πŸ”οΈ Trigonometry Peaks Β· Trigonometry

Graphs of Sine & Cosine

Unroll the unit circle into a wave: read the amplitude, period, midline and phase shift of y = a sin(b(x - h)) + k off its equation and its graph, and write the equation of a wave you are shown.

In short

  • Every sine and cosine graph is y = a sin(b(x - h)) + k: a sets the size of the swing, b sets how often it happens, h slides it sideways and k lifts it.
  • The amplitude is |a|, a distance measured from the midline, so it is never negative and it is never the highest point; the midline is y = k.
  • The period is 2pi/b, not b. Running it backwards, b = 2pi / period, and the pi always cancels.
  • A phase shift can only be read once the bracket is in the form b(x - h), so factor b out before reading anything: 2x - pi is 2(x - pi/2), a shift of pi/2.
  • From a graph, amplitude = (greatest - least) / 2 and midline = (greatest + least) / 2, and the period is measured peak to the next peak.
  • Modelling a wheel, a tide or a spring is the same four numbers, plus one decision: whether the situation starts at the top, at the bottom or on the midline.

Unrolling the circle into a wave

Climber Sora keeps a rope wound round a drum at the top of Sine Summit, and paints one red mark on the rim. Turn the drum steadily and watch the mark: it rises, hangs at the top, comes back down, drops to the bottom, and climbs again. Nothing else happens, ever. That is the whole of this skill.

On the unit circle a point at angle t sits at (cos t, sin t). Sine is the height of that point and cosine is how far across it is. Walk once round the circle and the height goes 0, up to 1, back to 0, down to -1, back to 0. Plot that height against the angle instead of against the circle and you have unrolled the circle into a wave.

  • y = sin(x) starts at 0, on the middle, and rises.
  • y = cos(x) starts at 1, at the top, and comes down.
  • Both finish a complete lap at x = 2pi and then do exactly the same thing again.

That "and then again" is what makes these functions worth having. A rope swinging, a tide, a wheel, a bouncing weight, daylight through a year β€” all of them come back to where they were and go round again, and all of them are described by a sine or a cosine that has been stretched, squashed, lifted and slid.

Everything on the rest of this page is one of those four moves, written into one equation:

y = a sin(b(x - h)) + k

Read it left to right and it tells you the whole story of the wave: a is how big the swing is, b decides how often it happens, h slides it sideways, and k lifts it. The same four letters work with cosine in place of sine. Note also that x here is an angle in radians, not in degrees β€” on this mountain everything is measured in radians.

Amplitude and midline: how big the swing, and where it sits

Two of the four numbers are easy, because both are read straight off the equation without any work.

The midline is the horizontal line the wave swings about β€” halfway between its highest point and its lowest point. Plain sine and cosine swing about 0, so the only thing that can move the midline is the number added on the outside. In y = a sin(b(x - h)) + k, the midline is the line y = k.

The amplitude is the distance from the midline up to the highest point β€” half the total swing. It is |a|, the number multiplying the sine or cosine. Two words in that sentence do a lot of work:

  • Distance. An amplitude is never negative. y = -3 sin(x) has amplitude 3; the minus sign flips the wave upside down about its midline, but it does not make the swing smaller.
  • From the midline. The amplitude is not the highest point. For y = 3 sin(x) + 2 the highest point is 5, and the amplitude is 3.

Put the two together and the greatest and least values fall out with no effort at all, because sin and cos never leave the range from -1 to 1 whatever is fed into them:

greatest value = k + |a| least value = k - |a|

For y = 4 cos(x) - 1: amplitude 4, midline y = -1, so the wave lives between -5 and 3.

And it runs backwards just as easily. If a graph rises to 11 and drops to 3, then

amplitude = (11 - 3) / 2 = 4 midline: k = (11 + 3) / 2 = 7

Subtract and halve for the amplitude; add and halve for the midline. Those two lines answer most "read the graph" questions on their own.

Period: what b really does

The period is how far along x you travel before the wave starts repeating. A plain sine or cosine has period 2pi, because that is once round the circle.

Now put a number in front of x. In y = sin(2x), the input runs twice as fast, so the wave gets through a whole lap in half the distance. In y = sin(x/3) the input crawls, and the wave takes three times as long. In general:

period = 2pi / b

b is not the period. b counts how many complete cycles fit into 2pi. That single confusion accounts for more wrong answers here than everything else combined, so say the formula out loud with the division in it: *two pi over b*.

  • y = cos(2x): b = 2, period = 2pi/2 = pi.
  • y = sin(x/3): b = 1/3, period = 2pi / (1/3) = 6pi.
  • y = sin((3/2)x): b = 3/2, period = 2pi / (3/2) = 4pi/3.

The formula runs backwards too, and rearranging it needs no new idea:

b = 2pi / period

So a wave with period 4pi has b = 2pi / (4pi) = 1/2, and a wave that repeats every pi has b = 2pi / pi = 2. Notice how the pi cancels: whenever the period is a multiple or a fraction of pi, b comes out as a plain number or a plain fraction with no pi left in it. If your b still has a pi in it, check the division.

On a graph, measure a period between two features that look identical: peak to the next peak, or trough to the next trough, or one rising crossing of the midline to the next rising crossing. Peak to the next trough is only half a period β€” a classic way to end up with an answer exactly half the size it should be.

Phase shift: factor b out first

The last move slides the wave sideways. In y = sin(x - h) every point of the wave moves h to the right, and in y = sin(x + h) every point moves h to the left. That distance is the phase shift.

The direction reads backwards from the sign, which feels wrong until you ask *which x now does the old job of x = 0*. In sin(x - pi/2), the bracket is 0 when x = pi/2, so the point that used to sit at 0 has moved right to pi/2. Minus means right.

y = sin(x - pi/4) shifted pi/4 to the RIGHT y = cos(x + pi/3) shifted pi/3 to the LEFT

Now the trap, and it is the whole reason this skill is worth a page of its own. Consider

y = cos(2x - pi)

It is tempting to read a shift of pi. It is not pi. The formula reads a shift from the form b(x - h), and 2x - pi is not yet in that form. Factor b out first:

2x - pi = 2(x - pi/2)

so h = pi/2 and the wave is shifted pi/2 to the right β€” half of what it first looked like. The rule is short: divide the constant by b. A shift is a distance along the x-axis, and b has already stretched that axis, so the constant has to be stretched back.

One more example, in the other direction. For y = sin((1/2)x + pi/4):

(1/2)x + pi/4 = (1/2)(x + pi/2)

so h = -pi/2 and the shift is pi/2 to the left β€” this time bigger than it looked, because b was less than 1. Multiplying back out is the check that costs five seconds and settles it.

Sine and cosine are the same wave a quarter turn apart, which is why a shift can turn one into the other: cos(x) = sin(x + pi/2). Two people can describe the same graph with two different equations, and both be right.

Writing an equation, modelling the world, and typing your answer

Building an equation from a description is the four moves in reverse, in a fixed order. Sora does it the way ropes get set: same order every time, no thinking required.

1. a β€” the amplitude, half the distance from the lowest point to the highest. 2. k β€” the midline, the average of the lowest and the highest. 3. b β€” from the period, using b = 2pi / period. 4. h β€” the shift, positive for a slide to the right.

For a wave with amplitude 2, period pi, midline y = -1 and no shift: a = 2, k = -1, b = 2pi/pi = 2, h = 0, giving y = 2 sin(2x) - 1.

Modelling something real is the same four steps with the numbers hidden inside a story. A Ferris wheel of radius 5 m whose centre is 8 m up and which turns once every 20 seconds gives a = 5, k = 8, and b = 2pi/20 = pi/10. The last decision is where the ride starts. A rider boarding at the bottom is at the lowest point at t = 0, and cos(0) = 1 puts a plain cosine at the top β€” so the cosine is turned over:

h(t) = 8 - 5 cos((pi/10)t)

Check it at the two easy times. At t = 0 the height is 8 - 5 = 3 m, the boarding platform. Ten seconds later, half a turn, cos(pi) = -1 and the height is 8 + 5 = 13 m, the very top. A model that survives those two checks is almost certainly right.

How to type your answer here.

  • An amplitude, a midline value k, and a greatest or least value are whole numbers: type 3, or -1. An amplitude is always positive.
  • A period, a phase shift, and the x-value of a peak are exact multiples of pi. Write pi as the two letters pi, and type the value the way you would write it: pi/4, 2pi/3, 3pi/2, 6pi. The checker works these out, so (2/3)pi passes too β€” but a rounded decimal such as 2.09 does not, because it is not the same number. When a question says "give an exact value using pi", a decimal will not be accepted.
  • Where an equation, a direction ("to the left") or a model is asked for, the options are listed and you pick one.
  • Angles here are always radians. There are no degree answers anywhere in this skill.

Worked examples

Example 1

State the amplitude, the period and the midline of y = 3 sin(2x) - 1.

  1. Match it against y = a sin(b(x - h)) + k: a = 3, b = 2, h = 0, k = -1.
  2. Amplitude = |a| = |3| = 3 β€” the distance from the midline to the top, not the top itself.
  3. Period = 2pi/b = 2pi/2 = pi. The wave gets through a whole cycle twice as fast as a plain sine.
  4. Midline: k = -1, so the wave swings about the line y = -1.
  5. As a check, the wave runs between -1 - 3 = -4 and -1 + 3 = 2.

Example 2

How far, and in which direction, is the graph of y = cos(2x - pi/3) shifted from the graph of y = cos(2x)?

  1. The shift can only be read from the form b(x - h), and 2x - pi/3 is not in that form yet.
  2. Factor b = 2 out of the bracket: 2x - pi/3 = 2(x - pi/6).
  3. Check by multiplying back out: 2 * x = 2x and 2 * (-pi/6) = -pi/3. It matches.
  4. The bracket is now x - h with h = pi/6.
  5. A positive h slides the graph to the right, so the shift is pi/6 to the RIGHT β€” half of the pi/3 it first looked like.

Example 3

Write the equation of a cosine wave with amplitude 4, period 6pi and midline y = 2, with no horizontal shift.

  1. Amplitude 4 gives a = 4.
  2. Midline y = 2 gives k = 2.
  3. The period is the only one that needs arithmetic: b = 2pi / period = 2pi / (6pi) = 1/3. The pi cancels, as it always does.
  4. No shift, so h = 0 and the bracket is just bx.
  5. Assemble: y = 4 cos((1/3)x) + 2. Check the extremes: it should run between 2 - 4 = -2 and 2 + 4 = 6.

Example 4

The tide at Percent Port is 11 m deep at high tide and 3 m deep at low tide, and one high tide follows another 12 hours later. Write a cosine model for the depth d(t), t in hours after a high tide, and find the depth after 6 hours.

  1. Amplitude = (11 - 3) / 2 = 4 metres: half the distance between the two.
  2. Midline = (11 + 3) / 2 = 7 metres, so k = 7.
  3. Period 12 hours, so b = 2pi / 12 = pi/6.
  4. At t = 0 the tide is HIGH, and cos(0) = 1 already puts a cosine at the top, so the amplitude keeps its plus sign: d(t) = 7 + 4 cos((pi/6)t).
  5. After 6 hours: (pi/6) * 6 = pi, and cos(pi) = -1, so d(6) = 7 + 4 * (-1) = 3 metres β€” low tide, exactly half a cycle in, which is the check that the model is right.

Practice problems, with solutions

Three problems of increasing difficulty, each with the full working. In the game these are generated fresh every time; these three are fixed so this page always shows the same ones.

Problem 1

Difficulty 1 of 5

Look at the wave y = 2 cos(x) + 4. It swings forever between a high point and a low point. What is its amplitude?

Answer: 2

  1. Compare y = 2 cos(x) + 4 with y = a cos(bx) + k.
  2. a = 2 and k = 4.
  3. Amplitude = |a| = |2| = 2.

Problem 2

Difficulty 3 of 5

Find the period of y = 7 cos((1/2)x) + 4 β€” the length of x it takes to complete one full cycle. Give an exact value using pi, for example pi/4 or 2pi/3 β€” not a decimal.

Answer: 4pi

  1. Compare y = 7 cos((1/2)x) + 4 with y = a cos(bx) + k: b = 1/2.
  2. period = 2pi/b = 2pi / (1/2)
  3. period = 4pi

Problem 3

Difficulty 4 of 5

The graph of y = cos(2x + 2pi) is the graph of y = cos(2x) slid sideways. Which way, and by how much?

  1. pi to the left
  2. 2pi to the left
  3. pi to the right
  4. 2pi to the right

Answer: A. pi to the left

  1. Factor b = 2 out of the bracket: 2x + 2pi = 2(x + pi).
  2. Now the bracket is x - h with h = -pi.
  3. The shift is pi to the left.

Common mistakes

  • Giving b as the period. In y = cos(2x) the period is 2pi/2 = pi, not 2. b counts the cycles that fit into 2pi; the period is what you get after dividing.
  • Reading the phase shift straight off an unfactored bracket. For y = cos(2x - pi) the shift is pi/2, not pi β€” factor the 2 out of the bracket first.
  • Sliding the graph the wrong way. x - h moves the wave h to the RIGHT and x + h moves it left, which is the opposite of how the sign looks.
  • Giving the amplitude as the highest point. For y = 3 sin(x) + 2 the highest point is 5 and the amplitude is 3; the amplitude is measured from the midline, not from the x-axis.
  • Giving the amplitude as the full distance from the lowest point to the highest. That distance is twice the amplitude β€” subtract, then halve.
  • Writing a negative amplitude. y = -3 cos(x) has amplitude 3; the minus sign turns the wave upside down without changing how far it swings.
  • Giving the midline as the greatest value instead of the average of the greatest and the least.
  • Answering with a decimal where an exact value is asked for. A period of 2pi/3 is not 2.09; type it as 2pi/3.

What you should be able to do

  • State the amplitude, period and midline of a sine or cosine function from its equation.
  • Find the phase shift and the direction of a reflection.
  • Write the equation of a sine or cosine wave from its features or its graph.
  • Model a periodic situation, such as a tide or a wheel, with a sinusoidal function.

Where this fits in the curriculum

Common Core

  • HSF-TF.B.5

    High school β€” Choose trigonometric functions to model periodic phenomena with specified amplitude, frequency and midline.

  • HSF-IF.C.7.E

    High school β€” Graph exponential and logarithmic functions, showing intercepts and end behaviour, and trigonometric functions, showing period, midline and amplitude.

  • HSF-TF.A.4

    High school β€” Use the unit circle to explain the symmetry and periodicity of the trigonometric functions.

    HSF-TF.A.4 is a (+) standard β€” beyond the college- and career-ready threshold, i.e. precalculus rather than Algebra II.

  • HSF-BF.B.3

    High school β€” Identify the effect on a graph of replacing f(x) by f(x) + k, kΒ·f(x), f(kx) and f(x + k).

SAT

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