IMAP^2 Fairfield , IA
Trig Function Explorations
June 2008 Notes by WS

EXAMPLE 1

The graphic below shows the two curves
 y1 = Sin[x] and 
 y2= (1/3) Sin [ 3 x ] 
 x ∈[-2π, 2π],  y ∈[-3,3 ]. 
 
    QUESTION: what does the curve y1 + y2 look like?  See next cell for an answer.

Plot[{Sin[x],(1/3)Sin[3*x]},{x,-2Pi,2Pi},PlotRange->{-3,3},PlotStyle->{Thickness[.0025]},ImageSize->600]

y1 = Sin[x] and 
 y2 = (1/3) Sin [ 3 x ] 
 x ∈[-2π, 2π],  y ∈[-3,3 ]. 
 ANSWER: y1 + y2 looks like this:

Plot[{Sin[x]+(1/3)Sin[3*x]},{x,-2Pi,2Pi},PlotRange->{-3,3},PlotStyle->{Thickness[.0025]},ImageSize->600]

Plot[{Sin[x],(1/3)Sin[3*x],Sin[x]+(1/3)Sin[3*x]},{x,-2Pi,2Pi},PlotRange->{-3,3},PlotStyle->{Thickness[.0025]},ImageSize->600]

The graphic below shows the curves
 (y1 = Sin[x] ) + (y2 = (1/3) Sin [ 3 x ] )
 y3 = (1/5) Sin [ 5 x ]  
 x ∈[-2π, 2π],  y ∈[-3,3 ]. 
 
    QUESTION: what does the curve y1 + y2 + y3 look like?

Plot[{Sin[x]+(1/3)Sin[3*x],(1/5)Sin[5*x]},{x,-2Pi,2Pi},PlotRange->{-3,3},PlotStyle->{Thickness[.0025]},ImageSize->600]

Plot[{Sin[x],(1/3)Sin[3*x],(1/5)Sin[5*x]},{x,-2Pi,2Pi},PlotRange->{-3,3},PlotStyle->{Thickness[.0025]},ImageSize->600]

EXAMPLE 2

The graphic below shows the two curves
 y1 = Cos[x] and 
 y2= (1/3^2) Cos [ 3 x ] 
 x ∈[-2π, 2π],  y ∈[-3,3 ]. 
 
    QUESTION: what does the curve y1 + y2 look like?  See next cell for an answer.

Plot[{Cos[x],(1/3^2)Cos[3*x]},{x,-2Pi,2Pi},PlotRange->{-1.5,1.5},PlotStyle->{Thickness[.0025]},ImageSize->600]

y1 = Cos[x] and 
 y2= (1/3^2) Cos [ 3 x ]
 x ∈[-2π, 2π],  y ∈[-3,3 ]. 
 ANSWER: y1 + y2 looks like this:

Plot[{Cos[x]+(1/3^2)Cos[3*x]},{x,-2Pi,2Pi},PlotRange->{-1.5,1.5},PlotStyle->{Thickness[.0025]},ImageSize->600]

Plot[{Cos[x],(1/3^2)Cos[3*x],Cos[x]+(1/3^2)Cos[3*x]},{x,-2Pi,2Pi},PlotRange->{-1.5,1.5},PlotStyle->{Thickness[.0025]},ImageSize->600]

The graphic below shows the curves
 (y1 = Cos[x] )+ (y2= (1/3^2) Cos [ 3 x ])
 and
 y3 = (1/5^2) Sin [ 5 x ]  
 x ∈[-2π, 2π],  y ∈[-3,3 ]. 
 
    QUESTION: what does the curve y1 + y2 + y3 look like?

Plot[{Cos[x]+(1/3^2)Cos[3*x],(1/5^2)Cos[5*x]},{x,-2Pi,2Pi},PlotRange->{-1.5,1.5},PlotStyle->{Thickness[.0025]},ImageSize->600]

Plot[{Cos[x]+(1/3^2)Cos[3*x]+(1/5^2)Cos[5*x]},{x,-2Pi,2Pi},PlotRange->{-1.5,1.5},PlotStyle->{Thickness[.0025]},ImageSize->600]