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10.3 Graphs of `y` = `a` sin( `bx` + `c`) and `y` = `a` cos(`bx` + `c`)

Phase Angle • Displacement • Graphs of $y = a \sin (b x + c)$ $y = a \sin (b x + c)$ and $y = a \cos (b x + c)$ $y = a \cos (b x + c)$ • Cycle

In the function $y = a \sin (b x + c),$ $y = a \sin (b x + c),$ c represents the phase angle. It is another very important quantity in graphing the sine and cosine functions. Its meaning is illustrated in the following example.

EXAMPLE 1 Sketch function with phase angle

Sketch the graph of $y = \sin (2 x + \frac{π}{4}) .$ $y = \sin (2 x + \frac{π}{4}) .$

Here, $c = π / 4 .$ $c = π / 4 .$ Therefore, in order to obtain values for the table, we assume a value for x, multiply it by 2, add $π / 4$ $π / 4$ to this value, and then find the sine of the result. The values shown are those for which $2 x + π / 4 = 0, π / 4, π,$ $2 x + π / 4 = 0, π / 4, π,$ 2, $3 π / 4, π,$ $3 π / 4, π,$ and so on, which are the important values for $y = \sin 2 x .$ $y = \sin 2 x .$

`x`	$- \frac{π}{8}$ $- \frac{π}{8}$	0	$\frac{π}{8}$ $\frac{π}{8}$	$\frac{π}{4}$ $\frac{π}{4}$	$\frac{3 π}{8}$ $\frac{3 π}{8}$	$\frac{π}{2}$ $\frac{π}{2}$	$\frac{5 π}{8}$ $\frac{5 π}{8}$	$\frac{3 π}{4}$ $\frac{3 π}{4}$	$\frac{7 π}{8}$ $\frac{7 π}{8}$	$π$ $π$
`y`	0	0.7	1	0.7	0	$- 0.7$ $- 0.7$	$- 1$ $- 1$	$- 0.7$ $- 0.7$	0	0.7

Solving $2 x + π / 4 = 0,$ $2 x + π / 4 = 0,$ we get $x = - π / 8,$ $x = - π / 8,$ which gives $y = \sin 0 = 0 .$ $y = \sin 0 = 0 .$ The other values for y are found in the same way. See Fig. 10.13.

A curve oscillates about y = 0 with amplitude 1, period pi, and maximum (pi over 8, 1). It rises and falls through points given in the previous table. — Fig. 10.13

Note from Example 1 that the graph of $y = \sin (2 x + \frac{π}{4})$ $y = \sin (2 x + \frac{π}{4})$ is precisely the same as the graph of $y = \sin 2 x,$ $y = \sin 2 x,$ except that it is shifted $π / 8$ $π / 8$ units to the left. In Fig. 10.14, a calculator view shows the graphs of $y = \sin 2 x$ $y = \sin 2 x$ and $y = \sin (2 x + \frac{π}{4}) .$ $y = \sin (2 x + \frac{π}{4}) .$ We see that the shapes are the same except that the graph of $y = \sin (2 x + \frac{π}{4})$ $y = \sin (2 x + \frac{π}{4})$ is $π / 8$ $π / 8$ units to the left of the graph of $y = \sin 2 x .$ $y = \sin 2 x .$

In general, the effect of c in the equation $y = a \sin (b x + c)$ $y = a \sin (b x + c)$ is to shift the curve of $y = a \sin b x$ $y = a \sin b x$ to the left if $c > 0,$ $c > 0,$ or shift the curve to the right if $c < 0 .$ $c < 0 .$ The amount of this shift is given by $- c / b .$ $- c / b .$ Due to its importance in sketching curves, the quantity $- c / b$ $- c / b$ is called the displacement (or phase shift).

We can see the reason that the displacement is $- c / b$ $- c / b$ by noting corresponding points on the graphs of $y = \sin b x$ $y = \sin b x$ and $y = \sin (b x + c) .$ $y = \sin (b x + c) .$ For $y = \sin b x,$ $y = \sin b x,$ when $x = 0,$ $x = 0,$ then $y = 0 .$ $y = 0 .$ For $y = \sin (b x + c),$ $y = \sin (b x + c),$ when $x = - c / b,$ $x = - c / b,$ then $y = 0 .$ $y = 0 .$ The point $(- c / b, 0)$ $(- c / b, 0)$ on the graph of $y = \sin (b x + c)$ $y = \sin (b x + c)$ is $- c / b$ $- c / b$ units horizontally from the point (0, 0) on the graph of $y = \sin x .$ $y = \sin x .$ In Fig. 10.14, $- c / b = - π / 8 .$ $- c / b = - π / 8 .$

Figure 10.14 Full Alternative Text

Therefore, we use the displacement combined with the amplitude and the period along with the other information from the previous sections to sketch curves of the functions $y = a \sin (b x + c)$ $y = a \sin (b x + c)$ and $y = a \cos (b x + c),$ $y = a \cos (b x + c),$ where $b > 0 .$ $b > 0 .$

By use of these quantities and the one-fourth period distance, the graphs of the sine and cosine functions can be readily sketched.

A general illustration of the graph of $y = a \sin (b x + c)$ $y = a \sin (b x + c)$ is shown in Fig. 10.15.

CAUTION

Note that the displacement is negative (to the left) for $c > 0$ $c > 0$ and positive (to the right) for $c < 0$ $c < 0$ as shown in Figs. 10.15(a) and (b), respectively.

Two coordinate systems. For each, ay > 0, and b > 0. — Fig. 10.15

Figure 10.15 Full Alternative Text

NOTE

[Note also that we can find the displacement for the graphs of $y = a \sin (b x + c)$ $y = a \sin (b x + c)$ by solving $b x + c = 0$ $b x + c = 0$ for x. We see that $x = - c / b .$ $x = - c / b .$ ]

This is the same as the horizontal shift for graphs of functions that is shown on page 105.

EXAMPLE 2 Sketching graph of `y` = `a` sin(`bx` + `c`)

Sketch the graph of $y = 2 \sin (3 x - π) .$ $y = 2 \sin (3 x - π) .$

First, note that $a = 2, b = 3,$ $a = 2, b = 3,$ and $c = - π .$ $c = - π .$ Therefore, the amplitude is 2, the period is $2 π / 3,$ $2 π / 3,$ and the displacement is $- (- π / 3) = π / 3 .$ $- (- π / 3) = π / 3 .$ (We can also get the displacement from $3 x - π = 0, x = π / 3 .$ $3 x - π = 0, x = π / 3 .$ )

Note that the curve “starts” at $x = π / 3$ $x = π / 3$ and starts repeating $2 π / 3$ $2 π / 3$ units to the right of this point. Be sure to grasp this point well. The period tells us the number of units along the x-axis between such corresponding points. One-fourth of the period is $\frac{1}{4} (\frac{2 π}{3}) = \frac{π}{6} .$ $\frac{1}{4} (\frac{2 π}{3}) = \frac{π}{6} .$

Important values are at $\frac{π}{3}, \frac{π}{3} + \frac{π}{6} = \frac{π}{2}, \frac{π}{3} + 2 (\frac{π}{6}) = \frac{2 π}{3},$ $\frac{π}{3}, \frac{π}{3} + \frac{π}{6} = \frac{π}{2}, \frac{π}{3} + 2 (\frac{π}{6}) = \frac{2 π}{3},$ and so on. We now make the table of important values and sketch the graph shown in Fig. 10.16.

`x`	0	$\frac{π}{6}$ $\frac{π}{6}$	$\frac{π}{3}$ $\frac{π}{3}$	$\frac{π}{2}$ $\frac{π}{2}$	$\frac{2 π}{3}$ $\frac{2 π}{3}$	$\frac{5 π}{6}$ $\frac{5 π}{6}$	$π \leftarrow \frac{1}{4} (period) = \frac{π}{6}$ $π \leftarrow \frac{1}{4} (period) = \frac{π}{6}$
`y`	0	$- 2$ $- 2$	0	2	0	$- 2$ $- 2$	0

Figure 10.16 Full Alternative Text

EXAMPLE 3 Sketching graph of `y` = `a` cos(`bx` + `c`)

Sketch the graph of the function $y = - \cos (2 x + \frac{π}{6}) .$ $y = - \cos (2 x + \frac{π}{6}) .$

First, we determine that

the amplitude is 1
the period is $\frac{2 π}{2} = π$ $\frac{2 π}{2} = π$
the displacement is $- \frac{π}{6} \div 2 = - \frac{π}{12}$ $- \frac{π}{6} \div 2 = - \frac{π}{12}$

We now make a table of important values, noting that the curve starts repeating $π$ $π$ units to the right of $- \frac{π}{12} .$ $- \frac{π}{12} .$

`x`	$- \frac{π}{12}$ $- \frac{π}{12}$	$\frac{π}{6}$ $\frac{π}{6}$	$\frac{5 π}{12}$ $\frac{5 π}{12}$	$\frac{2 π}{3}$ $\frac{2 π}{3}$	$\begin{array}{r} \frac{11 π}{12} & \leftarrow \frac{1}{4} (period) = \frac{π}{4} \end{array}$ $\begin{array}{r} \frac{11 π}{12} & \leftarrow \frac{1}{4} (period) = \frac{π}{4} \end{array}$
`y`	$- 1$ $- 1$	0	1	0	$- 1$ $- 1$

From this table, we sketch the graph in Fig. 10.17.

Figure 10.17 Full Alternative Text

Each of the heavy portions of the graphs in Figs. 10.16 and 10.17 is called a cycle of the curve. A cycle is any section of the graph that includes exactly one period.

EXAMPLE 4 Graph on calculator

View the graph of $y = 2 \cos (\frac{1}{2} x - \frac{π}{6})$ $y = 2 \cos (\frac{1}{2} x - \frac{π}{6})$ on a calculator.

From the values $a = 2, b = 1/2,$ $a = 2, b = 1/2,$ and $c = - π / 6,$ $c = - π / 6,$ we determine that

the amplitude is 2
the period is $2 π \div \frac{1}{2} = 4 π$ $2 π \div \frac{1}{2} = 4 π$
the displacement is $- (- \frac{π}{6}) \div \frac{1}{2} = \frac{π}{3}$ $- (- \frac{π}{6}) \div \frac{1}{2} = \frac{π}{3}$

This curve’s cycle starts at $x = \frac{π}{3} (\approx 1.05)$ $x = \frac{π}{3} (\approx 1.05)$ and ends at $\frac{π}{3} + 4 π = \frac{13 π}{3} (\approx 13.6) .$ $\frac{π}{3} + 4 π = \frac{13 π}{3} (\approx 13.6) .$ When choosing the window settings, we should make sure these two values are between the Xmin and Xmax. The amplitude determines the Ymin and Ymax. The graph is shown in Fig. 10.18.

A curve oscillates about y = 0 on a screen that is [negative 1, 14] by [negative 2, 2]. — Fig. 10.18

Example 5 Phase angle—ac voltage/current lead and lag

In alternating-current (ac) circuits, voltage (v) and current (i) can both be represented by sine waves with the same period. Depending on the existence of resistors, capacitors, or inductors in the circuit, these waves can be in phase (maximums and minimums occur at the same time) or one can lead or lag the other. See Fig. 10.19.

Three coordinate systems with two oscillating curves each. — Fig. 10.19

Figure 10.19 Full Alternative Text

If we assume that the sine wave for current passes through the origin, it can be represented as $i = I_{m} \sin (2 π f t),$ $i = I_{m} \sin (2 π f t),$ where f is the frequency (in cycles/s) and t is time (in s). The voltage can then be represented by $v = V_{m} \sin (2 π f t + ϕ),$ $v = V_{m} \sin (2 π f t + ϕ),$ where $ϕ$ $ϕ$ is the phase angle.

NOTE

[The phase angle, often written in degrees, tells us how many degrees (assuming 360° is one complete cycle) the voltage leads (if $ϕ$ $ϕ$ is positive) or lags (if $ϕ$ $ϕ$ is negative) the current by.]

For example, if $i = 15 \sin (120 π t)$ $i = 15 \sin (120 π t)$ and $v = 170 \sin (120 π t + 90 °),$ $v = 170 \sin (120 π t + 90 °),$ then voltage leads current by $90 °$ $90 °$ , or one-quarter of the period. See Fig. 10.20. It is somewhat unusual that the voltage function combines radians $(2 π f t)$ $(2 π f t)$ and degrees $(90 °)$ $(90 °)$ , but this is commonly done in this particular application. If we convert the phase angle to $\frac{π}{2}$ $\frac{π}{2}$ radians, we can then find the displacement, which gives the time shift in seconds between the two waves:

The graphs are voltage and current curves oscillating about y = 0. Voltage leads current by 90 degrees along the horizontal. — Fig. 10.20

Displacement = - \frac{c}{b} = - \frac{π /_{2}}{120 π} = - \frac{1}{240} s

$Displacement = - \frac{c}{b} = - \frac{π /_{2}}{120 π} = - \frac{1}{240} s$

Therefore, voltage leads current by $\frac{1}{240} s .$ $\frac{1}{240} s .$ This is one-quarter of the period, with the period equal to $\frac{2 π}{120 π} = \frac{1}{60} s .$ $\frac{2 π}{120 π} = \frac{1}{60} s .$

NOTE

[Note that in order to graph the voltage function on a calculator, the phase angle must be expressed in radians.]

Exercises 10.3

In Exercises 1 and 2, graph the function if the given changes are made in the indicated examples of this section.

In Example 3, if the sign before $π / 6$ $π / 6$ is changed, sketch the graph of the resulting function.
In Example 4, if the sign before $π / 6$ $π / 6$ is changed, sketch the graph of the resulting function.

In Exercises 3–26, determine the amplitude, period, and displacement for each function. Then sketch the graphs of the functions. Check each using a calculator.

$y = \sin (x - \frac{π}{6})$ $y = \sin (x - \frac{π}{6})$
$y = 3 \sin (x + \frac{π}{4})$ $y = 3 \sin (x + \frac{π}{4})$
$y = \cos (x + \frac{π}{6})$ $y = \cos (x + \frac{π}{6})$
$y = 2 \cos (x - \frac{π}{8})$ $y = 2 \cos (x - \frac{π}{8})$
$y = 0.2 \sin (2 x + \frac{π}{2})$ $y = 0.2 \sin (2 x + \frac{π}{2})$
$y = - \sin (3 x - \frac{π}{2})$ $y = - \sin (3 x - \frac{π}{2})$
$y = - \cos (2 x - π)$ $y = - \cos (2 x - π)$
$y = 0.4 \sin (3 x + \frac{π}{3})$ $y = 0.4 \sin (3 x + \frac{π}{3})$
$y = \frac{1}{2} \sin (\frac{1}{2} x - \frac{π}{4})$ $y = \frac{1}{2} \sin (\frac{1}{2} x - \frac{π}{4})$
$y = 2 \sin (\frac{1}{4} x + \frac{π}{2})$ $y = 2 \sin (\frac{1}{4} x + \frac{π}{2})$
$y = 30 \cos (\frac{1}{3} x + \frac{π}{3})$ $y = 30 \cos (\frac{1}{3} x + \frac{π}{3})$
$y = - \frac{1}{3} \cos (\frac{1}{2} x - \frac{π}{8})$ $y = - \frac{1}{3} \cos (\frac{1}{2} x - \frac{π}{8})$
$y = - \sin (π x + \frac{π}{8})$ $y = - \sin (π x + \frac{π}{8})$
$y = - 2 \sin (2 π x - π)$ $y = - 2 \sin (2 π x - π)$
$y = 0.08 \cos (4 π x - \frac{π}{5})$ $y = 0.08 \cos (4 π x - \frac{π}{5})$
$y = 25 \cos (3 π x + \frac{π}{4})$ $y = 25 \cos (3 π x + \frac{π}{4})$
$y = - 0.6 \sin (2 π x - 1)$ $y = - 0.6 \sin (2 π x - 1)$
$y = 1.8 \sin (π x + \frac{1}{3})$ $y = 1.8 \sin (π x + \frac{1}{3})$
$y = 40 \cos (3 π x + 1)$ $y = 40 \cos (3 π x + 1)$
$y = 360 \cos (6 π x - 3)$ $y = 360 \cos (6 π x - 3)$
$y = \sin (π^{2} x - π)$ $y = \sin (π^{2} x - π)$
$y = - \frac{1}{2} \sin (2 x - \frac{1}{π})$ $y = - \frac{1}{2} \sin (2 x - \frac{1}{π})$
$y = - \frac{3}{2} \cos (π x + \frac{π^{2}}{6})$ $y = - \frac{3}{2} \cos (π x + \frac{π^{2}}{6})$
$y = π \cos (\frac{1}{π} x + \frac{1}{3})$ $y = π \cos (\frac{1}{π} x + \frac{1}{3})$

In Exercises 27–30, write the equation for the given function with the given amplitude, period, and displacement, respectively. [Hint: Use Eqs. (10.1) to determine a, b, and c.]

sine, 4, $3 π, - π / 4$ $3 π, - π / 4$
cosine, 8, $2 π / 3, π / 3$ $2 π / 3, π / 3$
cosine, 12, 1/2, 1/8
sine, 18, 4, $- 1$ $- 1$

In Exercises 31 and 32, show that the given equations are identities. The method using a calculator is indicated in Exercise 31.

By viewing the graphs of $y_{1} = \sin x$ $y_{1} = \sin x$ and $y_{2} = \cos (x - π / 2),$ $y_{2} = \cos (x - π / 2),$ show that $\cos (x - π / 2) = \sin x .$ $\cos (x - π / 2) = \sin x .$
Show that $\cos (2 x - 3 π / 8) = \cos (3 π / 8 - 2 x) .$ $\cos (2 x - 3 π / 8) = \cos (3 π / 8 - 2 x) .$

In Exercises 33–40, solve the given problems.

What conclusion do you draw from the calculator graphs of

$y_{1} = 2 \sin (3 x + \frac{π}{6}) {and y}_{2} = - 2 \sin [- (3 x + \frac{π}{6})] ?$ $y_{1} = 2 \sin (3 x + \frac{π}{6}) {and y}_{2} = - 2 \sin [- (3 x + \frac{π}{6})] ?$
What conclusion do you draw from the calculator graphs of

$y_{1} = 2 \cos (3 x + \frac{π}{6}) {and y}_{2} = - 2 \cos [- (3 x + \frac{π}{6})] ?$ $y_{1} = 2 \cos (3 x + \frac{π}{6}) {and y}_{2} = - 2 \cos [- (3 x + \frac{π}{6})] ?$
The current in an alternating-current circuit is given by $i = 12 \sin (120 π t) .$ $i = 12 \sin (120 π t) .$ Find a function for the voltage v if the amplitude is 18V and voltage lags current by $60 °$ $60 °$ . Then find the displacement. See Example 5.
The current in an alternating-current circuit is given by $i = 8.50 \sin (120 π t) .$ $i = 8.50 \sin (120 π t) .$ Find a function for the voltage v if the amplitude is 28V and voltage leads current by $45 °$ $45 °$ . Then find the displacement. See Example 5.
A wave traveling in a string may be represented by the equation $y = A \sin 2 π (\frac{t}{T} - \frac{x}{λ}) .$ $y = A \sin 2 π (\frac{t}{T} - \frac{x}{λ}) .$ Here, A is the amplitude, t is the time the wave has traveled, x is the distance from the origin, T is the time required for the wave to travel one wavelength $λ$ $λ$ (the Greek letter lambda). Sketch three cycles of the wave for which $A = 2.00 cm, T = 0.100 s, λ = 20.0 cm,$ $A = 2.00 cm, T = 0.100 s, λ = 20.0 cm,$ and $x = 5.00 cm .$ $x = 5.00 cm .$
The electric current i (in $μ A$ $μ A$ ) in a certain circuit is given by $i = 3.8 \cos 2 π (t + 0.20),$ $i = 3.8 \cos 2 π (t + 0.20),$ where t is the time in seconds. Sketch three cycles of this function.
39. A certain satellite circles Earth such that its distance y, in miles north or south (altitude is not considered) from the equator, is $y = 4500 \cos (0.025 t - 0.25),$ $y = 4500 \cos (0.025 t - 0.25),$ where t is the time (in min) after launch. Use a calculator to graph two cycles of the curve.
In performing a test on a patient, a medical technician used an ultrasonic signal given by the equation $I = A \sin (ω t + ϕ) .$ $I = A \sin (ω t + ϕ) .$ Use a calculator to view two cycles of the graph of I vs. t if $A = 5 nW / m^{2}, ω = 2 \times 10^{5} rad / s,$ $A = 5 nW / m^{2}, ω = 2 \times 10^{5} rad / s,$ and $ϕ = 0.4 .$ $ϕ = 0.4 .$ Explain how you chose your calculator’s window settings.

In Exercises 41–44, give the specific form of the equation by evaluating a, b, and c through an inspection of the given curve. Explain how a, b, and c are found.

$y = a \sin (b x + c)$ $y = a \sin (b x + c)$ Fig. 10.21
$y = a \cos (b x + c)$ $y = a \cos (b x + c)$ Fig. 10.21

Fig. 10.21
$y = a \cos (b x + c)$ $y = a \cos (b x + c)$ Fig. 10.22
$y = a \sin (b x + c)$ $y = a \sin (b x + c)$ Fig. 10.22

Fig. 10.22

Answers to Practice Exercises

$amp . = 8, per . = π, disp . = π / 6$ $amp . = 8, per . = π, disp . = π / 6$
$v = 170 \sin (120 π t - 45 °)$ $v = 170 \sin (120 π t - 45 °)$

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Table of Contents for 10.3 Graphs of y&#8201;=&#8201;a sin(&#8201;bx&#8201;+&#8201;c) and y&#8201;=&#8201;a cos(bx&#8201;+&#8201;c) Graphs of y = a sin( bx + c) and y = a cos(bx + c)

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10.3 Graphs of y = a sin( bx + c) and y = a cos(bx + c) Graphs of y = a sin( bx + c) and y = a cos(bx + c)