Results 1 to 6 of 6

Math Help - Rigorous Sine and Cosine

  1. #1
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10

    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    TD!
    TD! is offline
    Senior Member
    Joined
    Jan 2006
    From
    Brussels, Belgium
    Posts
    405
    Thanks
    3
    I'd use the definitions which extend the arguments to the complex case.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    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 02:19 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    TD!
    TD! is offline
    Senior Member
    Joined
    Jan 2006
    From
    Brussels, Belgium
    Posts
    405
    Thanks
    3
    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 />
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    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 )
    Follow Math Help Forum on Facebook and Google+

  6. #6
    TD!
    TD! is offline
    Senior Member
    Joined
    Jan 2006
    From
    Brussels, Belgium
    Posts
    405
    Thanks
    3
    That is indeed easy to show by using sin & cos
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. cosine and sine
    Posted in the Trigonometry Forum
    Replies: 6
    Last Post: May 7th 2009, 07:14 AM
  2. Cosine/Sine
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: January 5th 2009, 06:10 PM
  3. Sine Law and Cosine Law
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: June 2nd 2008, 08:18 PM
  4. sine and cosine
    Posted in the Trigonometry Forum
    Replies: 9
    Last Post: March 27th 2008, 10:59 PM
  5. Cosine/Sine
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: October 25th 2006, 09:53 AM

Search Tags


/mathhelpforum @mathhelpforum