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Math Help - problem with sinus :)

  1. #1
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    problem with sinus :)

    Hey guys.
    cos(n*x) when n is natural and x=pi equals to (-1)^n as I wrote in the pic.
    What I'm looking for is the same thing in sinus, lets say again that n is natural, is there a x that can make sinus to behave something like cos? and by that I mean an expression that can be written in on line like (-1)^n (x can be anything).
    I hope that was understood.
    Thanks in advance.
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  2. #2
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    x=\frac{\pi}{2}
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  3. #3
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    Quote Originally Posted by asi123 View Post
    Hey guys.
    cos(n*x) when n is natural and x=pi equals to (-1)^n as I wrote in the pic.
    What I'm looking for is the same thing in sinus, lets say again that n is natural, is there a x that can make sinus to behave something like cos? and by that I mean an expression that can be written in on line like (-1)^n (x can be anything).
    I hope that was understood.
    Thanks in advance.
    there's no such x because if \sin x = -1, then \cos x =0 and hence \sin(2x)=2\sin x \cos x =0 \neq 1 =(-1)^2.
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    well, lets say x=pi/2, is there a way to write the result in one line? like cos(n*pi) = (-1)^n.
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by asi123 View Post
    well, lets say x=pi/2, is there a way to write the result in one line? like cos(n*pi) = (-1)^n.
    \begin{aligned}\cos(n\pi)&=\sin\left(\frac{\pi}{2}-n\pi\right)\\<br />
&=\sin\left(\pi\left(\frac{1}{2}-n\right)\right)\\<br />
&=\sin\left(\pi\cdot\frac{1-2n}{2}\right)\\<br />
&=-\sin\left(\pi\cdot\frac{2n-1}{2}\right)\\<br />
&=(-1)^n\end{aligned}

    Actually more aestheically pleasing we can use \cos(n\pi)=\sin\left(\frac{\pi}{2}+n\pi\right) to arrive alternatively at (-1)^n=\sin\left(\pi\cdot\frac{2n+1}{2}\right)
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  6. #6
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    Quote Originally Posted by Mathstud28 View Post
    \begin{aligned}\cos(n\pi)&=\sin\left(\frac{\pi}{2}-n\pi\right)\\<br />
&=\sin\left(\pi\left(\frac{1}{2}-n\right)\right)\\<br />
&=\sin\left(\pi\cdot\frac{1-2n}{2}\right)\\<br />
&=-\sin\left(\pi\cdot\frac{2n-1}{2}\right)\\<br />
&=(-1)^n\end{aligned}

    Actually more aestheically pleasing we can use \cos(n\pi)=\sin\left(\frac{\pi}{2}+n\pi\right) to arrive alternatively at (-1)^n=\sin\left(\pi\cdot\frac{2n+1}{2}\right)
    Hey Mathstud28.
    This is not what I meant (thanks a lot due ).
    I gave the cos just as an example.
    I have the function f(x) = sin(nx), and I need to find any x that will give me the ability to write the result in one line.
    Lets take again x = pi/2 f(x) = sin(n*p/2) = {0 when n=0, 1 when n=1, 0 when n=2, -1 when n=3 and so on...} is there a way to write this expression in one line that will be depend in n? (just like cos(nx) = (-1)^n)

    Again, thanks a lot.
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  7. #7
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    Hello,
    Quote Originally Posted by asi123 View Post
    Hey Mathstud28.
    This is not what I meant (thanks a lot due ).
    I gave the cos just as an example.
    I have the function f(x) = sin(nx), and I need to find any x that will give me the ability to write the result in one line.
    Lets take again x = pi/2 f(x) = sin(n*p/2) = {0 when n=0, 1 when n=1, 0 when n=2, -1 when n=3 and so on...} is there a way to write this expression in one line that will be depend in n? (just like cos(nx) = (-1)^n)

    Again, thanks a lot.
    No, but in two lines it's possible :
    \sin \left(\frac{n\pi}{2}\right)=\left\{\begin{array}{l  l} 0 \text{ if n is even} \\<br />
(-1)^{(n-1)/2} \text{ if n is odd} \end{array} \right.
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  8. #8
    MHF Contributor Mathstud28's Avatar
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    \sin\left(\frac{n\pi}{2}\right)=\frac{1}{2}(-1)^{\frac{n(n-1)}{2}}\left[(-1)^{n-1}+1\right]

    Hows that?
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  9. #9
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    Quote Originally Posted by Mathstud28 View Post
    \sin\left(\frac{n\pi}{2}\right)=\frac{1}{2}(-1)^{\frac{n(n+1)}{2}}\left[(-1)^n+1\right]

    Hows that?
    Thanks a lot.
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  10. #10
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by asi123 View Post
    Thanks a lot.
    Note! I made a typo...the thing you quoted gives the values of \sin\left(\frac{\pi(n+1)}{2}\right)...to get the values of \sin\left(\frac{n\pi}{2}\right) see my revised post.
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