Example 1: Identifying the Period of a Sine or Cosine Function

Determine the period of the function [latex]f(x) = \sin(\frac{π}{6}x)\\[/latex].

Solution

Let's begin by comparing the equation to the general form [latex]y=Asin(Bx)[/latex]. In the given equation, [latex]B =\frac{π}{6}[/latex], so the period will be

[latex]\begin{array}P=\frac{\frac{2}{\pi}}{|B|}\hfill \\ =\frac{2\pi}{\frac{x}{6}}\hfill \\ =2\pi\times \frac{6}{\pi}\hfill \\ =12\hfill \end{array}\\[/latex]

Example 2: Identifying the Amplitude of a Sine or Cosine Function

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

Example 3: Identifying the Phase Shift of a Function

Determine the direction and magnitude of the phase shift for [latex]f(x)=\sin(x+\frac{π}{6})−2\\[/latex].

Solution

Let’s begin by comparing the equation to the general form [latex]y=A\sin(Bx−C)+D\\[/latex]. In the given equation, notice that B = 1 and [latex]C=−\frac{π}{6}\\[/latex]. So the phase shift is

[latex]\begin{array}\frac{C}{B}=−\frac{\frac{x}{6}}{1}\hfill \\ =−\frac{\pi}{6}\hfill \end{array}\\[/latex]

or [latex]\frac{\pi}{6}\\[/latex] units to the left.

Example 4: Identifying the Vertical Shift of a Function

Determine the direction and magnitude of the vertical shift for [latex]f(x)=\cos(x)−3\\[/latex].

Solution

Let's begin by comparing the equation to the general form [latex]y=A\cos(Bx−C)+D\\[/latex]

Example 5: Identifying the Variations of a Sinusoidal Function from an Equation

Determine the midline, amplitude, period, and phase shift of the function [latex]y=3\sin(2x)+1\\[/latex].

Solution

Let’s begin by comparing the equation to the general form [latex]y=A\sin(Bx−C)+D\\[/latex]. A = 3, so the amplitude is |A| = 3. Next, B = 2, so the period is [latex]\text{P}=\frac{2π}{|B|}=\frac{2π}{2}=π\\[/latex]. There is no added constant inside the parentheses, so C = 0 and the phase shift is [latex]\frac{C}{B}=\frac{0}{2}=0\\[/latex]. Finally, D = 1, so the midline is y = 1.

Example 6: Identifying the Equation for a Sinusoidal Function from a Graph

Determine the formula for the cosine function in Figure 15.

Figure 15

Solution

Example 7: Identifying the Equation for a Sinusoidal Function from a Graph

Determine the equation for the sinusoidal function in Figure 17.

Figure 17

Solution

With the highest value at 1 and the lowest value at−5, the midline will be halfway between at −2. So D = −2. The distance from the midline to the highest or lowest value gives an amplitude of |A|=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 the lowest points. Therefore, [latex]\text{P}=\frac{2\pi}{|B|}=6[/latex]. Using the positive value for B, we find that

[latex]B=\frac{2π}{P}=\frac{2π}{6}=\frac{π}{3}\\[/latex]

So far, our equation is either [latex]y=3\sin(\frac{\pi}{3}x−C)−2\\[/latex] or [latex]y=3\cos(\frac{\pi}{3}x−C)−2\\[/latex]. For the shape and shift, we have more than one option. We could write this as any one of the following:

• a cosine shifted to the right
• a negative cosine shifted to the left
• a sine shifted to the left
• a negative sine shifted to the right

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

[latex]y=3\cos(\frac{π}{3}x−\frac{π}{3})−2\\[/latex] or [latex]y=−3\cos(\frac{π}{3}x+\frac{2π}{3})−2\\[/latex]

Again, these functions are equivalent, so both yield the same graph.

Example 8: Graphing a Function and Identifying the Amplitude and Period

Sketch a graph of [latex]f(x)=−2\sin(\frac{πx}{2})\\[/latex].

Solution

Let’s begin by comparing the equation to the form [latex]y=A\sin(Bx)\\[/latex]. Step 1. We can see from the equation thatA=−2,so the amplitude is 2.

|A| = 2

Step 2. The equation shows that [latex]B=\frac{π}{2}\\[/latex], so the period is

[latex] \begin{array} \text{P}=\frac{2\pi}{\frac{\pi}{2}}\\=2\pi\times\frac{2}{\pi}\\=4 \end{array}\\[/latex]

Step 3. Because A is negative, the graph descends as we move to the right of the origin. Step 4–7. The x-intercepts are at the beginning of one period, x = 0, the horizontal midpoints are at x = 2 and at the end of one period at x = 4. The quarter points include the minimum at x = 1 and the maximum at x = 3. A local minimum will occur 2 units below the midline, at x = 1, and a local maximum will occur at 2 units above the midline, at x = 3. Figure 19 shows the graph of the function.

Example 9: Graphing a Transformed Sinusoid

Sketch a graph of [latex]f(x)=3\sin(\frac{π}{4}x−\frac{π}{4})\\[/latex].

Solution

Step 1. The function is already written in general form: [latex]f(x)=3\sin(\frac{π}{4}x−\frac{π}{4})\\[/latex]. 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. Step 3. Since [latex]|B|=|\frac{π}{4}|=\frac{π}{4}\\[/latex], we determine the period as follows.

[latex]\text{P}=\frac{2π}{|B|}=\frac{2π}{\frac{π}{4}}=2π\times\frac{4}{π}=8\\[/latex]

The period is 8. Step 4. Since [latex]\text{C}=\frac{π}{4}\\[/latex], the phase shift is

[latex]\frac{C}{B}=\frac{\frac{\pi}{4}}{\frac{\pi}{4}}=1\\[/latex].

The phase shift is 1 unit. Step 5. Figure 20 shows the graph of the function.

Example 10: Identifying the Properties of a Sinusoidal Function

Given [latex]y=−2\cos(\frac{\pi}{2}x+\pi)+3\\[/latex], determine the amplitude, period, phase shift, and horizontal shift. Then graph the function.

Solution

Begin by comparing the equation to the general form and use the steps outlined in Example 9. [latex-display]y=A\cos(Bx−C)+D\\[/latex-display] Step 1. The function is already written in general form. Step 2. Since A = −2, the amplitude is|A| = 2. Step 3. [latex]|B|=\frac{\pi}{2}\\[/latex], so the period is [latex]\text{P}=\frac{2π}{|B|}=\frac{2\pi}{\frac{\pi}{2}}\times2\pi=4\\[/latex]. The period is 4. Step 4. [latex]C=−\pi\\[/latex], so we calculate the phase shift as [latex]\frac{C}{B}=\frac{−\pi}{\frac{\pi}{2}}=−\pi\times\frac{2}{\pi}=−2\\[/latex]. The phase shift is −2. Step 5. D = 3, so the midline is y = 3, and the vertical shift is up 3. Since A is negative, the graph of the cosine function has been reflected about the x-axis. Figure 21 shows one cycle of the graph of the function.

Example 11: 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 [latex]y=r\sin(x)[/latex], so in this case, we get the equation [latex]y(x)=3\sin(x)[/latex]. 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 22.

Example 12: 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 23. 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.

Figure 23

Solution

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 24. 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

[latex]y=−3\cos(x)+4[/latex]

Example 13: 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 [latex]B=\frac{2\pi}{30}=\frac{\pi}{15}[/latex]
• Shape: −cos(t)

An equation for the rider’s height would be

[latex]y=−67.5\cos(\frac{\pi}{15}t)+69.5[/latex]

where t is in minutes and y is measured in meters.