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Precalculus · F.TF.B.5

Amplitude and reflection

A guitar string swings farther from rest when it is plucked harder, and half a vibration later it bulges the other way. How can one number in front of sine describe both?

You need
Graphing sine and cosinePrecalculusStretches and compressionsAlgebra 2
Unlocks
Writing an equation from a wavePrecalculusCombining a sine and a cosine into one wavePrecalculus
Where you are 4 in Amplitude, period and phase shift
Trigonometry7
Trigonometric functions and their graphs8
Amplitude, period and phase shift4
Amplitude and reflection
Finding the period from the equationMidline and vertical shiftPhase shift

Why this exists

The graphs of y = \sin x and y = \cos x (Graphing sine and cosine) never rise above 1 or fall below -1, because they record the coordinates of a point on a circle of radius 1. Most things that repeat swing by other amounts. When a guitar string vibrates, its middle may move 2 millimeters from rest or a tenth of that, depending on how hard it was plucked, and half a vibration after the string bulges up it bulges down by the same amount. To describe such motion with sine and cosine, a wave needs a size other than 1, and it needs a way to be turned upside down. Both come from one number multiplied onto the output: y = a\sin x and y = a\cos x. This lesson says what that number does to the graph and names the size it controls, the amplitude.

Definition

The graph of a periodic function such as y = a\sin x swings between a highest value M and a lowest value m. Its amplitude is half the distance between them:

\text{amplitude} = \frac{M - m}{2}

It is the distance from the line halfway between the highest and lowest values, the midline, up to a highest point, and equally down to a lowest point. For the waves in this lesson the midline is the x-axis, y = 0; moving it is the subject of Midline and vertical shift. Amplitude is a distance, so it is never negative.

For any real number a \ne 0, the graphs of

y = a\sin x \qquad\text{and}\qquad y = a\cos x

have amplitude |a|. When a < 0, each graph is the graph with |a| in place of a reflected across the x-axis: every point (x, y) moves to (x, -y).

Intuition

Multiplying a function’s whole output by a is a vertical stretch or compression: every height is multiplied by a and every input is left alone. On a wave that has three visible effects.

  • Heights at the key points. The five key points of y = \sin x on [0, 2\pi] have heights 0, 1, 0, -1, 0. On y = a\sin x they become 0, a, 0, -a, 0. The tops and bottoms move to heights a and -a, so the wave rises |a| above the axis and falls |a| below it.
  • What stays put. Zero times a is still zero, so the wave crosses the axis at the same places. The highest and lowest points move straight up or down, keeping their x-values, and the cycle still repeats every 2\pi. Only the vertical size changes.
  • A negative a. Multiplying by a negative number changes the sign of every height. Where y = \sin x is above the axis, y = -\sin x is the same distance below, so arches become troughs and troughs become arches: the graph is flipped over the x-axis.

The figure shows the stretch and the flip together for \displaystyle y = -\frac{5}{2}\sin x. Each of its heights is \displaystyle \frac{5}{2} times the height of y = \sin x, with the sign changed.

\displaystyle y = -\frac{5}{2}\sin x
y = \sin x
-2\pi
\displaystyle -\frac{3\pi}{2}
-\pi
\displaystyle -\frac{\pi}{2}
\displaystyle \frac{\pi}{2}
\pi
\displaystyle \frac{3\pi}{2}
2\pi
1
-1
\displaystyle \frac{5}{2}
\displaystyle -\frac{5}{2}
x
y
Figure 1Every height of y = \sin x (dashed) multiplied by \displaystyle -\frac{5}{2}. The zeros stay put, the wave is \displaystyle \frac{5}{2} times as tall, and it is flipped over the x-axis.

When |a| > 1 the wave is stretched away from the axis; when 0 < |a| < 1 it is compressed toward it; and |a| = 1 gives a wave of the same size, flipped when a = -1.

Results

For a real number a \ne 0, each of y = a\sin x and y = a\cos x has

  • amplitude |a|, range \left[-|a|, |a|\right], highest value |a| and lowest value -|a|;
  • the same zeros and period as the unstretched graph: zeros of a\sin x at x = n\pi and of a\cos x at \displaystyle x = \frac{\pi}{2} + n\pi, for every integer n, and period 2\pi;
  • when a > 0, its highest points where the unstretched graph has its highest points; when a < 0, its highest points where the unstretched graph has its lowest points, and the other way round.

For sine the flip over the x-axis is also a mirror image left to right, since -\sin x = \sin(-x). For cosine it is not (Example 6).

Why it is true

Start with -1 \le \sin x \le 1, true for every x because \sin x is the height of a point on the unit circle. If a > 0, multiplying all three parts by a keeps the order:

-a \le a\sin x \le a

If a < 0, multiplying by a reverses both inequalities:

a \le a\sin x \le -a

In both cases this says -|a| \le a\sin x \le |a|. Both ends are reached: \sin x takes the values 1 and -1 at \displaystyle \frac{\pi}{2} and \displaystyle \frac{3\pi}{2}, so a\sin x takes a and -a there, which are |a| and -|a| in some order. So M = |a| and m = -|a|, and the amplitude is

\frac{M - m}{2} = \frac{|a| - (-|a|)}{2} = |a|

The argument for cosine is the same with \cos in place of \sin; its extremes are at 0 and \pi.

For the reflection, let a < 0, so a = -|a|. A point (x, y) is on y = |a|\sin x exactly when -y = -|a|\sin x = a\sin x, that is, exactly when (x, -y) is on y = a\sin x. The move (x, y) \to (x, -y) is reflection across the x-axis, so the two graphs are mirror images.

Worked examples

Example 1

The lead wave

For \displaystyle y = -\frac{5}{2}\sin x, give the amplitude and range, the five key points on [0, 2\pi], and the highest and lowest points on [0, 2\pi].

Solution

Here \displaystyle a = -\frac{5}{2}, so the amplitude is \displaystyle \left|-\frac{5}{2}\right| = \frac{5}{2} and the range is \displaystyle \left[-\frac{5}{2}, \frac{5}{2}\right]. Since |a| > 1 the wave is stretched, and since a < 0 it is reflected.

Multiply each key height of y = \sin x, which are 0, 1, 0, -1, 0, by \displaystyle -\frac{5}{2}:

(0, 0),\ \left(\tfrac{\pi}{2}, -\tfrac{5}{2}\right),\ (\pi, 0),\ \left(\tfrac{3\pi}{2}, \tfrac{5}{2}\right),\ (2\pi, 0)

The highest point on [0, 2\pi] is \displaystyle \left(\frac{3\pi}{2}, \frac{5}{2}\right), where \sin x is lowest, and the lowest point is \displaystyle \left(\frac{\pi}{2}, -\frac{5}{2}\right), where \sin x is highest. The cycle opens with a trough instead of an arch, as the figure above shows. Check with the definition: \displaystyle \frac{M - m}{2} = \frac{\frac{5}{2} - \left(-\frac{5}{2}\right)}{2} = \frac{5}{2}.

Example 2

A compressed cosine

For \displaystyle y = \frac{3}{4}\cos x, give the amplitude, range and key points on [0, 2\pi], and find every x in [0, 2\pi] where \displaystyle y = \frac{3}{8}.

Solution

The amplitude is \displaystyle \frac{3}{4} and the range is \displaystyle \left[-\frac{3}{4}, \frac{3}{4}\right]. Since 0 < a < 1, the wave is compressed toward the axis and not reflected. The cosine key heights 1, 0, -1, 0, 1 become

\left(0, \tfrac{3}{4}\right),\ \left(\tfrac{\pi}{2}, 0\right),\ \left(\pi, -\tfrac{3}{4}\right),\ \left(\tfrac{3\pi}{2}, 0\right),\ \left(2\pi, \tfrac{3}{4}\right)

For the equation, divide both sides of \displaystyle \frac{3}{4}\cos x = \frac{3}{8} by \displaystyle \frac{3}{4}, allowed because it is not zero: \displaystyle \cos x = \frac{1}{2}. The compressed wave is at height \displaystyle \frac{3}{8} exactly where the plain cosine is at height \displaystyle \frac{1}{2}. On [0, 2\pi] that is \displaystyle x = \frac{\pi}{3}, in Quadrant I, and \displaystyle x = \frac{5\pi}{3}, in Quadrant IV with reference angle \displaystyle \frac{\pi}{3}. Multiplying by a changes the heights, but not where along the axis the wave reaches a given fraction of its full height.

Example 3

Reading a from a graph

Asahi has the graph below and knows it is either y = a\sin x or y = a\cos x. Find its amplitude, decide which form it is, and find a.

\displaystyle \frac{\pi}{2}
\pi
\displaystyle \frac{3\pi}{2}
2\pi
2
-2
x
y
Figure 2The graph in Example 3. Its five key points are dotted.
Solution

The highest value is M = 2, at x = \pi, and the lowest is m = -2, at x = 0 and x = 2\pi. The amplitude is \displaystyle \frac{2 - (-2)}{2} = 2.

The graph passes through (0, -2). Every graph y = a\sin x passes through (0, 0), since \sin 0 = 0, so this is not one of them. For y = a\cos x, the point (0, -2) gives a\cos 0 = a = -2. Check the other key points: \displaystyle -2\cos\frac{\pi}{2} = 0, -2\cos\pi = 2 and \displaystyle -2\cos\frac{3\pi}{2} = 0, matching the graph. So the graph is y = -2\cos x. Its amplitude is 2, and it is reflected, which is why it opens at a lowest point where y = \cos x opens at a highest.

Example 4

Mio’s guitar string

Mio’s guitar string is pinned at both ends. Measure position along it in a unit chosen so the string runs from x = 0 to x = \pi, and measure displacement y from the resting position in millimeters, up positive. Treat it as an ideal string, perfectly flexible and evenly stretched. When it vibrates in its simplest way, its fundamental mode, its shape at any instant is exactly half a sine wave along its length: y = a\sin x for 0 \le x \le \pi, where the number a changes with time as the string vibrates.

y = 1.8\sin x
y = -1.8\sin x
\displaystyle \frac{\pi}{2}
\pi
1.8
-1.8
x
y
Figure 3Mio’s string in Example 4, at two instants half a vibration apart. Each shape is the reflection of the other across the rest line.

(a) At one instant the middle of the string is 1.8 millimeters above rest. Find the shape, and the displacement at \displaystyle x = \frac{\pi}{6}. (b) Half a vibration later the middle is 1.8 millimeters below rest. Find the shape and the displacement at \displaystyle x = \frac{\pi}{6}. (c) Later still the middle is 0.6 millimeters below rest. Find the displacement at \displaystyle x = \frac{\pi}{4}, to the nearest hundredth of a millimeter.

Solution

The middle of the string is at \displaystyle x = \frac{\pi}{2}, where \sin x = 1, so the displacement of the middle is \displaystyle a\sin\frac{\pi}{2} = a. The number a is the signed displacement of the middle, and |a| is the amplitude of the shape.

(a) a = 1.8, so the shape is y = 1.8\sin x. At \displaystyle x = \frac{\pi}{6}, \displaystyle y = 1.8 \cdot \frac{1}{2} = 0.9: the point is 0.9 millimeters above rest.

(b) a = -1.8, so the shape is y = -1.8\sin x, the reflection of the shape in (a) across the rest line. Every point is now the same distance on the other side: at \displaystyle x = \frac{\pi}{6}, y = -0.9, so 0.9 millimeters below rest.

(c) a = -0.6, so y = -0.6\sin x, and at \displaystyle x = \frac{\pi}{4}, \displaystyle y = -0.6 \cdot \frac{\sqrt{2}}{2} = -0.3\sqrt{2} \approx -0.42. The point is about 0.42 millimeters below rest. The ends stay pinned at every instant, since a\sin 0 = a\sin\pi = 0 for every a.

Example 5

Finding a from one point

The graph of y = a\cos x passes through \displaystyle \left(\frac{3\pi}{4}, 3\right). Find a and the amplitude. Could the point \displaystyle \left(\frac{\pi}{2}, 0\right) have been used instead?

Solution

Substitute the point: \displaystyle 3 = a\cos\frac{3\pi}{4} = a\left(-\frac{\sqrt{2}}{2}\right). Divide by \displaystyle -\frac{\sqrt{2}}{2}:

a = \frac{3}{-\frac{\sqrt{2}}{2}} = -\frac{6}{\sqrt{2}} = -3\sqrt{2}

So \displaystyle a = -3\sqrt{2}, and the amplitude is \displaystyle 3\sqrt{2} \approx 4.24. The sign makes sense: at \displaystyle \frac{3\pi}{4} the plain cosine is negative, yet this graph is above the axis there, so the wave must be flipped. Check: \displaystyle -3\sqrt{2} \cdot \left(-\frac{\sqrt{2}}{2}\right) = \frac{3 \cdot 2}{2} = 3.

The point \displaystyle \left(\frac{\pi}{2}, 0\right) is no use. It is a zero of \cos x, and a \cdot 0 = 0 for every a, so every graph y = a\cos x passes through it. Only a point off the axis pins down a.

Example 6

Two ways to flip

Aoi reflects the graph of y = \sin x across the y-axis, and Asahi reflects it across the x-axis. Show that they get the same graph. Then show that for y = \cos x the two reflections differ.

Solution

Reflecting across the y-axis sends (x, y) to (-x, y), so it turns the graph of y = f(x) into the graph of y = f(-x). Reflecting across the x-axis turns it into y = -f(x).

For sine, Aoi’s graph is y = \sin(-x) and Asahi’s is y = -\sin x. The terminal point of -x is the reflection across the x-axis of the terminal point of x, so its height is the opposite: \sin(-x) = -\sin x (Unit circle symmetry and even-odd identities). The two graphs are the same.

For cosine, Aoi’s graph is y = \cos(-x). Those same two terminal points share their x-coordinate, so \cos(-x) = \cos x, and reflecting across the y-axis gives back y = \cos x unchanged. Asahi’s graph is y = -\cos x, which has height -1 at x = 0 where y = \cos x has height 1. The two reflections differ, so “reflect the wave” must say which axis. A negative a in y = a\cos x always means across the x-axis.

Boundaries

  • a \ne 0. With a = 0 the graph is the line y = 0. It does not swing at all, the formula |a| gives amplitude 0, and there is nothing to reflect.
  • The midline is the x-axis only here. Amplitude is measured from the midline. Once a constant is added, as in y = 2\sin x + 5, the highest value is no longer the amplitude, though \displaystyle \frac{M - m}{2} still is (Midline and vertical shift).
  • Only a number multiplying the whole output. A number multiplying x inside, as in \sin 3x, changes the period and not the height (Finding the period from the equation).
  • A constant a. If the factor in front changes with x, as in y = x\sin x, the swings grow without bound and the graph has no amplitude at all.

Where this goes next

Amplitude is one of the numbers that set the shape of a wave. Finding the period from the equation reads how wide a cycle is from the number multiplying x, a phase shift slides the wave sideways once the inside is factored, and Midline and vertical shift lifts the midline off the axis. Together they let you write an equation from a wave, where reading the amplitude as \displaystyle \frac{M - m}{2} comes first. A sine wave plus a cosine wave of the same period is a single wave whose amplitude is neither coefficient (Combining a sine and a cosine into one wave). And amplitude is the size of the swing in every model of something that repeats, from the string in Example 4 to Simple harmonic motion.

Exercises

Reading the equation

For each graph, give the amplitude and range, say whether it is reflected, and say whether it is stretched or compressed compared with y = \sin x or y = \cos x.

  1. y = 6\sin x
  2. \displaystyle y = -\frac{1}{3}\cos x
  3. y = \pi\cos x
  4. \displaystyle y = -\frac{7}{4}\sin x

Key points and extremes

  1. Give the five key points of y = -4\cos x on [0, 2\pi].
  2. Give the five key points of \displaystyle y = \frac{5}{3}\sin x on [0, 2\pi].
  3. Find the highest points of y = -3\sin x on [-2\pi, 2\pi].
  4. Find the lowest points of \displaystyle y = -\frac{1}{2}\cos x on [0, 4\pi].
  5. Give the exact point of the graph of y = -5\cos x at \displaystyle x = \frac{2\pi}{3}.

Finding a

  1. The graph of y = a\sin x has a highest point at \displaystyle \left(\frac{3\pi}{2}, 7\right). Find a.
  2. The graph of y = a\cos x passes through \displaystyle \left(\frac{\pi}{3}, 4\right). Find a and the amplitude.
  3. The graph of y = a\sin x passes through \displaystyle \left(\frac{5\pi}{6}, -3\right). Find a and the amplitude.
  4. The graph of y = a\sin x passes through \displaystyle \left(\frac{\pi}{4}, -2\right). Find a and the amplitude.

The guitar string

Use the setup of Example 4: the shape of the string is y = a\sin x for 0 \le x \le \pi, with y in millimeters.

  1. At one instant the shape is y = -2.4\sin x. How far from rest is the point at \displaystyle x = \frac{\pi}{3}, and on which side, to the nearest hundredth of a millimeter?
  2. Asahi photographs the string when the point at \displaystyle x = \frac{\pi}{6} is 0.7 millimeters above rest. Find the shape, and how far the middle is from rest.
  3. When the shape is y = 2.1\sin x, which points of the string are at least 1.05 millimeters above rest?

Solving with a stretched wave

  1. Find every x in [0, 2\pi] with -2\sin x = 1.
  2. Where on [0, 2\pi] is y = -3\cos x increasing?

Reasoning

  1. In y = a\sin x the number a may be negative, but the amplitude never is. Explain why, using the definition of amplitude.
  2. Show that -\cos x = \cos(x - \pi) for every x, and say what it means for the graph of y = -\cos x.
  3. For a > 0, find the straight-line distance between the points \displaystyle \left(\frac{\pi}{2}, a\right) and \displaystyle \left(\frac{3\pi}{2}, -a\right) of y = a\sin x. Mio says doubling a doubles this distance. Is she right?
  4. Explain why every graph y = a\sin x with a \ne 0 crosses the x-axis at the same places.
  5. The graphs of y = a\sin x and y = c\sin x have the same amplitude but are different graphs. How are a and c related, and how are the graphs related?
Answers
  1. Amplitude 6, range [-6, 6]; not reflected, stretched by a factor of 6.
  2. Amplitude \displaystyle \frac{1}{3}, range \displaystyle \left[-\frac{1}{3}, \frac{1}{3}\right]; reflected, and compressed, since \displaystyle \left|-\frac{1}{3}\right| < 1.
  3. Amplitude \pi, range [-\pi, \pi]; not reflected, stretched by a factor of \pi \approx 3.14.
  4. Amplitude \displaystyle \frac{7}{4}, range \displaystyle \left[-\frac{7}{4}, \frac{7}{4}\right]; reflected, and stretched, since \displaystyle \frac{7}{4} > 1.
  5. (0, -4), \displaystyle \left(\frac{\pi}{2}, 0\right), (\pi, 4), \displaystyle \left(\frac{3\pi}{2}, 0\right), (2\pi, -4). The cosine heights 1, 0, -1, 0, 1, each times -4.
  6. (0, 0), \displaystyle \left(\frac{\pi}{2}, \frac{5}{3}\right), (\pi, 0), \displaystyle \left(\frac{3\pi}{2}, -\frac{5}{3}\right), (2\pi, 0).
  7. \displaystyle \left(-\frac{\pi}{2}, 3\right) and \displaystyle \left(\frac{3\pi}{2}, 3\right). The highest value 3 comes where \sin x = -1, at \displaystyle x = \frac{3\pi}{2} + 2\pi n for integers n.
  8. \displaystyle \left(0, -\frac{1}{2}\right), \displaystyle \left(2\pi, -\frac{1}{2}\right) and \displaystyle \left(4\pi, -\frac{1}{2}\right). The lowest value \displaystyle -\frac{1}{2} comes where \cos x = 1.
  9. \displaystyle \left(\frac{2\pi}{3}, \frac{5}{2}\right). Here \displaystyle \cos\frac{2\pi}{3} = -\frac{1}{2}, and \displaystyle -5 \cdot \left(-\frac{1}{2}\right) = \frac{5}{2}.
  10. a = -7. From \displaystyle a\sin\frac{3\pi}{2} = -a = 7. A highest point where \sin x is lowest means the wave is reflected.
  11. a = 8, amplitude 8. From \displaystyle a\cos\frac{\pi}{3} = \frac{a}{2} = 4.
  12. a = -6, amplitude 6. From \displaystyle a\sin\frac{5\pi}{6} = \frac{a}{2} = -3.
  13. \displaystyle a = -2\sqrt{2}, amplitude \displaystyle 2\sqrt{2} \approx 2.83. From \displaystyle a \cdot \frac{\sqrt{2}}{2} = -2, \displaystyle a = -\frac{4}{\sqrt{2}} = -2\sqrt{2}.
  14. About 2.08 millimeters below rest. \displaystyle -2.4 \cdot \frac{\sqrt{3}}{2} = -1.2\sqrt{3} \approx -2.08.
  15. y = 1.4\sin x, and the middle is 1.4 millimeters above rest. From \displaystyle a \cdot \frac{1}{2} = 0.7, a = 1.4, and the displacement of the middle is a.
  16. The points with \displaystyle \frac{\pi}{6} \le x \le \frac{5\pi}{6}. The condition 2.1\sin x \ge 1.05 means \displaystyle \sin x \ge \frac{1}{2}, which on [0, \pi] holds between the two crossings \displaystyle \frac{\pi}{6} and \displaystyle \frac{5\pi}{6}.
  17. \displaystyle x = \frac{7\pi}{6}, \frac{11\pi}{6}. Dividing by -2 gives \displaystyle \sin x = -\frac{1}{2}: Quadrants III and IV, reference angle \displaystyle \frac{\pi}{6}.
  18. On [0, \pi]. Multiplying by -3 turns falling into rising, and \cos x falls on [0, \pi].
  19. The amplitude is \displaystyle \frac{M - m}{2}, and the highest value M is at least the lowest value m, so it is never negative. For y = a\sin x it equals |a|, which is -a when a < 0; the sign of a goes into the reflection, not into the size.
  20. The angle x - \pi is half a turn from x, so its terminal point is the opposite point (-\cos x, -\sin x), and \cos(x - \pi) = -\cos x. The graph of y = -\cos x is the cosine graph slid \pi to the right: for cosine, the reflection across the x-axis and that slide give the same wave.
  21. The distance is \displaystyle \sqrt{\pi^2 + 4a^2}: the points are \pi apart across and 2a apart up and down. Mio is wrong. With 2a the distance is \displaystyle \sqrt{\pi^2 + 16a^2}, which is less than \displaystyle 2\sqrt{\pi^2 + 4a^2} = \sqrt{4\pi^2 + 16a^2}. Only the up-and-down gap doubles; the gap across stays \pi.
  22. With a \ne 0, a\sin x = 0 holds exactly when \sin x = 0, since a product of two numbers is zero only when one of them is. So the zeros are x = n\pi, for every integer n, whatever a is.
  23. |a| = |c| and a \ne c, so c = -a. Then c\sin x = -(a\sin x) for every x, so each graph is the other reflected across the x-axis.
kg1128 · updated 2026-10-20 · CC-BY-SA-4.0