5.5: Graphs of the Sine and Cosine Functions (2024)

PERIOD OF SINE AND COSINE

The periods of the sine function \(y = \sin (x), \) and cosine function \(y = \cos(x), \) are both 2\(\pi \).

DOMAIN AND RANGE OF SINE AND COSINE

The domain of sine and cosine is all real numbers, \((-\infty ,\infty )\).

The range of sine and cosine is the interval [-1, 1].

NEGATIVE ANGLE IDENTITIES

The sine is an odd function, symmetric about the origin, so \[\sin (-\theta )=-\sin (\theta ) \nonumber \]

The cosine is an even function, symmetric about the y-axis, so \[\cos (-\theta )=\cos (\theta ) \nonumber \]

CHARACTERISTICS OF SINE AND COSINE FUNCTIONS

The sine and cosine functions have several distinct characteristics:

• They are periodic functions with a period of \(2\pi\).
• The domain of each function is \((−\infty,\infty)\) and the range is \([ −1,1 ]\).
• The graph of \(y=\sin\space x\) is symmetric about the origin, because it is an odd function.
• The graph of \(y=\cos\space x\) is symmetric about they- \(y\)-axis, because it is an even function.

AMPLITUDE OF SINUSOIDAL FUNCTIONS and vertical stretching

The amplitudeof the sinusoidal functions\(y=A\sin(Bx−C)+D\) and\(y=A\cos(Bx−C)+D\) is \(|A|\), the absolute value of the parameter\(A\).

The amplitudecan be directly observedon a graph of a sinusoidal function. It is half the height between the top and the bottom of the curve and can be calculated using the formula:

\(|A| = \text{amplitude} = \dfrac{1}{2}∣ \text{maximum} − \text{minimum} | \)

Amplitude can also be determined by using the fact that it isthe vertical height from either the top or bottom of the curve to the horizontal line (called the midline)that divides the curve horizontally in half.

Equation parameter \(A\) represents the vertical stretch factor.

• If \(| A |>1\), the function is vertically stretched.
• If \(| A |<1\), the function is vertically compressed.
• Reflection over the \(x\)-axis occurs when \(A\) is negative\(^{*}\).

\(^{*}\)Before preparingto graph the function,if \(B\) is negative, rewrite the equation using the Negative Angle Identities so the equation under analysis has\(B >0\). (The argument \(Bx-C\) becomes \(-Bx+C\)and if the sine function is being used,the sign of \(A\) changes.)

• When parameter \(B\) is positive, then if parameter \(A\) is negative, the result is a reflection across the \(x\)-axis.

Example \(\PageIndex{1}\): Identify the Amplitude of a Sine or Cosine Function

What is the amplitude of the sinusoidal function \(f(x)=−4\sin(x)\)? Is the function stretched or compressed vertically?

Solution

Let’s begin by comparing the function to the simplified form \(y=A\sin(Bx)\).

In the given function, \(A=−4\), so the amplitude is \(| A |=| −4 |=4\). The function is stretched vertically because \(|A|>1\).

The negative value of \(A\) results in a reflection across the \(x\)-axis of the sine function, as shown in the figure to the right.

Try It\(\PageIndex{1a}\)

What is the amplitude of the sinusoidal function \(f(x)=\dfrac{1}{2}\sin(x)\)? Is the function stretched or compressed vertically?

Answer

The amplitude is \(\dfrac{1}{2}\). The graph is vertically compressed.
\(f(x)\) is in dark blue in the graph below. The basic sine function is in light grey.

Try It \(\PageIndex{1b}\)

What is the amplitude of the function \(f(x)=7\cos(x)\)? Sketch a graph of this function.

Answer

A graph of \(f(x)=7\cos(x)\). Amplitude is \(7\)

Definition: MIDLINE

The midline of a sinusoidal function is the center value of the function -- the \(y\)-value that the function oscillates above and below.

The midline of the basic sine and cosine functions, \(y = \sin (x)\) and \( y = \cos (x), \) is the line \(y=0\).

VERTICAL SHIFT OF SINUSOIDAL FUNCTIONS

The vertical shiftof the sinusoidal functions\(y=A\sin(Bx−C)+D\) and\(y=A\cos(Bx−C)+D\) is indicated by the value of the parameter \(D\).

The equation for the midline of a sinusoidal function is \(y=D\)

• If \(D > 0\) the graph has been shifted up \(D\) units
• If \(D < 0\) the graph has been shifted down\(D\) units

The vertical shiftcan be directly observedon a graph of a sinusoidal function. It is the vertical distance the midline has moved up or down from the \(x\)-axis.

• Local maxima are a distance \(|A|\) above the vertical midline of the graph, and have \(y\) coordinates of\(y=D + |A|\).
• Local minima will be the same distance below the midline,and have \(y\) coordinates of\(y=D - |A|\).
• The range of sinusoidal functions in interval notation is \( [D-|A|, D+|A| ] \)

Example \(\PageIndex{2}\): Identifythe Vertical Shift of a Sinusoidal Function

Given the function \(f(\theta )=\cos (\theta )-4\),

1. What is the midline of the function \(f(\theta )=\cos (\theta )-4\)?
2. Determine the direction and magnitude of the transformation neededon the basic cosine function to obtain \(f(\theta )\).
3. Sketch a graph of the function.

Solution

(1). Because the parameter \(D=−4\), themidline is \(y = -4\). \(\qquad\)
(2). Shift \(y = \cos (x)\) down 4 units to obtain \(f(\theta )\).\(\qquad\)
(3). The graph below shows the basic function \(y = \cos (x)\) ,the midline for \(f\), and\(f(\theta )\).

Try It\(\PageIndex{2}\)

Determine the direction and magnitude of the vertical shift for \(f(x)=\sin(x)+2\). State the equation for the midline and sketch the graph.

Answer

The midline is \(y = 2\). Shift \(y = \sin (x)\) up 2units to obtain \(f(x)\).
The graph below shows\(f(x)\), the midline \(y = 2\), and the basic function \(y = \sin (x)\).

PERIOD OF SINUSOIDAL FUNCTIONS and horizontal stretching

The period \(P\) of the sinusoidal functions

\(y=A\sin(Bx−C)+D\) and \(y=A\cos(Bx−C)+D\) is \(\quad\) \(P = \dfrac{2\pi}{|B|}\).

The period \(P\) can be directly observedon a graph of a sinusoidal function. It is the horizontal length of one complete cycle. \(P\) is always positive.

Notice that the parameter \(B\) isnot directly visible from a graph, but it can be calculated once the period \(P\) IS obtained by looking at thegraph.The period is inversely proportionalto the parameter \(|B|\).

• If \(|B|>1\), then the period is less than \(2\pi\) and the function undergoes a horizontal compression
• if \(| B |<1\), then the period is greater than \(2\pi\) and the function undergoes a horizontal stretch
• If \(B\) is negative, before graphing the function,rewrite the equation using the Negative Angle Identities (also known as the Even/Odd Properties)so the equation under analysis has\(B >0\) andreflection will be done properly.

When graphing a sinusoidal function, the period is broken down intoquarters.When \(C=0\), these quarter points are at the beginning of a cycle \( \color{Green}{x=0}\), \(\color{Green}{x=\dfrac{1}{4}P}\),\(\color{Green}{x=\dfrac{1}{2}P}\),\(\color{Green}{x=\dfrac{3}{4}P}\), and at the end of a cycle \( \color{Green}{x=P}\). These quarter points, using the shape of the parent function as a guide,are the locations, inone cycle, of thelocal minimum, local maximum, and intersection with the midline.

Example \(\PageIndex{3}\): Identify the Period of a Sine or Cosine Function

Determine the period of the function \(f(x)=\sin\left(\dfrac{\pi}{6}x\right)\). Then sketch the function.

Solution

Let’s begin by comparing the equation to the general form \(y=A\sin(Bx)\).

In the given equation, \(B=\dfrac{\pi}{6}\), so the period will be \(\quad\)\( P =\dfrac{2\pi}{|B|} =\dfrac{2\pi}{\tfrac{\pi}{6}} =2\pi ⋅ \dfrac{6}{\pi} =12 \)

To graph the function, the sketch will take on the general shape of a sine function, starting a cycle at \(x=0\) with quarter points at \(x=3\), \(x=6\), \(x=9\), and \(x=12\) to complete one cycle.

Try It \(\PageIndex{3}\)

Determine the period of the function \(g(x)=\cos \left( \dfrac{x}{3} \right) \). Then sketch the function.

Answer

\(6\pi\).To graph the function, the sketch will take on the general shape of a cosine function, starting a cycle at \(x=0\) with quarter points at \(x=1.5\pi\), \(x=3\pi\), \(x=4.5\pi\), and \(x=6\pi\) to complete one cycle.

TRANSFORMATIONS OF SINE AND COSINE

Howto: Sketch the graph of a sinusoidal function that does not have aphase shift.

Before doing anything else, if\(B\), the coefficient of \(x\) in the argument is negative, use the Even/Odd Properties to rewrite the sinusoidal equation so that \(B\)is positive (This could change the sign of \(A\)).

Given sinusoidal functions of the form

\(y=A\sin(Bx)+D \quad \) or \(\quad y=A\cos(Bx)+D\)

1. Identify the amplitude, \(| A |\). If \(B\) is negative, rewrite the sinusoidal equation so that \(B\) is positive (This rewrite could change the sign of \(A\)).
2. Use the equation and its sign for \(A\)to determine the shape of the graph.The four options are shown in the graphsto the right.
3. Identify the vertical shift from the parameter \(D\). The midline is the line \(y=D\). The graph willoscillate \(| A | \) units above and below this line, so the maximum is \(y=D+|A|\), and the minimum is \(y=D-|A|\).
4. Identify the period, \(P=\dfrac{2\pi}{| B |}\).
5. Divide the period into quarters, These quarter points are at the beginning of a cycle (\(\color{Green}{x=0}\)), \(\color{Green}{x=\tfrac{1}{4}P}\),\(\color{Green}{x=\tfrac{1}{2}P}\),\(\color{Green}{x=\tfrac{3}{4}P}\), and at the end of a cycle (\(\color{Green}{x=P}\)). These quarter points, using the shape of the parent function as a guide,are the locations, inone cycle, of thelocal minimum, local maximum, and intersection with the midline. (See accompanying graphs)

Quarter points correspond to intervals of equal length each representing \(1/4\) of the period. They indicate the horizontal location of the maximum and minimumvalues and where the midline is crossed for one period of the function. The green dots in the figures to the right show where the quarter points are for the four basic sinusoidal functions.

Example \(\PageIndex{4a}\): Graphing a Function and Identifying the Amplitude and Period

Sketch the graph of

1. \(f(x)=−2\sin\left(\dfrac{\pi x}{2}\right)\).
2. \(g(x) =2\sin\left(\dfrac{-\pi x}{2}\right)\).

Solution

a. Let’s begin by comparing the equation to the form \(y=A\sin(Bx)+D\).

• Step 1. We can see from the equation that \(A=−2\), so the amplitude is 2. \(\quad\) \(|A|=2 \)
• Step 2. Because \(A\) is negative, the graph is a reflection over the \(x\)-axis of the basic sine function.The shape of the graph is like the graph for \(y = - \sin (x)\), starting on the \(y\)-axis at the midline and going down.
• Step 3. The parameter \(D\) is zero so the midline is along \(y=D \Rightarrow y=0\) or the \(x\)-axis, and, since the amplitude is \(2\), the graph oscillates between \(2\) and \(-2\).
• Step 4. The equation shows that \(B=\dfrac{\pi}{2}\), so the period is \(\quad\) \( P =\dfrac{2\pi}{|B|} =\dfrac{2\pi}{\dfrac{\pi}{2}}=2\pi \cdot \dfrac{2}{\pi}=4 \)
• Step5. The quarter pointsare at \(x = 0, 1, 2, 3, 4\). One cycle starts at the midline and goes down. The graph intersects the midline at \(x = 0\), \(x=2\), and \(x=4\). A localminimum will occur at \(2\) units below the midline, at \(x=1\). A localmaximum will occur at \(2\) units above the midline, at \(x=3\). The graph of the function is below.

b.

• Preparation.Because \(B\) is negative, the equation needs to be rewritten using the Negative Angle Identity. Since \(\sin(-x) = -\sin (x)\), then \(g(x) = −2\sin\left(\dfrac{\pi x}{2}\right),\) which is the same as \(f(x)\) in part (a).
• Step 1. We can see from the equation that \(|A|=2\), so the amplitude is 2.
• Step 2. Using the modified version of \(g\), \(A\) is negative, so the shape of the graph for\(g\) is like the shape of\(y = - \sin (x)\). [Notice that if the Negative Angle Identity is not used, the shape chosen would be that of \(y = \sin(x)\) ,which would result in a graph that would not becorrect].
• Step 3. The midline is \(y=0\).
• Step 4. The period is \(P = 4\).
• Step 5. The quarter points areat \(x = 0, 1, 2, 3, 4\), and the graph is the same as that in the graph above.

Example \(\PageIndex{4b}\)

Determine the midline, amplitude, and period of the function \(y=3\sin (2x)+1\). Then sketch the function.

Solution

Step 1. \(A=3\), so the amplitude is \(| A |=3\). Step 2. The graph has the same shape as the graph for \(y=\sin(x)\), starting on the \(y\)-axis at the midline and going up. Step 3. \(D=1\), so the midline is \(y=1\), and the graph will oscillate between \(D-|A|\) and |\(D+|A|\) or between \(y=-2\) and \(y=4\). Step 4. \(B=2\), so the period is \(P=\dfrac{2\pi}{| B |}=\dfrac{2\pi}{2}=\pi\). Step 5. The quarter points are at \(x=0, \dfrac{\pi}{4}, \dfrac{\pi}{2},\dfrac{3}{4}, \pi\). One cycle starts at the midline and goes up. One cycle of the function appears in the graph on the right.

Try It\(\PageIndex{4}\)

Sketch a graph of \( g(x)=−0.8\cos(2x)\). Determine the midline, amplitude, and period.

Answer

amplitude: \(| A |=0.8\);

shape: \(y=-\cos(x)\);

midline: \(y=0\), minimum \(y=-.8\), maximum: \(y=.8\);

period: \(P=\dfrac{2\pi}{| B |}=\pi\);

quarter points: \(x=0, \dfrac{\pi}{4}, \dfrac{\pi}{2},\dfrac{3\pi}{4}, \text{ and }\pi\).

Howto: Construct an equation fromthe graph of a sinusoidal function.

Given a sinusoidal graph, construct one ofthe sinusoidal functions:\(y=A\sin(Bx)+D\) and\(y=A\cos(Bx)+D\)

1. Identify the amplitude, \(| A |\), half the distance between the highest and lowest values of \(y\) on the graph.
2. Identify the midline, the line halfway between thehighest and lowest values of \(y\) on the graph. This is parameter \(D\).
3. Look at the shape of the graph at the point where it intersects the y-axis.
   • If the point is on the midline and the graph is going up, use the sine functionand make parameter\(A\) positive. The shape is the same as \(y = \sin x\).
   • If the point is on the midline and the graph is going down, use the sine function and make parameter\(A\) negative. The shape is the same as \(y = -\sin x\).
   • If the point is a maximum, use the cosine function and make parameter\(A\) positive. The shape is the same as \(y = \cos x\).
   • If the point is a minimum, use the cosine function and make parameter\(A\) negative. The shape is the same as \(y = -\cos x\).
4. Identify the period \(P\),the horizontal distance between peaks on the graph. Calculate parameter \(B\) from the equation\(B=\dfrac{2\pi}{P}\).
5. Use the values of the parameters obtained above to construct a possible sinusoidal equation for the graph.

Example \(\PageIndex{5a}\): Identifying the Equation for a Sinusoidal Function from a Graph

Determine the formula for the function in Figure \(\PageIndex{27}\).

Solution

To determine the equation, we need to identify each parameter value in the general form of a sinusoidal function.

\(y=A\sin (Bx)+D \qquad \qquad y=A\cos (Bx)+D\)

The amplitude is halfthe difference between the height of local maxima and minima. This difference is \(1\), so \(| A |=\frac{1}{2}=0.5\). Another way to obtain the amplitude is to identify the midline and observe the greatest distance above and below the midline, which is the amplitude.

In general, the shape of the graph indicates a sinusoidal function. When \(x=0\), the graph has an extreme point, \((0,0)\). Since the reflected cosine function has an extreme (minimum) point for \(x=0\), let us write our equation in terms of a reflected cosine function. Therefore, the parameter \(A\) must be negative, and\(A=−0.5\).

We can see that the graph rises and falls an equal distance above and below \(y=0.5\). This value, which is the midline, is \(D\) in the equation, so \(D=0.5\).

The period is \(P=2\pi\), so the graph is not horizontally stretched or compressed, so \(B=\dfrac{2\pi}{P}=1\).

Putting this all together,

\(g(x)=−0.5\cos (x)+0.5\)

Try It\(\PageIndex{5a}\)

Determine the formula for the function in the graph below

Answer

amplitude is half the differencebetween maximum and minimum so \(|A| = 1\); midline is \(y=2\) so \(D=2\). When \(x=0\), the graph is on the midline and is going up, so the shape is the same as for the function \(y=sin (x)\), so write the equation in terms of a sine function, and \(A\) is positive, so \(A=1\). The period is \(P = 2\pi\), so there has been no horizontal stretching, so \(B=\dfrac{2\pi}{P}=1\). Putting it all together, the equation is:\(f(x)=\sin(x)+2\)

Example \(\PageIndex{5b}\)

Find a formula for the sinusoidal function graphed here.

Solution

The amplitude is half the distance between the highest value and lowest valueof the graph. So the amplitude is \(\dfrac{3-(-1)}{2} = 2\).

The graph oscillates from a low of -1 to a high of 3, putting the midline at \(y = 1\), halfway between.

At \(t = 0\), the graph is at the midline value, which tells us the graph can most easily be represented as a sine function. Since the graph then decreases, this must be a vertical reflection of the sine function. Therefore the shape is the same as that of \(y=-\sin(t)\) so \(A\) is negative, and thus \(A=-2\).

The period of the graph is 8. We can measure this from the first peak at \(x = -2\) to the second at \(x = 6\). Since the period is 8, the stretch/compression factor we will use will be

\[B=\dfrac{2\pi }{P} =\dfrac{2\pi }{8} =\dfrac{\pi }{4}\nonumber\]

Putting this all together,

\[f(t)=-2\sin \left(\dfrac{\pi }{4} t\right)+1\nonumber\]

Try It\(\PageIndex{5b}\)

If a sinusoidal function has a maximum valueat point (0,5), has an amplitude of 2, and a period of 4, write a formula for the function. Sketch the graph.

Answer

We are given \(|A|=2\) and there is a maximum is on the \(y\)-axis. Therefore the shape is the same as that of \(y=cos (x)\), so \(A\) is positive, and thus \(A=2\). The midline is at \(y=3\), so \(D=3\).\(B= \dfrac{2\pi}{4}=\dfrac{\pi}{2}\).

Putting it all together, a possible equation would be \(f(x)=2\cos \left(\dfrac{\pi }{2} x\right)+3\). Its graph is below.

Howto: Given a sinusoidal function with a phase shift and a vertical shift, sketch its graph

1. Express the function in the general form \(y=A\sin(Bx−C)+D\) or \(y=A\cos(Bx−C)+D\). Rewrite the equation so \(B\) is positive and determine the basic shape of the graph.
2. Identify the amplitude, \(| A |\) and the midline \(y=D\), and maximum and minimum \(y\) values.
3. Identify the period, \(P=\dfrac{2\pi}{| B |}\) and the value of \(x\) at the quarter points.
4. Identify the phase shift, \(\dfrac{C}{B}\) and adjust the quarter points accordingly.
5. Draw the graph of \(f(x)=A\sin(Bx)\) or \(f(x)=A\cos(Bx)\) shifted to the right or left by \(\dfrac{C}{B}\) units and up or down by \(D\) units.

Example \(\PageIndex{7a}\): Grapha Sinusoidal Equation

Sketch a graph of \(f(x)=3\sin\left(\dfrac{\pi}{4}x−\dfrac{\pi}{4}\right)\).

Solution

• Step 1. The function is already written in general form: \(f(x)=3\sin\left(\dfrac{\pi}{4}x−\dfrac{\pi}{4}\right)\).This graph will have the shape of a sine function, starting at the midline and increasing to the right.
• Step 2. \(| A |=| 3 |=3\). The amplitude is \(3\). \(D=0\) so the midline is \(y=0\) and the graph has not been shifted vertically. The maximum and minimun values for \(y\) are \(3\) and \(-3\).
• Step 3. Since \(| B |=| \dfrac{\pi}{4} |=\dfrac{\pi}{4}\), we determine the period as follows.

  \( P=\dfrac{2\pi}{|B|}=\dfrac{2\pi}{\dfrac{\pi}{4}}=2\pi \cdot \dfrac{4}{\pi}=8 \)

  The period is \(8\). The quarter points are \(x=0,\; 2,\; 4,\; 6, \text{ and } 8\).

• Step 4. Since \(C=\dfrac{\pi}{4}\), the phase shift is\( \quad \dfrac{C}{B}=\dfrac{\dfrac{\pi}{4}}{\dfrac{\pi}{4}}=1\).

  The phase shift is \(1\) unit. This means \(1\) is added to each of the quarter points making them\(x=1,\; 3,\; 5,\; 7, \text{ and } 9\).

• Step 5.The graph of the function is below.

Example \(\PageIndex{7b}\)

Given \(y=−2cos\left(\dfrac{\pi}{2}x+\pi\right)+3\), determine the amplitude, period, phase shift, and horizontal shift. Then graph the function.

Solution

Begin by comparing the equation to the general form.

\(y=A\cos(Bx−C)+D\)

• Step 1. The function is already written in general form. It will have the shape of a reflected cosine function, starting at a minimum and going up to the right.
• Step 2. Since \(A=−2\), the amplitude is \(| A |=2\). \(D=3\) so the midline is \(y=3\) . The maximum and minimum values for \(y\) are \(y=3+2=5\) and \(y=3-2=1\).
• Step 3. \(| B |=\dfrac{\pi}{2}\), so the period is \(P=\dfrac{2\pi}{| B |}=\dfrac{2\pi}{\dfrac{pi}{2}}=2\pi⋅\dfrac{2}{\pi}=4\).
  The period is 4. The quarter points are \(x=0,\; 1,\; 2,\; 3, \text{ and } 4\).
• Step 4. \(C=−\pi\), so we calculate the phase shift as \(\dfrac{C}{B}=−\dfrac{\pi}{\dfrac{\pi}{2}}=−\pi⋅\dfrac{2}{\pi}=−2\).
  The phase shift is \(−2\) or two units to the left. This changes the quarter points to \(x=-2,\; -1,\; 0,\; 1, \text{ and } 2\).

One complete cycle starts at \(x=-2\) and since \(A\) is negative, the graph will start a a local minimum (\(y=1\)) and go up (like the shape of the function \(y=-\cos(x)\). Another cycle will start at \(x=2\) and this is the cycle indicated inthe graph of the function illustrated in Figure \(\PageIndex{23}\).

Try It\(\PageIndex{7}\)

Draw a graph of \(g(x)=−2\cos\left(\dfrac{\pi}{3}x+\dfrac{\pi}{6}\right)\). Determine the midline, amplitude, period, and phase shift.

Answer

midline: \(y=0\); amplitude: \(| A |=2\); period: \(P=\dfrac{2\pi}{| B |}=6\); phase shift: \(\dfrac{C}{B}=−\dfrac{1}{2}\) or \(\dfrac{1}{2}\) unit to the left.

Howto: Construct an equation fromthe graph of a sinusoidal function.

From a sinusoidal graph, construct one ofthe sinusoidal functions:\(y=A\sin(Bx−C)+D\) and\(y=A\cos(Bx−C)+D\)

Determine the values of the parameters used in the

1. Identify the amplitude, \(| A |\), half the distance between the highest and lowest values of \(y\) on the graph.
2. Identify the midline, the line halfway between thehighest and lowest values of \(y\) on the graph. This is parameter \(D\).
3. Identify the period \(P\),the horizontal distance between peaks on the graph. Calculate parameter \(B\) from the equation\(B=\dfrac{2\pi}{P}\).
4. Look for a highest point, a lowest point, or a point on the midline that is on (ideally - then there is no phase shift) or close to the \(y\) axis and determine the horizontal distance, \(H\), this point is from the \(y\)-axis. This is the phase shift and marks the beginning of one cycle of the function. Calculate parameter \(C\) from the equation \(C=B \cdot H\).
5. Identify thefunction to be used and the parameter \(A\) from the shape of the graph beginning at the starting point for one cycle,determined in the previous step.

\( \begin{array}{|c|c||c|c|}
\hline\text{Starting point}& \text{Graph direction}& \text{Function}& \text{parameter }A\\
\hline\text{a Maximum}& \text{Down}& \cos& |A|\\
\hline\text{on Midline}& \text{Up} & \sin& |A| \\[2pt]
\hline \text{on Midline} & \text{Down}& \sin & -|A| \\[2pt]
\hline \text{a Minimum} & \text{Up} & \cos & -|A| \\[2pt]
\hline \end{array} \)

Use the values of the parameters obtained above to construct a possible sinusoidal equation for the graph.

Example \(\PageIndex{8}\)Identifying the Equation for a Sinusoidal Function from a Graph

Determine the equation for the sinusoidal function below.

Solution

Step 1. Half the distance between the highest or lowest values gives an amplitude of \(| A |=3\).

Step 2. With the highest value at \(1\) and the lowest value at \(−5\), the midline will be halfway between at \(−2\). So \(D=−2\).

Step 3. The period of the graph is \(6\), which can be measured from the peak at \(x=1\) to the next peak at \(x=7\), or from the distance between adjacent lowest points. Therefore, \(P=\dfrac{2\pi}{| B |}=6\). Using the positive value for \(B\), we find that

\( B =\dfrac{2\pi}{P} =\dfrac{2\pi}{6} =\dfrac{\pi}{3} \)

So far, our equation is either \(y=3\sin\left(\dfrac{\pi}{3}x−C\right)−2\) or \(y=3\cos\left(\dfrac{\pi}{3}x−C\right)−2\).

Steps 4 and 5. For the shape and horizontal shift, we have more than one option. We could write this as any one of the following:

• a cosine shifted to the right \(\qquad \) Starting at \(x = 1 = H = C/B\)
• a negative cosine shifted to the left \(\qquad \) Starting at \(x = -2\)
• a sine shifted to the left\(\qquad \) Starting at \(x \approx 1/2\)
• a negative sine shifted to the right\(\qquad \) Starting at \(x \approx2 \; 1/2\)

While any of these would be correct, the cosine shifts are easier to work with than the sine shifts in this case because they involve integer values. So our function becomes either\(y=3\cos\left(\dfrac{\pi}{3}x−C\right)−2\) or\(y=-3\cos\left(\dfrac{\pi}{3}x−C\right)−2\). The horizontal displacement when \(A=3\) to the highest point on the graph is one unit to the right, so the equation used to solve for \(C\) is \(\dfrac{C}{B} = 1 \RightarrowC=B = \dfrac{\pi}{3}\). Alternatively, we could have chosen to use the equation in which \(A = -3\) (indicating reflection over the \(x\)-axis).In this case, the horizontal displacement to the lowest point on the graph is two units to the left, and the equation used to solve for \(C\) is \(\dfrac{C}{B} = -2 \RightarrowC=-2B = -\dfrac{2\pi}{3}\).

So, two possibilities for our equation are \(y=3\cos\left(\dfrac{\pi}{3}x−\dfrac{\pi}{3}\right)−2\) and \(y=-3\cos\left(\dfrac{\pi}{3}x+\dfrac{2\pi}{3}\right)−2\).Again, these functions are equivalent, so both yield the same graph.

Try It\(\PageIndex{8}\)

Write a formula for the function graphed below.

Answer

Two possible equations are: \(y=4\sin\left(\dfrac{\pi}{5}x−\dfrac{\pi}{5}\right)+4\) or \(y=−4\sin\left(\dfrac{\pi}{5}x+\dfrac{4\pi}{5}\right)+4\)

Example \(\PageIndex{9}\): Finding the Vertical Component of Circular Motion

A point rotates around a circle of radius \(3\) centered at the origin. Sketch a graph of the \(y\)-coordinate of the point as a function of the angle of rotation.

Solution

Recall that, for a point on a circle of radius \(r\), the \(y\)-coordinate of the point is \(y=r \sin(x)\), so in this case, we get the equation \(y(x)=3 \sin(x)\). The constant \(3\) causes a vertical stretch of the \(y\)-values of the function by a factor of \(3\), which we can see in the graph in Figure \(\PageIndex{24}\).

Analysis

Notice that the period of the function is still \(2\pi\); as we travel around the circle, we return to the point \((3,0)\) for \(x=2\pi,4\pi,6\pi,\)....Because the outputs of the graph will now oscillate between \(–3\) and \(3\), the amplitude of the sine wave is \(3\).

Example \(\PageIndex{10}\): Finding the Vertical Component of Circular Motion

A circle with radius \(3\) ft is mounted with its center \(4\) ft off the ground. The point closest to the ground is labeled \(P\), as shown in Figure \(\PageIndex{26}\). Sketch a graph of the height above the ground of the point \(P\) as the circle is rotated; then find a function that gives the height in terms of the angle of rotation.

Solution

For a point rotating on a circle of radius \(r\), the height is the \(y\)-coordinate of the point defined by the equation \(y=r \sin(\theta)\) so \(y=3 \sin(\theta)\). The amplitude is \(|A| = 3\).

Sketching the height, we note that it will start \(1\) ft above the ground, then increase up to \(7\) ft above the ground, and continue to oscillate \(3\) ft above and below the center value of \(4\) ft, as shown in Figure \(\PageIndex{27}\).

Although we could use a transformation of either the sine or cosine function, we start by looking for characteristics that would make one function easier to use than the other. Let’s use a cosine function because it starts at the highest or lowest value, while a sine function starts at the middle value. A standard cosine starts at the highest value, and this graph starts at the lowest value, so we need to incorporate a vertical reflection.

Second, we see that the graph oscillates \(3\) above and below the center, while a basic cosine has an amplitude of \(1\), so this graph has been vertically stretched by \(3\), as in the last example.

Finally, to move the center of the circle up to a height of \(4\), the graph has been vertically shifted up by \(4\). Putting these transformations together, we find that

\(y=−3\cos(x)+4\)

Example \(\PageIndex{11}\)

A point completes 1 revolution every 2 minutes around a circle of radius 5. Find the \(x\) coordinate of the point as a function of time, if it starts at (5, 0).

Solution

For a point moving on a rotating circle of radius \(r\), the \(x\)-coordinate of the point is \(x=r \cos(\theta)\) so \(x=5 \cos(\theta)\). The amplitude is \(|A| = 5\).

We are told the starting point is at a maximum point, so this is a cosine function.

Simultaneously, from a different perspective of time rather than rotational angle, the same motion completes one revolution every \(t=2\) minutes. A cosine function with the same amplitude buta period \(P=2\) has equation parameter \(B\) defined by the formula\(P = \dfrac{2\pi}{B}\). So \(B = \pi\) and the equation in terms of time is\(x=5\cos(\pi t)\).

Example \(\PageIndex{12}\): Determining a Rider’s Height on a Ferris Wheel

The London Eye is a huge Ferris wheel with a diameter of \(135\) meters (\(443\) feet). It completes one rotation every \(30\) minutes. Riders board from a platform \(2\) meters above the ground. Express a rider’s height above ground as a function of time in minutes.

Solution

With a diameter of \(135\) m, the wheel has a radius of \(67.5\) m. The height will oscillate with amplitude \(67.5\) m above and below the center.

Passengers board \(2\) m above ground level, so the center of the wheel must be located \(67.5+2=69.5\) m above ground level. The midline of the oscillation will be at \(69.5\) m.

The wheel takes \(30\) minutes to complete \(1\) revolution, so the height will oscillate with a period of \(30\) minutes.

Lastly, because the rider boards at the lowest point, the height will start at the smallest value and increase, following the shape of a vertically reflected cosine curve.

• Amplitude: \(67.5\), so \(A=67.5\)
• Midline: \(69.5\), so \(D=69.5\)
• Period: \(30\), so \(B=\dfrac{2\pi}{30}=\dfrac{\pi}{15}\)
• Shape: \(−\cos(t)\)

An equation for the rider’s height would be

\(y=−67.5\cos\left(\dfrac{\pi}{15}t\right)+69.5\)

where \(t\) is in minutes and \(y\) is measured in meters.

Try It \(\PageIndex{12}\)

The Ferris wheel at the Puyallup Fair (Photo by photogirl7.1, www.flickr.com/photos/kitkaph...86205/sizes/z/, CC-BY) has a diameter of about 70 feet and takes 3 minutes to complete a full rotation. Passengers board from a platform 10 feet above the ground. Write an equation for a rider’s height above ground over time.

Answer

\[h(t)=-35\cos \left(\dfrac{2\pi }{3} t\right)+45\nonumber\]

Example \(\PageIndex{13}\)

A bicycle wheel with radius 14 inches has the bottom-most point on the wheel marked in red. The wheel then begins rolling down the street. Write a formula for the height above ground of the red point after the bicycle has travelled \(x\) inches.

Solution

For a point moving on a rotating circle of radius \(r\), the height of the point above the ground is the \(y\)-coordinate of the point, \(y=r \sin(\theta)\). Therefore\(y=14 \cos(\theta)\), so the amplitude is \(|A| = 14\). The height oscillates between \(0\) inches and \(28\) inches above ground, so the vertical displacement is \(D=14\).

We are told the starting point is at a minimum point, so this is a reflected cosine function.

Simultaneously, from a different perspective of distance \(d\)travelled rather than rotational angle, the same motion completes one revolution every \(d=28\pi\) inches. A cosine function with the same amplitude and vertical displacement, buta period \(P=28\pi\) has equation parameter \(B\) defined by the formula\(P = \dfrac{2\pi}{B}\). So \(B = \dfrac{\pi}{14}\) and the equation for the height of the point in terms of distance travelled is

\(y=-14\cos\left(\dfrac{\pi}{14}d\right)+14)\).

Example \(\PageIndex{14}\): Modeling an Equation and Sketching a Sinusoidal Graph to Fit Criteria

The average monthly temperatures for a small town in Oregon are given inthe table below. Find a sinusoidal function of the form \(y=A \sin (Bt−C)+D\) that fits the data (round to the nearest tenth) and sketch the graph.

Solution

Recall that amplitude is found using the formula\(A=\tfrac{\text{largest value −smallest value}}{2}\).Thus, the amplitude is

\(|A| = \tfrac{69−42.5}{2} =13.25 \)

The data covers a period of 12 months, so \(\frac{2π}{B}=12\) which gives

\(B=\frac{2π}{12}=\frac{π}{6}\).

The vertical shift is found using the equation\(D=\tfrac{\text{highest value+lowest value}}{2}\).Thus, the vertical shift is

\(D = \tfrac{69+42.5}{2} =55.8 \)

So far, we have the equation \(y=13.3 \sin (\frac{π}{6}x−C)+55.8\).To find the horizontal shift, we input the \(x\) and \(y\) values for the first month and solve for \(C\).

\(\begin{array} {rll} 42.5 & =13.3 \sin (\frac{π}{6}(1)−C)+55.8 \\ −13.3 & =13.3 \sin (\frac{π}{6}−C) \\ −1 & =\sin (\frac{π}{6}−C) \qquad\sin θ=−1→ θ=−\frac{π}{2} \\ −\frac{π}{2} &=\frac{π}{6}−C \qquad \qquad\quad C = \frac{π}{6}+\frac{π}{2} → C =\frac{2π}{3} \end{array}\)

We have the equation \(y=13.3 \sin (\frac{π}{6}x−\frac{2π}{3})+55.8\). See the graph in Figure \(\PageIndex{14}\).

Try It\(\PageIndex{15}\)

A weight is attached to a spring that is then hung from a board, as shown in Figure \(\PageIndex{28}\). As the spring oscillates up and down, the position \(y\) of the weight relative to the board ranges from \(–1\) in. (at time \(t=0\)) to \(–7\) in. (at time \(t=π\)) below the board. Assume the position of \(y\) is given as a sinusoidal function of time \(t\). Sketch a graph of the function, and then find a function that gives the position \(y\) in terms of \(t\).

Answer

\(y=3\cos(x)−4\)