With your calculator set in radian mode, graph the three functions in the ZTrig window (press ZOOM 7). The graphs are shown below. All three graphs have the same period ([latex]2\pi[/latex]) and midline ([latex]y=0[/latex]), but the graph of [latex]f[/latex] has amplitude 2, and the graph of [latex]g[/latex] has amplitude 0.5.

With your calculator set in radian mode, graph [latex]f(x)=\cos 2x[/latex] and [latex]y=\cos x[/latex] in the same window, as shown below. Both graphs have the same amplitude ([latex]1[/latex]) and midline ([latex]y=0[/latex]), but the graph of [latex]f[/latex] completes two cycles from [latex]0[/latex] to [latex]2\pi[/latex] instead of one. The period of [latex]f(x)=\cos {2}x[/latex] is [latex]\dfrac{2\pi}{{2}}=\pi{.}[/latex]
Now graph [latex]g(x)=\cos \dfrac{1}{3}x[/latex] and [latex]y=\cos x[/latex] in the same window. Set Xmin [latex]=0[/latex] and Xmax [latex]=6\pi{.}[/latex]
The graph of [latex]g(x)=\cos {\dfrac{1}{3}}x[/latex] completes one cycle between [latex]0[/latex] and [latex]6\pi{.}[/latex] Its period is [latex]\dfrac{2\pi}{{\dfrac{1}{3}}}=6\pi{.}[/latex]

Graph both functions in the ZTrig window. The graphs are shown below. Each point on the graph of [latex]f(x)=2+\sin x[/latex] has [latex]y[/latex]-coordinate 2 units higher than the corresponding point on the graph of [latex]y=\sin x{.}[/latex] Thus, the graph of [latex]f(x)=2+\sin x[/latex] is shifted vertically by 2 units relative to the graph of [latex]y=\sin x{.}[/latex] In particular, the midline of [latex]f(x)={2}+\sin x[/latex] is the line [latex]y={2}{.}[/latex]

1. The amplitude of the graph is 3, its midline is [latex]y=2{,}[/latex] and its period is [latex]\dfrac{2\pi}{4}=\dfrac{\pi}{2}{.}[/latex]
2. One way to make a quick sketch of a sinusoidal graph is to use a table of values. The trick is to choose convenient values for the input variable. In the table below, notice that we choose the quadrantal angles as the input values for the trigonometric function.
   

   [latex]t[/latex]	[latex]4t[/latex]	[latex]\cos 4t[/latex]	[latex]3\cos 4t[/latex]	[latex]y=2+\cos 4t[/latex]
   [latex]\hphantom{0000}[/latex]	[latex]{0}[/latex]	[latex]1[/latex]	[latex]\hphantom{0000}[/latex]	[latex]\hphantom{0000}[/latex]
   [latex]\hphantom{0000}[/latex]	[latex]{\dfrac{\pi}{2}}[/latex]	[latex]0[/latex]	[latex]\hphantom{0000}[/latex]	[latex]\hphantom{0000}[/latex]
   [latex]\hphantom{0000}[/latex]	[latex]{\pi}[/latex]	[latex]-1[/latex]	[latex]\hphantom{0000}[/latex]	[latex]\hphantom{0000}[/latex]
   [latex]\hphantom{0000}[/latex]	[latex]{\dfrac{3\pi}{2}}[/latex]	[latex]0[/latex]	[latex]\hphantom{0000}[/latex]	[latex]\hphantom{0000}[/latex]
   [latex]\hphantom{0000}[/latex]	[latex]{2\pi}[/latex]	[latex]1[/latex]	[latex]\hphantom{0000}[/latex]	[latex]\hphantom{0000}[/latex]

   Now we work backward from [latex]4t[/latex] to find the values of [latex]t{,}[/latex] and forward from [latex]\cos 4t[/latex] to find the values of [latex]y{.}[/latex]

   [latex]t[/latex]	[latex]4t[/latex]	[latex]\cos 4t[/latex]	[latex]3\cos 4t[/latex]	[latex]y=2+\cos 4t[/latex]
   [latex]0[/latex]	[latex]0[/latex]	[latex]1[/latex]	[latex]3[/latex]	[latex]5[/latex]
   [latex]{\dfrac{\pi}{8}}[/latex]	[latex]\dfrac{\pi}{2}[/latex]	[latex]0[/latex]	[latex]0[/latex]	[latex]{2}[/latex]
   [latex]{\dfrac{\pi}{4}}[/latex]	[latex]\pi[/latex]	[latex]-1[/latex]	[latex]-3[/latex]	[latex]{-1}[/latex]
   [latex]{\dfrac{3\pi}{8}}[/latex]	[latex]\dfrac{3\pi}{2}[/latex]	[latex]0[/latex]	[latex]0[/latex]	[latex]{2}[/latex]
   [latex]{\dfrac{\pi}{2}}[/latex]	[latex]2\pi[/latex]	[latex]1[/latex]	[latex]3[/latex]	[latex]{5}[/latex]

   Notice from the table that the graph completes one cycle from [latex]t=0[/latex] to [latex]t=\dfrac{\pi}{2}{,}[/latex] which confirms that the period is [latex]\dfrac{\pi}{2}{.}[/latex] Finally, we plot the points [latex](t,y)[/latex] from the table and use them as “guide points” to sketch a sinusoidal graph, as shown below.
   

We would like a function of the form [latex]y=k+a\sin Bt{,}[/latex] so we must find the values of the parameters [latex]A,~B[/latex] and [latex]k{.}[/latex]

• The midline of the graph is [latex]y=\dfrac{70+110}{2}=90{,}[/latex] and the amplitude is [latex]110-90=20{,}[/latex] so [latex]A=20[/latex] and [latex]k=90{.}[/latex]
• The graph repeats 60 times per minute, so the period is [latex]\dfrac{1}{60}[/latex] minute, and [latex]B=\dfrac{2\pi}{\dfrac{1}{60}}=120\pi{.}[/latex]

Thus,

[latex]y=90+20\sin 120 \pi t[/latex]

The graph of the function is shown below.

1. Recall that the period of the tangent function is [latex]\pi{.}[/latex] We make a table of values for one cycle of the function, choosing multiples of [latex]\dfrac{\pi}{4}[/latex] as the inputs for the tangent function. Then we plot the guide points and sketch a tangent function through them. The graph is shown below.
   

   [latex]x[/latex]	[latex]2x[/latex]	[latex]\tan 2x[/latex]	[latex]1+3\tan 2x[/latex]
   [latex]0[/latex]	[latex]{0}[/latex]	[latex]0[/latex]	[latex]1[/latex]
   [latex]\dfrac{\pi}{8}[/latex]	[latex]{\dfrac{\pi}{4}}[/latex]	[latex]1[/latex]	[latex]4[/latex]
   [latex]\dfrac{\pi}{4}[/latex]	[latex]{\dfrac{\pi}{2}}[/latex]	[latex]---[/latex]	[latex]---[/latex]
   [latex]\dfrac{3\pi}{8}[/latex]	[latex]{\dfrac{3\pi}{4}}[/latex]	[latex]-1[/latex]	[latex]-2[/latex]
   [latex]\dfrac{\pi}{2}[/latex]	[latex]{\pi}[/latex]	[latex]0[/latex]	[latex]1[/latex]

   

2. Writing the formula as[latex]y=A\tan Bx+k=3\tan 2x+1[/latex]we see that the graph is stretched vertically by a factor of [latex]A=3{.}[/latex] The midline is [latex]y=1{,}[/latex] so the graph is shifted up by 1 unit. Finally, the coefficient [latex]B=2[/latex] compresses the graph horizontally by a factor of 2, so the period of the graph is [latex]\dfrac{\pi}{2}{,}[/latex] and there are four cycles between [latex]0[/latex] and [latex]2\pi[/latex]