Math
trigonometrymedium~20 min

Graphing the Sine Function

You will be able to sketch the parent sine wave from its five key points, read the amplitude, period, phase shift, and midline straight from an equation of the form y = A sin(B(x - C)) + D, write an equation from a described or drawn graph, find the maximum and minimum values of any shifted wave without graphing, and connect the wave back to the unit circle.

Introduction

The sine function produces the most recognizable shape in mathematics after the straight line: a smooth wave that rises to a crest, falls through a trough, and repeats forever in both directions. Every sine wave you will ever be asked about is the parent wave y = sin(x) stretched, squeezed, or slid, and exactly four numbers control those changes. Once you can name what each number does, a question that looks like a wall of symbols becomes a matter of reading off four values.

The two tests treat this topic differently. The digital SAT keeps trigonometry light: it asks you to convert between radians and degrees, to use the unit circle to find a sine or cosine value, and occasionally to recognize a translated trig graph, but it rarely asks you to compute a period. The ACT asks directly. Expect an ACT question that hands you an equation like y = 3 sin(2x) - 1 and asks for the amplitude or the period, or one that shows a wave and asks which equation produced it. That makes this lesson a guaranteed point or two on the ACT and a reliable one on the SAT.

The common failure here is treating the four transformation numbers as four unrelated facts to memorize. They are not. Each one answers a physical question about the wave, namely how tall it is, how long one cycle takes, where the cycle starts, and where its center line sits. Ask those four questions of any equation and the graph draws itself.

Game plan

How to attack these questions on test day.
  1. 1

    Read the four numbers in a fixed order

    Take every sine equation in the same order: amplitude first, then period, then phase shift, then vertical shift. Write y = A sin(B(x - C)) + D next to the given equation and match each piece. Working in a fixed order means you never skip the period because you were staring at the shift, and it makes the equation-from-graph questions run in reverse with no new thinking.

  2. 2

    Turn the max and min into amplitude and midline

    When a question gives or draws a wave with a highest and lowest value, compute two things immediately: the amplitude is half the distance between them, (max - min) / 2, and the midline is their average, (max + min) / 2. Those two numbers are A and D. The ACT loves to show a wave that oscillates between, say, 1 and 7 and ask for the amplitude, and the trap answer is always the full distance 6 rather than half of it.

  3. 3

    Count cycles to get the period, then divide into 2π

    From an equation, the period is 2π divided by the number multiplying x, not the number itself. From a graph, find the distance along the x-axis between two consecutive crests, or between a crest and the next trough doubled. If a graph shows k full cycles across an interval of length L, the period is L / k. Only after you have the period do you go looking for B, and B = 2π / period.

  4. 4

    Factor before you read the phase shift

    In y = sin(2x - π/2) the phase shift is not π/2. Factor the coefficient out of the parentheses first: 2x - π/2 = 2(x - π/4), so the shift is π/4 to the right. The shift is the number subtracted from x after the coefficient has been pulled out. If a question does not involve the phase shift at all, which is common on both tests, skip this step entirely rather than risk a sign error for nothing.

  5. 5

    Check with one point you can evaluate in your head

    After you settle on an equation, substitute one easy x-value. At the phase shift C, the argument is zero and sin(0) = 0, so the wave must sit on its midline y = D. A quarter period later the wave must be at its maximum, D + A. If either check fails, your period or your shift is wrong. On the digital SAT the built-in graphing calculator will draw the wave for you, so type it in and compare against the figure.

Theory

Start with the parent function y = sin(x). Its outputs are never larger than 1 or smaller than -1, and it repeats every 2π units, so one full cycle runs from x = 0 to x = 2π. Five points define that cycle and you should know them cold: the wave starts on the axis at (0, 0), rises to a crest at (π/2, 1), returns to the axis at (π, 0), falls to a trough at (3π/2, -1), and comes back to the axis at (2π, 0). Plot those five points, join them with a smooth curve, and copy the shape left and right forever. In decimal terms, a full cycle is about 6.28 units long, so the crest sits near x = 1.57 and the trough near x = 4.71.

Why those values? A sine is a coordinate on the unit circle, the circle of radius 1 centered at the origin. Rotate counterclockwise from the positive x-axis through an angle θ, and the point you reach has coordinates (cos θ, sin θ). The sine is the height of that point. At θ = 0 the point is on the x-axis, so the height is 0; at θ = π/2 you are at the top of the circle, height 1; at θ = π you are back on the axis, height 0; at θ = 3π/2 you are at the bottom, height -1. Since a full turn is 2π radians, the heights repeat every 2π, which is exactly why the period of sin(x) is 2π. The same picture explains the SAT's radian and degree questions: 2π radians is one full turn, so 2π radians equals 360 degrees and π radians equals 180 degrees, and the sine of an angle is the same number whether you name the angle in degrees or radians.

The general sine function is y = A sin(B(x - C)) + D, and each letter answers one question about the wave. The amplitude is |A|, the distance from the center line up to a crest or down to a trough. It is a distance, so it is never negative; a negative A flips the wave upside down, so it starts by falling instead of rising, but the height of the wave is unchanged. The full vertical span of the wave, from trough to crest, is 2|A|. So y = 2 sin(x) rises to 2 and falls to -2, twice as tall as the parent wave, while y = (1/2) sin(x) stays between -1/2 and 1/2.

The period is the horizontal length of one full cycle, and it is controlled by B through the formula period = 2π / B. The reason is that the parent wave completes a cycle when its argument runs from 0 to 2π, so B(x - C) must grow by 2π, which takes a change in x of 2π / B. A larger B squeezes more cycles into the same space and shortens the period. So y = sin(2x) has period 2π / 2 = π and completes two full cycles between 0 and 2π, while y = sin(x / 2) has period 2π / (1/2) = 4π and needs twice as much room as the parent. Reading in the other direction, if you know the period from a graph, then B = 2π / period.

The phase shift is C, and it slides the whole wave horizontally: C units to the right if the equation reads (x - C), and C units to the left if it reads (x + C). The sign feels backward, and the way to trust it is to ask where the cycle begins. The parent wave starts a cycle where its argument equals zero, and B(x - C) equals zero when x = C, so the cycle now begins at x = C instead of at x = 0. When an equation is written as y = A sin(Bx + E) + D with the coefficient not factored, the shift is -E / B, which is the same thing as factoring B out of the parentheses first. On both tests a phase shift usually appears only in the multiple-choice matching questions, where you must tell sin(x - π/2) from sin(x + π/2), so it is worth being precise about the sign.

The vertical shift is D, and it moves the center line of the wave, called the midline, from y = 0 to y = D. Amplitude and midline together give the maximum and minimum of any sine wave with no graphing at all: the maximum value is D + |A| and the minimum value is D - |A|. The reason is that sin of anything lies between -1 and 1, so A sin(anything) lies between -|A| and |A|, and adding D slides that whole interval up or down. Running the formulas backward is how you read a graph: the midline is halfway between the maximum and minimum, D=max+min2D = \frac{\text{max} + \text{min}}{2}, and the amplitude is half the distance between them, A=maxmin2|A| = \frac{\text{max} - \text{min}}{2}.

In the interactive graph below, drag the amplitude slider and watch the crests and troughs move away from the x-axis while the crossing points stay fixed. Drag the period slider and watch the crossings spread apart or bunch together while the height never changes, which is the visual proof that amplitude and period are independent. Drag the phase shift and notice that the entire wave slides sideways as a rigid shape, and drag the vertical shift and watch the midline rise and fall carrying the wave with it. The app writes the sine as A sin(2π x / period + phase) + shift, so the period slider is the actual length of one cycle in x-units, and the parent wave corresponds to a period of about 6.28.

Worked figures

The parent wave y = sin(x) and its five key points

One full cycle runs from x = 0 to x = 2π, about 6.28 units. The wave starts on the axis, crests at a height of 1 a quarter of the way through, crosses the axis at the halfway point, bottoms out at -1 three quarters of the way through, and returns to the axis to start over. The dashed lines mark the maximum and minimum, which the wave touches but never exceeds.

Figure 1
-1.5-1-0.500.511.501234567y = 1y = -1(0, 0)(π/2, 1)(π, 0)(3π/2, -1)(2π, 0)x (radians)y
  • y = sin(x)

Key values of one cycle, in radians and degrees

These are the five points every sketch is built from. The unit-circle column explains each value: the sine is the height of the point on the circle of radius 1 after rotating through the given angle. The SAT expects you to move between the radian and degree columns freely, using π radians = 180 degrees.

Table 1
x (radians)x (degrees)sin(x)Position on the unit circlePlace in the cycle
000on the positive x-axis, height 0start, on the midline heading up
π/2901top of the circle, height 1crest, one quarter through
π1800on the negative x-axis, height 0midline, halfway through, heading down
3π/2270-1bottom of the circle, height -1trough, three quarters through
3600back on the positive x-axis, height 0end of the cycle, ready to repeat

Amplitude changes the height, not the timing

Both waves cross the x-axis at exactly the same places, because the period has not changed. The taller wave reaches 2 and -2 instead of 1 and -1. Amplitude is measured from the midline to a crest, so doubling A doubles the height of the crest.

Figure 2
-2-101201234567y = 2y = -2(π/2, 1)(π/2, 2)(3π/2, -1)(3π/2, -2)x (radians)y
  • y = sin(x), amplitude 1
  • y = 2 sin(x), amplitude 2

Period changes the timing, not the height

Doubling the coefficient of x halves the period. y = sin(2x) has period 2π / 2 = π, so it finishes one full cycle by x = π and squeezes two complete cycles into the space where y = sin(x) completes one. Both waves still rise only to 1 and fall only to -1.

Figure 3
-1.5-1-0.500.511.501234567x = πx = 2πsin(2x) completes a cycle at x = πsin(x) completes a cycle at x = 2πx (radians)y
  • y = sin(x), period 2π
  • y = sin(2x), period π

A vertical shift moves the midline, and amplitude then sets the max and min

y = 2 sin(x) + 1 has midline y = 1 and amplitude 2, so it oscillates between 1 + 2 = 3 and 1 - 2 = -1. The dashed midline is the new center of the wave; the crest and trough sit exactly 2 units above and below it. Notice the wave no longer crosses the x-axis at the key points, so read amplitude from the midline, never from the x-axis.

Figure 4
-2-10123401234567midline y = 1y = 3y = -1maximum (π/2, 3)minimum (3π/2, -1)(0, 1)x (radians)y
  • y = 2 sin(x) + 1

Where sine comes from: a height on the unit circle

Rotate through an angle θ from the positive x-axis to reach a point on the circle of radius 1. That point has coordinates (cos θ, sin θ), so the sine is its height above the x-axis and the cosine is its distance to the right. Here the point is (0.8, 0.6), which gives sin θ = 0.6 and cos θ = 0.8 with a hypotenuse of exactly 1, since 0.8² + 0.6² = 0.64 + 0.36 = 1. As θ sweeps around the circle the height traces out one cycle of the sine wave.

Figure 5
-0.4-0.20.20.40.60.811.21.41.6-0.4-0.20.20.40.60.811.2radius 1sin θ = 0.6cos θ = 0.8θ(0.8, 0.6)O

Try it yourself

Amplitude (A, height of the wave)1
Period (length of one cycle)6.283
Phase shift (slides the wave sideways)0
Vertical shift (D, moves the midline)0
Equation
y = 1 · sin(2πx / 6.28)

Worked examples

Try each one before opening the solution.

Example 1: Read all four features from an equation

State the amplitude, period, phase shift, and midline of y = 3 sin(2x - π/2) + 1, and give its maximum and minimum values.

y = 3 sin(2x - π/2) + 1
-3-2-101234501234567midline y = 1x = π/4cycle starts at (π/4, 1)maximum (π/2, 4)minimum (π, -2)x (radians)y
  • y = 3 sin(2x - π/2) + 1
Show solution
  1. Amplitude: the number multiplying the sine is A = 3, so the amplitude is |3| = 3.

    Amplitude is a distance from the midline, so it is always |A|. Nothing inside the parentheses affects it.

  2. Period: the coefficient of x is B = 2, so the period is 2π / B = 2π / 2 = π.

    The period is 2π divided by B, not B itself. A coefficient of 2 makes the wave repeat twice as often, so the period shrinks.

  3. Phase shift: factor the coefficient out of the argument, 2x - π/2 = 2(x - π/4), so C = π/4 and the wave is shifted π/4 units to the right.

    Reading the shift as π/2 straight from the unfactored form is the most common error on this question type.

  4. Midline: the constant added at the end is D = 1, so the midline is the line y = 1.
  5. Maximum and minimum: max = D + |A| = 1 + 3 = 4 and min = D - |A| = 1 - 3 = -2.

    Sine never exceeds 1 or drops below -1, so 3 sin(anything) stays between -3 and 3, and adding 1 slides that to between -2 and 4.

  6. Check with a point: at x = π/4 the argument is 2(π/4) - π/2 = 0 and sin(0) = 0, so y = 3(0) + 1 = 1, which is on the midline, exactly where a cycle should begin.

    A quarter period later, at x = π/4 + π/4 = π/2, the argument is π/2 and y = 3(1) + 1 = 4, the maximum. Both checks pass.

Answer: Amplitude 3, period π, phase shift π/4 to the right, midline y = 1, maximum 4, minimum -2.

Example 2: Write an equation from a described graph

A sine wave has a maximum value of 5 and a minimum value of -1, completes exactly one full cycle every 4 units along the x-axis, and passes upward through its midline at x = 0. Write an equation for the wave in the form y = A sin(Bx) + D.

y = 3 sin((π/2) x) + 2
-2-10123456012345678midline y = 2period 4(0, 2)maximum (1, 5)minimum (3, -1)(4, 2), one cycle donexy
  • y = 3 sin((π/2) x) + 2
Show solution
  1. Amplitude: half the distance from minimum to maximum, A = (5 - (-1)) / 2 = 6 / 2 = 3.

    The full span from -1 to 5 is 6, but amplitude is measured from the midline, so it is half of that.

  2. Midline: the average of the maximum and minimum, D = (5 + (-1)) / 2 = 4 / 2 = 2, so the midline is y = 2.

    Check: 2 + 3 = 5 and 2 - 3 = -1 recover the given maximum and minimum.

  3. Period: one cycle every 4 units means the period is 4, so B = 2π / period = 2π / 4 = π/2.

    B is not the period. It is what you multiply x by so that the argument grows by 2π over one cycle.

  4. Phase shift: the parent sine also crosses its midline going upward at x = 0, so no horizontal shift is needed and C = 0.
  5. Assemble the equation: y = 3 sin((π/2) x) + 2.
  6. Check a crest: a quarter period is 1 unit, so at x = 1 the argument is π/2, sin(π/2) = 1, and y = 3(1) + 2 = 5, the given maximum. At x = 3 the argument is 3π/2, sin(3π/2) = -1, and y = 3(-1) + 2 = -1, the given minimum.

Answer: y = 3 sin((π/2) x) + 2

Example 3: Find the maximum, minimum, and period of a flipped and shifted wave

Let f(x) = -4 sin(3x + π) - 2. What are the maximum value of f, the minimum value of f, and the period of its graph?

Show solution
  1. Amplitude: A = -4, so the amplitude is |-4| = 4.

    The negative sign reflects the wave across its midline so it starts by falling instead of rising. It does not change how far the wave reaches.

  2. Midline: D = -2, so the wave is centered on the line y = -2.
  3. Maximum: D + |A| = -2 + 4 = 2.
  4. Minimum: D - |A| = -2 - 4 = -6.

    The trap is computing the maximum as A + D = -4 + (-2) = -6, which is actually the minimum. Always measure from the midline using |A|, and keep the sign of A as a separate fact about direction.

  5. Period: B = 3, so the period is 2π / 3.

    The + π inside the parentheses is a phase shift of 3x + π = 3(x + π/3), which is π/3 to the left. It changes where the cycle starts but not the maximum, the minimum, or the period.

  6. Check by bounding: sin(3x + π) lies between -1 and 1, so -4 sin(3x + π) lies between -4 and 4, and subtracting 2 gives a range from -6 to 2, matching the minimum and maximum found above.

Answer: Maximum 2, minimum -6, period 2π/3.

Practice

Check your understanding 1

What is the period of the graph of y = 5 sin(4x) - 1?

Check your understanding 2

The graph of a sine function oscillates between a minimum value of 1 and a maximum value of 7. What is the amplitude of the function?

Check your understanding 3

Which equation represents the graph of y = sin(x) shifted π/3 units to the right and 2 units up?

Check your understanding 4

What is the maximum value of the function y = -3 sin(x/2 + 1) + 4?

Check your understanding 5

The graph of a function of the form y = A sin(Bx) has amplitude 2 and completes exactly 3 full cycles between x = 0 and x = 2π. Which of the following is the equation of the function?

Common mistakes

Using the coefficient of x as the period. In y = sin(4x) the period is 2π / 4 = π/2, not 4. The coefficient tells you how many cycles fit into 2π, so a bigger number means a shorter period. Every time you write down a period, make sure you have divided into 2π, and sanity check that a coefficient bigger than 1 gave a period smaller than 2π.

Taking the full height of the wave as the amplitude. A wave that runs from -1 to 5 has amplitude 3, not 6, because amplitude is measured from the midline to a crest, not from trough to crest. Compute (max - min) / 2 every time, and pair it with (max + min) / 2 for the midline; the two numbers together should rebuild the original maximum and minimum.

Reading the phase shift without factoring. In y = sin(2x - π/2) the shift is π/4, not π/2, because 2x - π/2 = 2(x - π/4). Factor the coefficient of x out of the parentheses before you read C. Then remember the sign is backward: (x - π/4) is a shift to the right, and (x + π/4) is a shift to the left.

Letting a negative amplitude change the maximum and minimum. In y = -4 sin(x) + 1 the amplitude is 4, the midline is 1, and the wave still reaches 1 + 4 = 5 and 1 - 4 = -3. The negative sign only reflects the wave across the midline so it falls first instead of rising. Use |A| in the max and min formulas and treat the sign as a separate fact about direction.

Mixing amplitude and vertical shift. Amplitude controls how tall the wave is and vertical shift controls where its center line sits; a wave can be tall and low, or short and high, and the two numbers are independent. When you read a graph, find the midline first, because a shifted wave no longer crosses the x-axis at its key points and measuring amplitude from the x-axis instead of from the midline will give the wrong answer.

Forgetting that radians and degrees describe the same angle. The period of sin(x) is 2π when x is in radians and 360 when x is in degrees, and sin(π/2) and sin(90 degrees) are both 1. On the SAT, convert using π radians = 180 degrees before you compare an angle to the key points, and make sure your calculator is in the mode the problem uses.