Advanced Mathematics
lesson 22
The unit circle and sine, cosine

A unit circle is a circle with radius of one unit.  It is helpful in graphing the sine and cosine trignometric functions to look at the unit circle. The sine equals the ratio of the opposite side to the hypotenuse.  This equals the length of the opposite side since the hypotenuse is one.

circ = ParametricPlot[{Cos[t], Sin[t]}, {t,0,6.282}, AspectRatio->Automatic,Ticks->None,  DisplayFunction->Identity,PlotRange->{-1,1}]
plt = Plot[Sin[x- Pi],{x,Pi,3Pi},DisplayFunction -> Identity]
Do[Show[circ,plt,Graphics[{{PointSize[.03],Point[{Cos[(1/6)n 3.142], Sin[(1/6)n 3.142]}]},{PointSize[.03], RGBColor[1,0,0],Point[{(1/6)n 3.142+3.142,-Sin[(1/6)n 3.142 +3.142]}]},
Line[{{0,0},{Cos[(1/6)n 3.142], Sin[(1/6)n 3.142]}}],{Thickness[.01], RGBColor[1,0,0],
Line[{{(1/6)n 3.142+3.142,-Sin[(1/6)n 3.142 +3.142]},{(1/6)n 3.142+3.142,0}}],{Thickness[.01], RGBColor[1,0,0]},Line[{{Cos[(1/6)n 3.142],0},{Cos[(1/6)n 3.142],Sin[(1/6)n 3.142]}}]},{Thickness[.01],RGBColor[0,1,0],Line[{{0,0},{Cos[(1/6)n 3.142],0}}]}}],
DisplayFunction ->$DisplayFunction],{n,0,12}]

Using the same unit circle graphing the cosine would be the length of the adjacent side of the triangle.  Since the cosine equals the ratio of the adjacent side to the hypotenuse and the hypotenuse is one.

circ = ParametricPlot[{Cos[t], Sin[t]}, {t,0,6.282}, AspectRatio->Automatic,Ticks->None,  DisplayFunction->Identity,PlotRange->{-1,1}]
plt = Plot[Cos[x- Pi],{x,Pi,3Pi},DisplayFunction -> Identity]
Do[Show[circ,plt,Graphics[{{PointSize[.03],RGBColor[1,0,0],Point[{Cos[(1/6)n 3.142], Sin[(1/6)n 3.142]}]},{PointSize[.03], RGBColor[0,1,0],Point[{(1/6)n 3.142+3.142,-Cos[(1/6)n 3.142 +3.142]}]},
Line[{{0,0},{Cos[(1/6)n 3.142], Sin[(1/6)n 3.142]}}],{Thickness[.01], RGBColor[0,1,0],
Line[{{(1/6)n 3.142+3.142,-Cos[(1/6)n 3.142 +3.142]},{(1/6)n 3.142+3.142,0}}]},{Thickness[.01],RGBColor[1,0,0],Line[{{Cos[(1/6)n 3.142],0},{Cos[(1/6)n 3.142],Sin[(1/6)n 3.142]}}]},{Thickness[.01],RGBColor[0,1,0],Line[{{0,0},{Cos[(1/6)n 3.142],0}}]}}],
DisplayFunction ->$DisplayFunction],{n,0,12}]

The standard form of the sine function is f[x]=C + K Sine[x].  In the preceding examples K=1 and C=0, what would happen to the graph of the sine if the constant K was something other than 1?

Clear[f,x,k]
 p1 = Plot[2Sin[x],{x,-Pi,2Pi},PlotLabel->"f[x]=2Sin[x]",PlotStyle->RGBColor[1,0,0],DisplayFunction->Identity]
 p2 = Plot[4Sin[x],{x,-Pi,2Pi},PlotLabel->"f[x]=4Sin[x]",PlotStyle->RGBColor[1,0,1],DisplayFunction->Identity]
 dp1 = Plot[-2Sin[x],{x,-Pi,2Pi},PlotLabel->"f[x]=-2Sin[x]",PlotStyle->RGBColor[0,1,0],DisplayFunction->Identity]
 dp2 = Plot[-4Sin[x],{x,-Pi,2Pi},PlotLabel->"f[x]=-4Sin[x]",PlotStyle->RGBColor[0,1,1],DisplayFunction->Identity]
 
 Show[GraphicsArray[{{p1,p2},{dp1,dp2}}],
 DisplayFunction->$DisplayFunction]

The same thing happens to the graph of the cosine function when the cosine is multiplied by a constant.

Clear[f,x,c,k]
 p1 = Plot[2Cos[x],{x,-Pi,2Pi},PlotLabel->"f[x]=2Cos[x]",PlotStyle->RGBColor[1,0,0],DisplayFunction->Identity];
 p2 = Plot[(1/2)Cos[x],{x,-Pi,2Pi},PlotLabel->"f[x]=(1/2)Cos[x]",PlotStyle->RGBColor[1,0,1],DisplayFunction->Identity];
 dp1 = Plot[-2Cos[x],{x,-Pi,2Pi},PlotLabel->"f[x]=-2Cos[x]",PlotStyle->RGBColor[0,1,0],DisplayFunction->Identity];
 dp2 = Plot[-4Cos[x],{x,-Pi,2Pi},PlotLabel->"f[x]=-4Cos[x]",PlotStyle->RGBColor[0,1,1],DisplayFunction->Identity];
 
 Show[GraphicsArray[{{p1,p2},{dp1,dp2}}],
 DisplayFunction->$DisplayFunction];

What do you think would happen to the graphs of the sine and cosine if you added a constant to the equations? Take a look.

Clear[f,x,c,k]
 p1 = Plot[2+2Sin[x],{x,-Pi,2Pi},PlotLabel->"f[x]=2+2Sin[x]",PlotStyle->RGBColor[1,0,0],DisplayFunction->Identity];
 p2 = Plot[-4-2Sin[x],{x,-Pi,2Pi},PlotLabel->"f[x]=-4-2Sin[x]",PlotStyle->RGBColor[1,0,1],DisplayFunction->Identity];
 dp1 = Plot[2-2Cos[x],{x,-Pi,2Pi},PlotLabel->"f[x]=2-2Cos[x]",PlotStyle->RGBColor[0,1,0],DisplayFunction->Identity];
 dp2 = Plot[-2+4Cos[x],{x,-Pi,2Pi},PlotLabel->"f[x]=-2+4Cos[x]",PlotStyle->RGBColor[0,1,1],DisplayFunction->Identity];
 
 Show[GraphicsArray[{{p1,p2},{dp1,dp2}}],
 DisplayFunction->$DisplayFunction];

We call  these shifts Vertical  Translations.

Check your understanding

In the equation f[x]=c + k sinx
1.  What does c indicate on the graph?

2.  How does k change the graph?

3.  How can you tell by looking at the graph of a sinusoid whether it is the graph of a sine or cosine function?

Test yourself

Try to write the equations of the following sinusoids.

1.

2.

3.

Try sketching the graphs of the following sinusoids.
4.  f[x]=3-4 Sin x
5.  f[x]=-4+2 Cos x

Lesson 36

As we have seen in lesson 22 we can tell the centerline and amplitude of a sinusoid by looking at the equation or the graph.  What do you think would happen to the graph if we added an angle to the x for instance f[x]= Sin (x-Pi/4)?  This angle is called a phase angle.

Clear[f,x,c,k]
 p1 = Plot[Sin[x-Pi/4],{x,-Pi,2Pi},PlotLabel->"f[x]=Sin[x-Pi/4]",PlotStyle->RGBColor[1,0,0],DisplayFunction->Identity];
 p2 = Plot[Sin[x+Pi/2],{x,-Pi,2Pi},PlotLabel->"f[x]=Sin[x+Pi/2]",PlotStyle->RGBColor[1,0,1],DisplayFunction->Identity];
 dp1 = Plot[Cos[x+Pi/3],{x,-Pi,2Pi},PlotLabel->"f[x]=Cos[x+Pi/3]",PlotStyle->RGBColor[0,1,0],DisplayFunction->Identity];
 dp2 = Plot[Cos[x-Pi/6],{x,-Pi,2Pi},PlotLabel->"f[x]=Cos[x-Pi/6]",PlotStyle->RGBColor[0,1,1],DisplayFunction->Identity];
 
 Show[GraphicsArray[{{p1,p2},{dp1,dp2}}],
 DisplayFunction->$DisplayFunction];

We note that the phase angle for a sine curve equals the angle nearest the origin where the curve crosses the centerline on the way up.  The phase angle for a cosine curve equals the angle nearest the origin where the curve reaches its highest point.  Arguments are written as x- the phase angle.  For example an argument (x+Pi/4)indicates a -45 degree phase shift.

Sinusoids are periodic functions because the f[x] values repeat each time the value of the argument changes 360 degrees or 2Pi radians.  A phase angle will not change the period of a function it will shift the whole curve left or right.  The period of a function is the number of degrees from one highest point to the next highest point.  When the variable X is multiplied by a nonzero number, the value of the argument is changed for every value of X and the period is changed.  When the variable is multiplied by 2 the value of the argument changes twice as fast as the change in the variable.

Clear[f,x,c,k]
 p1 = Plot[Sin[6x],{x,-Pi,2Pi},PlotLabel->"f[x]=Sin[6x]",PlotStyle->RGBColor[1,0,0],DisplayFunction->Identity];
 p2 = Plot[Sin[x],{x,-Pi,2Pi},PlotLabel->"f[x]=Sin[x]",PlotStyle->RGBColor[1,0,1],DisplayFunction->Identity];
 dp1 = Plot[Cos[2x],{x,-Pi,2Pi},PlotLabel->"f[x]=Cos[2x]",PlotStyle->RGBColor[0,1,0],DisplayFunction->Identity];
 dp2 = Plot[Cos[4x],{x,-Pi,2Pi},PlotLabel->"f[x]=Cos[4x]",PlotStyle->RGBColor[0,1,1],DisplayFunction->Identity];
 
 Show[GraphicsArray[{{p1,p2},{dp1,dp2}}],
 DisplayFunction->$DisplayFunction];

What kind of equation could we set up to find the coefficient or the period of the function?  When the coefficient is 2 the period is Pi or 180 degrees.  When the coefficient is 4 the period is Pi/2 or 90 degrees.  When the coefficient is 6 the period is Pi/3 or 60 degrees.  When the coefficient is 1 the period is 2Pi or 360 degrees.  Using these facts and that one revolution of theunit circle is 2Pi or 360 degrees.  Can you make an equation using 2,Pi or 180, and 2Pi or360.  Correct 2=360/180 or 2=2Pi/Pi.  This leads us to the fact that the coefficient equals 360 degrees divided by the period.

Try it yourself

Try putting some different values in the following equation and press enter and see what the graph looks like.

Plot[2+4Sin[2x],{x,-Pi,2Pi},
PlotStyle->RGBColor[Random[],Random[],Random[]]];