Graphs of the Sine and Cosine Function: Learn It 3
Phase Shift of [latex]y=\sin x[/latex] and [latex]y=\cos x[/latex]
Now that we understand how A and B relate to the general form equation for the sine and cosine functions, we will explore the variables C and D. Recall the general form:
[latex]y = A \sin(Bx−C)+D[/latex] and [latex]y=A\cos(Bx−C)+D[/latex]
or
[latex]y=A\sin(B(x−\frac{C}{B}))+D[/latex] and [latex]y=A\cos(B(x−\frac{C}{B}))+D[/latex]
The value [latex]\frac{C}{B}[/latex] for a sinusoidal function is called the phase shift, or the horizontal displacement of the basic sine or cosine function. If C > 0, the graph shifts to the right. If C < 0,the graph shifts to the left. The greater the value of |C|, the more the graph is shifted. Figure 11 shows that the graph of [latex]f(x)=\sin(x−π)[/latex] shifts to the right by π units, which is more than we see in the graph of [latex]f(x)=\sin(x−\frac{π}{4})[/latex], which shifts to the right by [latex]\frac{π}{4}[/latex]units.
Figure 11
While C relates to the horizontal shift, D indicates the vertical shift from the midline in the general formula for a sinusoidal function. The function [latex]y=\cos(x)+D[/latex] has its midline at [latex]y=D[/latex].
Figure 12
Any value of D other than zero shifts the graph up or down. Figure 13 compares [latex]f(x)=\sin x[/latex] with [latex]f(x)=\sin (x)+2[/latex], which is shifted 2 units up on a graph.
phase shift and vertical shift for sine and cosine
Given an equation in the form [latex]f(x)=A\sin(Bx−C)+D[/latex] or [latex]f(x)=A\cos(Bx−C)+D[/latex], [latex]\frac{C}{B}[/latex]is the phase shift and D is the vertical shift.
Determine the direction and magnitude of the phase shift for [latex]f(x)=\sin(x+\frac{π}{6})−2[/latex].
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
or [latex]\frac{\pi}{6}[/latex] units to the left.
Analysis of the Solution
We must pay attention to the sign in the equation for the general form of a sinusoidal function. The equation shows a minus sign before C. Therefore [latex]f(x)=\sin(x+\frac{π}{6})−2[/latex] can be rewritten as [latex]f(x)=\sin(x−(−\frac{π}{6}))−2[/latex]. If the value of C is negative, the shift is to the left.
Determine the direction and magnitude of the phase shift for [latex]f(x)=3\cos(x−\frac{\pi}{2})[/latex].
[latex]\frac{π}{2}[/latex]; right
Determine the direction and magnitude of the vertical shift for [latex]f(x)=\cos(x)−3[/latex].
Let’s begin by comparing the equation to the general form [latex]y=A\cos(Bx−C)+D[/latex]. In the given equation, [latex]D=-3[/latex], so the shift is 3 units downward.
Determine the direction and magnitude of the vertical shift for [latex]f(x)=3\sin(x)+2[/latex].
2 units up
How To: Given a sinusoidal function in the form [latex]f(x)=A\sin(Bx−C)+D[/latex], identify the midline, amplitude, period, and phase shift.
Determine the amplitude as [latex]|A|[/latex].
Determine the period as [latex]P=\frac{2π}{|B|}[/latex].
Determine the phase shift as [latex]\frac{C}{B}[/latex].
Determine the midline as [latex]y = D[/latex].
Determine the midline, amplitude, period, and phase shift of the function [latex]y=3\sin(2x)+1[/latex].
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]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.
Analysis of the Solution
Inspecting the graph, we can determine that the period is π, the midline is y = 1,and the amplitude is 3. See Figure 14.
Figure 14
Determine the midline, amplitude, period, and phase shift of the function [latex]y=\frac{1}{2}\cos(\frac{x}{3}−\frac{π}{3})[/latex].
