# 4.5 | Graphs of Sine and Cosine Functions

The graphs of $$y = a \sin(bx-c) +d$$ and $$y = a \cos(bx-c)+d$$ have the following characteristics. (Assume $$b>0.$$)

• The amplitude is given by $$|a|$$ and represents half the distance between the maximum and minimum values of the function.
• The period is given by $$\displaystyle \frac{2\pi}{b}.$$
• The phase shift is given by $$c/b$$.
• The center amplitude is given by $$d.$$ The corresponding midline is given by $$y = d.$$
• The left and right endpoints of a one-cycle interval can be determined by solving the equations $$bx-c = 0$$ and $$bx -c = 2\pi.$$

Sketch the graph of
$$y = 4 \sin\left(x – \frac{\pi}{4} \right) +4.$$
(Include two full periods.) Label $$x$$ and $$y$$ intercepts (if any). Find the amplitude, period, phase shift, and midline.

From the form of the equation we can have that $$a = 4,$$ $$b = 1,$$ $$c = \frac{\pi}{4},$$ and $$d = 4.$$ So that:

• The left and right endpoints of a two cycle interval can be determined by solving the equations $$x – \frac{\pi}{4} = 0$$ and $$x – \frac{\pi}{4} = 4 \pi$$. Thus, our two cycle interval is $$[\pi/4, 17\pi/4]$$.
• The amplitude is $$4.$$
• The period is $$\frac{2\pi}{b} = \frac{2\pi}{1} = 2\pi.$$
• The phase shift is $$\pi/4.$$
• The midline is $$y = 4.$$
• The $$y$$-intercept:
The two cycles we are sketched over the interval $$[\pi/4,17\pi/4]$$ which does not include $$x=0$$. So our graph will not include a $$y$$-intercept.
• The $$x$$-intercepts:
Set $$y = 0$$. Then
\begin{aligned} & y = 4 \sin\left(x – \frac{\pi}{4} \right) +4 \\ \Rightarrow & 0 = 4 \sin\left(x – \frac{\pi}{4} \right) +4 \\ \Rightarrow & -1 = \sin\left(x – \frac{\pi}{4} \right) \\ \Rightarrow & x- \frac{\pi}{4} = \frac{3\pi}{2} + n 2\pi\\ \Rightarrow & x = \frac{3\pi}{2}+ \frac{\pi}{4} + n 2\pi\\ \Rightarrow & x = \frac{7\pi}{4} + n 2\pi\\ \end{aligned}
So the $$x$$-intercepts over the interval $$[\pi/4,17\pi/4]$$ are $$\left(\frac{7\pi}{4},0\right)$$ and $$\left(\frac{15\pi}{4},0\right)$$.

Sketch the graph of
$$y = -1 + 2 \cos \frac{\pi t}{12}.$$
(Include two full periods.) Label $$t$$ and $$y$$ intercepts. Find the amplitude, period, phase shift, and midline.

First rewrite the equation in standard form
$$y = 2 \cos \left(\frac{\pi}{12} t \right) -1$$
So that $$a = 2,$$ $$b = \frac{\pi}{12},$$ $$c = 0,$$ and $$d = -1.$$ Now we have the following:

• The left and right endpoints of a two cycle interval can be determined by solving the equations $$\frac{\pi}{12} t = 0$$ and $$\frac{\pi}{12} t = 4 \pi$$. Thus, our two cycle interval is $$[0, 48]$$.
• The amplitude is $$2.$$
• The period is $$\frac{2\pi}{b} = \frac{2\pi}{\frac{\pi}{12}} = 24.$$
• The phase shift is $$0.$$
• The midline is $$y = -1.$$
• The $$y$$-intercept:
Set $$t = 0$$. Then
\begin{aligned} y & = -1 + 2 \cos \frac{\pi t}{12} \\ & = -1 + 2 \cos \frac{\pi 0}{12} \\ & = -1 + 2 \cos 0 \\ & = -1 + 2 = -1\\ \end{aligned}
So the $$y$$-intercept is $$\left( 0,-2 \right)$$
• The $$t$$-intercepts:
Set $$y = 0$$. Then
\begin{aligned} & y = -1 + 2 \cos \frac{\pi t}{12} \\ \Rightarrow & 0 = -1 + 2 \cos \frac{\pi t}{12} \\ \Rightarrow & \frac{1}{2} = \cos \frac{\pi t}{12} \\ \Rightarrow & \frac{\pi t}{12} = \pm\frac{\pi}{3} +n 2\pi \\ \Rightarrow & t = \pm 4 +24 n \\ \end{aligned}
So the $$t$$-intercepts over the interval $$[0,48]$$ are
$$(4,0),(20,0),(28,0),(44,0).$$