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Thread: find boundaries of sine and cos without graphing

  1. #1
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    find boundaries of sine and cos without graphing

    find boundaries of sine and cos without graphing-screenshot-2018-02-23-07.10.28.png

    Hi!

    Is there a way to find the boundaries without graphing?

    I think 3sinx=2sin^2 may work, but i'm not sure how to solve for zero. Could some show me how?

    Here is what i have:

    3sinx=2sin^2
    3sin - 2sin^2 - 0
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  2. #2
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    Re: find boundaries of sine and cos without graphing

    Factor: $\sin x (3-2\sin x)=0$

    Now you have either $\sin x=0$ or $3-2\sin x=0$. The latter has no solutions over the reals. So the intersections occur at $x=0,\pi,2\pi $

    Next plug in $\dfrac{\pi}{2} ,\dfrac{3\pi}{2} $ to figure out which function is bigger in each interval. This gives you

    $\displaystyle A=\int_0^\pi(3\sin x-2\sin^2 x)dx+\int_\pi^{2\pi}(2\sin^2 x-3\sin x)dx $
    Last edited by SlipEternal; Feb 23rd 2018 at 05:51 AM.
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    Re: find boundaries of sine and cos without graphing

    Thanks for the response, SlipEternal!!

    In the Area, why did you write 3sinx-2sin^2x twice? And why are they reversed in the second part? Also why are the boundaries different in each of the cases?
    Quote Originally Posted by SlipEternal View Post
    Factor: $\sin x (3-2\sin x)=0$

    Now you have either $\sin x=0$ or $3-2\sin x=0$. The latter has no solutions over the reals. So the intersections occur at $x=0,\pi,2\pi $

    Next plug in $\dfrac{\pi}{2} ,\dfrac{3\pi}{2} $ to figure out which function is bigger in each interval. This gives you

    $\displaystyle A=\int_0^\pi(3\sin x-2\sin^2 x)dx+\int_\pi^{2\pi}(2\sin^2 x-3\sin x)dx $
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  4. #4
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    Re: find boundaries of sine and cos without graphing

    Quote Originally Posted by RBB22 View Post
    Thanks for the response, SlipEternal!!

    In the Area, why did you write 3sinx-2sin^2x twice? And why are they reversed in the second part? Also why are the boundaries different in each of the cases?
    As slip eternal said, the two curves, 3sin(x) and $2 sin^2(x)$ cross at x= 0, $x= \pi$, and $x= 2\pi$. He then said "Next, plug in $\frac{\pi}{2}$ and $\frac{3\pi}{2}$ to figure out which function is bigger in each interval". When $x= \frac{\pi}{2}$, which is between 0 and $\pi$, $3sin(x)= 3sin(\pi/2)= 3$ while $2 sin^2(x)= 2 sin^2(\pi/2)= 2$. Since 3sin(x) is greater than $2sin^2(x)$ at $x= \frac{\pi}{2}$ (and both functions are continuous so the only place they can change order is where they are equal) $3sin(x)$ is larger than $2sin^2(x)$ for all x between 0 and $\pi$. So the area between them is $\int_0^{\pi} 3sin(x)- 2sin^2(x) dx$.

    At $x= \frac{3\pi}{2}$, $3 sin(x)= 3sin(3\pi/2)= -3$ while $2sin^2(x)= 2 sin^2(3\pi/2)= 2$ so now $2 sin^2(x)$ is larger. The area between them is $\int_{\pi}^{2\pi} 2sin^2(x)- 3 sin(x) dx$. The total area is the sum of those two.
    Last edited by HallsofIvy; Feb 23rd 2018 at 04:44 PM.
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  5. #5
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    Re: find boundaries of sine and cos without graphing

    @RBB22 ,
    One question: what does "without graphing" have to do with this question?
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  6. #6
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    Re: find boundaries of sine and cos without graphing

    RBB22 probably had used some graphing utility to graph the two functions and identified where they crossed from that. Now he was asking how to get the "boundary" if you don't have such a utility.
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