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Graphing Sin and Cosine Functions

Date: 5/24/96 at 4:22:51
From: Oniga M si Basu C
Subject:Graphing Sin and Cosine Functions 


I am a pre-college student (18 years) studying Mathematical Analysis. 
I will have my final exams at the end of June and I really need some 
help in finding the graphs of the following functions:

f:D->R,where D is the maximum domain of definition of each function

a) f(x) = pi*x + sin(pi*x), where pi=3.141592654...
b) f(x) = (sinx-cosx)/cos(2x)
c) f(x) = [tg(3x)]/(tgx*tgx*tgx) OR f(x) = [tan(3x)]/(tanx*tanx*tanx)
d) f(x) = (sinx*sinx) - 2(sinx*sinx*sinx*sinx)
Also, it will be of real help if you can find all the characteristic 
elements of the functions. 

Thank you very much, indeed!

Date: 5/24/96 at 11:3:54
From: Doctor Ethan
Subject: Graphing Sin and Cosine Functions 

Okay, let's get to it:


The first thing to do is determine the domain. Both x and sin(x)  have 
the same domain and it is all of R so our domain is all of R.

Now to go about graphing. To get some idea we can plug in a few points

f(0) = 0 + sin(0) = 0
f(pi) = pi^2 + sin(pi) = pi^2
f(2pi) = 2pi^2 + sin(2pi) = 2pi^2

So what we have here is a line of slope pi except that it is moved up 
or down by a value of sin(x). To graph, I recommend that you lightly 
draw a line with a slope of pi, marking the points at pi intervals and 
then draw a tilted sin curve along the line.



Here the big domain issue is that the denominator can't be zero. So we 
need to check when does cos(2x) = 0. That is when 2x = (2n+1)pi/2
or x = (2n+1)pi/4. So the domain is all of R except for the points of 
the form (2n+1)pi/4.

Since all of the parts of this equation have a period of pi or less we 
can just examine the section from 0 to pi. I recommend that you just 
plug in a bunch of points to see what it looks like. This same 
strategy will work for c) and d).

Why don't you try them out and write back if you need more help. 

-Doctor Ethan,  The Math Forum
Associated Topics:
High School Equations, Graphs, Translations
High School Functions

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