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Areas Bounded by Trigonometric Functions.

  1. Jul 31, 2013 #1
    I will do my best to describe the problem I am working on. The problem is not from a textbook or anything but something I am working on independently to strengthen my first year calculus knowledge.

    What I did is I took sin(x) and -sin(x) and graphed them together. Sin(x) and -sin(x) intersect at 0 and pi/2. These intersections will create a shape that resembles a ellipse and that is what i'm working with. What I did is I calculated the area of that shape. To do this I just found the area under the curve for sin(x) between 0 and pi/2 and added the absolute value of the area under the curve for -sin(x) from 0 to pi/2. I wanted to see how the area would increase as I increased/decreased the amplitude and period of the functions. I did this and found a very simple equation that is the same for increasing/decreasing amplitude and period. If I increase/decrease the period or amplitude of both functions equally, such as making the amplitude of both functions 3 or increasing period to 2pi I found an equation for the nth area, where n is any integer. An=4n. So if the amplitude of the function is 5, that is my n. I then did tangent and the same equation arises. So my question is, would this be expected? I was also surprised that the numbers would be so simple. Just a little collaboration on the problem would be appreciated. I'm looking for relations and things like that involving areas of trigonometric functions and put these relations into their most general form. Thanks.
  2. jcsd
  3. Jul 31, 2013 #2
    Let's start here.

    Let ##x\in[0,2\pi)##. There are two values of ##x## for which ##\sin(x)=-\sin(x)##. One of them is ##x=0##. What's the other value?

    Hint: ##\sin\frac{\pi}{2}=1\neq -1=-\sin\frac{\pi}{2}##. :wink:
  4. Jul 31, 2013 #3


    Staff: Mentor

    Maybe kevinnn meant ##\pi## instead of what he wrote: ##\pi/2##. The graphs of y = sin(x) and y = -sin(x) intersect at integer multiples of ##\pi##.
  5. Jul 31, 2013 #4
    I'm sorry that was a bad typo. I meant to say pi. I did all the calculations using pi because I wanted the shape created by the two graphs to be closed.
  6. Aug 2, 2013 #5
    That is wonderful that you are experimenting and trying to discover things. That is an excellent way to gain a deeper understanding of calculus.

    One thing that may be of benefit is the following property of integrals:
    [itex]\int c f(x) dx [/itex] = [itex]c \int f(x) dx [/itex]

    - Junaid Mansuri
    Last edited: Aug 2, 2013
  7. Aug 2, 2013 #6
    Thanks for the reply Junaid. I used this property a lot during simplification processes. I am now trying to figure out the results of the problem. The final equation is so simple and it is interesting to me that the tangent function would have the same equation for finding areas as the sine function. The mathematician I was talking to earlier was interested in the problem and wanted to look deeper into it but we got kicked out of the building because it was closing time. I figured talking on here would be a good thing because there is not just one mathematician I can talk to but many.
  8. Aug 2, 2013 #7
    hehe, very cool. I'm curious what interval you are using to find the area bounded by the tangent function since it doesn't define a closed region in and of itself.
  9. Aug 3, 2013 #8
    For the tangent function I'm using the interval [0,pi/4]. It just looked like the most logical interval because like you said, I can't do [0,pi/2) or a similar interval. It just appeared to me that working with intervals that result in nice answers, such as tan(pi/4)=1 would result in nice equations and numbers to work with. Would you have done something different?
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