⁡(12)\arcsin(\frac{1}{2}) asks which angle has a sine of 12\frac{1}{2}. The catch is that infinitely many angles do, such as π6\frac{\pi}{6}, 5π6\frac{5\pi}{6}, and 13π6\frac{13\pi}{6}, because sine repeats. A function can only have one output, so each inverse picks answers from a restricted range where the original function is one-to-one.

arcsin⁡\arcsin returns angles in [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}], arccos⁡\arccos returns angles in [0,π][0, \pi], and arctan⁡\arctan returns angles in (−π2,π2)(-\frac{\pi}{2}, \frac{\pi}{2}). Arcsin and arccos only accept inputs from −1-1 to 11, while arctan accepts any real number. The graph of y=arctan⁡xy = \arctan x has horizontal asymptotes at y=±π2y = \pm\frac{\pi}{2}.

The notation sin⁡−1x\sin^{-1} x means arcsin, not 1sin⁡x\frac{1}{\sin x}. That reciprocal is csc⁡x\csc x.

Worked example. Evaluate arccos⁡(−12)\arccos(-\frac{1}{2}). Cosine equals −12-\frac{1}{2} at 2π3\frac{2\pi}{3} and 4π3\frac{4\pi}{3}, but only 2π3\frac{2\pi}{3} is in [0,π][0, \pi], so the answer is 2π3\frac{2\pi}{3}. Compare that with arcsin⁡(−12)=−π6\arcsin(-\frac{1}{2}) = -\frac{\pi}{6}, which lands in quadrant IV instead of at 7π6\frac{7\pi}{6}.

Worked example. Find sin⁡(arccos⁡35)\sin(\arccos\frac{3}{5}). Let θ=arccos⁡35\theta = \arccos\frac{3}{5}, so cos⁡θ=35\cos\theta = \frac{3}{5} and θ\theta is in quadrant I. Draw a right triangle with adjacent side 3 and hypotenuse 5. The opposite side is 25−9=4\sqrt{25 - 9} = 4, so sin⁡θ=45\sin\theta = \frac{4}{5}.

  • Restricted range: The interval of angles an inverse trig function can output. It matches a restricted domain on which the original trig function is one-to-one.
  • One-to-one: Describes a function where no two inputs share the same output, which is required for an inverse to exist.
  • Arctangent: The inverse of tangent, which accepts any real number and returns an angle in (−π2,π2)(-\frac{\pi}{2}, \frac{\pi}{2}).

Try this. What is arcsin⁡(sin⁡5π6)\arcsin(\sin\frac{5\pi}{6})? (Answer: π6\frac{\pi}{6}. The inside equals 12\frac{1}{2}, and the angle in [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}] with sine 12\frac{1}{2} is π6\frac{\pi}{6}, not 5π6\frac{5\pi}{6}.)

FunctionAllowed inputsOutput rangePossible quadrants; axis angles are also possible
$\arcsin x$$[-1, 1]$$[-\frac{\pi}{2}, \frac{\pi}{2}]$I and IV
$\arccos x$$[-1, 1]$$[0, \pi]$I and II
$\arctan x$All real numbers$(-\frac{\pi}{2}, \frac{\pi}{2})$I and IV

Key terms

TermDefinition
Periodic functionA function with a positive number $P$ such that, for every $x$ in its domain, $x + P$ is also in its domain and $f(x + P) = f(x)$.
Sinusoidal functionAny function with the smooth wave shape of sine or cosine, written as $y = a\sin(b(x - c)) + d$ or with cosine.
AmplitudeHalf the distance between a sinusoid's maximum and minimum values, equal to $|a|$.
PeriodThe length of one full cycle. For functions in the form $y = a f(b(x - c)) + d$ with nonzero $a$ and $b$, it is $\frac{2\pi}{|b|}$ for sine, cosine, secant, and cosecant and $\frac{\pi}{|b|}$ for tangent and cotangent.
Phase shiftThe horizontal slide of a trig graph, equal to $c$ once the function is written as $b(x - c)$ inside.
MidlineThe horizontal line $y = d$ that a sinusoid oscillates around.
Vertical asymptoteA line $x = c$ where the function approaches positive or negative infinity as $x$ approaches $c$ from at least one side. For the parent tangent, cotangent, secant, and cosecant functions, these lines occur where the denominator is zero.
Inverse trigonometric functionA function such as arcsin, arccos, or arctan that takes a ratio as input and returns an angle from a restricted range.

Mistakes to avoid in Periodic Functions

Using $b$ as the period

In y=sin⁡(4x)y = \sin(4x) the period is not 4. The parent period gets divided by ∣b∣|b|, so the period is 2π4=π2\frac{2\pi}{4} = \frac{\pi}{2}. A bigger ∣b∣|b| means a shorter period and more cycles packed in. For tangent and cotangent, divide π\pi instead of 2π2\pi.

Reading the phase shift before factoring out $b$

For y=cos⁡(2x−π)y = \cos(2x - \pi), it's tempting to say the shift is π\pi. Factor first to get cos⁡(2(x−π2))\cos(2(x - \frac{\pi}{2})), so the shift is π2\frac{\pi}{2} to the right. Also watch the sign, since sin⁡(x+3)\sin(x + 3) shifts 3 units left.

Treating $\sin^{-1} x$ as $\frac{1}{\sin x}$

The −1-1 here means inverse function, not an exponent. sin⁡−1(12)\sin^{-1}(\frac{1}{2}) is the angle π6\frac{\pi}{6}, while 1sin⁡x\frac{1}{\sin x} is csc⁡x\csc x. The confusion comes from sin⁡2x\sin^2 x, where the exponent really does mean squaring.

Giving an inverse trig answer outside the restricted range

Your unit circle shows several angles with the same sine or cosine, so it's easy to pick the wrong one. Before you write an answer, check it against the range: [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}] for arcsin, [0,π][0, \pi] for arccos, and (−π2,π2)(-\frac{\pi}{2}, \frac{\pi}{2}) for arctan. For example, arctan⁡(−1)=−π4\arctan(-1) = -\frac{\pi}{4}, not 3π4\frac{3\pi}{4}.

Study unit 6 other ways

Ready to review Unit 6?Start with the notes, check the topic cards, and use the practice or resource links when they are available for this course.