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Math Help - Rigorous Sine and Cosine

  1. #1
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    Rigorous Sine and Cosine

    The definition on a rigorous level for sine and cosine is:
    \sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-...
    \cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-...

    From here can you prove that,
    \forall k\in\mathbb{Z},\sin (\pi k)=0

    If \sin x=0 then, x=\pi k, \exists k\in\mathbb{Z}

    And \sin(x+y)=\sin x\cos y+\cos x\sin y.

    I do not know how to start this one.
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  2. #2
    TD!
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    I'd use the definitions which extend the arguments to the complex case.
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    Quote Originally Posted by TD!
    I'd use the definitions which extend the arguments to the complex case.
    You mean the power series for e^z for complex numbers. From here we can use the fact,
    e^{zi}=\cos z+i\sin z then you can easily denomstarte that,
    \sin(z_1+z_2)=\sin z_1\cos z_2+\cos z_1\sin z_2.

    But what about the other two facts, the necessary and suffienct conditions for the zero's of the sine function?

    I was thinking, show that there exists a non-zero number \pi on some interval which is a zero of the sine function by the use of the indetermediate value theorem. Then, demonstate that is \pi is a zero then so is \pi k but that does not prove the "only if" part.
    Last edited by ThePerfectHacker; April 5th 2006 at 01:19 PM.
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    TD!
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    You can use that identity of course, but I was referring to the definition:

    \sin z  \equiv  \frac{{e^{iz}  - e^{ - iz} }}{{2i}}

    The exponential function is periodic, with period 2\pi i. We can now use this to find the zeroes.

    <br />
\sin z = 0 \Leftrightarrow \frac{{e^{iz}  - e^{ - iz} }}{{2i}} = 0 \Leftrightarrow e^{iz}  - e^{ - iz}  = 0 \Leftrightarrow e^{iz}  = e^{ - iz} <br />

    Now we use the periodicy of e^z.

    <br />
e^{iz}  = e^{ - iz}  \Leftrightarrow iz =  - iz + 2k\pi i \Leftrightarrow 2zi = 2k\pi i \Leftrightarrow z = k\pi <br />
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    Quote Originally Posted by TD!
    You can use that identity of course, but I was referring to the definition:

    \sin z  \equiv  \frac{{e^{iz}  - e^{ - iz} }}{{2i}}

    The exponential function is periodic, with period 2\pi i. We can now use this to find the zeroes.

    <br />
\sin z = 0 \Leftrightarrow \frac{{e^{iz}  - e^{ - iz} }}{{2i}} = 0 \Leftrightarrow e^{iz}  - e^{ - iz}  = 0 \Leftrightarrow e^{iz}  = e^{ - iz} <br />

    Now we use the periodicy of e^z.

    <br />
e^{iz}  = e^{ - iz}  \Leftrightarrow iz =  - iz + 2k\pi i \Leftrightarrow 2zi = 2k\pi i \Leftrightarrow z = k\pi <br />
    How do you show the exponential function is periodic. (You better not tell me that because sine and cosine are periodic )
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  6. #6
    TD!
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    That is indeed easy to show by using sin & cos
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