# Rigorous Sine and Cosine

• Apr 4th 2006, 07:23 PM
ThePerfectHacker
Rigorous Sine and Cosine
The definition on a rigorous level for sine and cosine is:
$\displaystyle \sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-...$
$\displaystyle \cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-...$

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

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

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

I do not know how to start this one.
• Apr 5th 2006, 09:46 AM
TD!
I'd use the definitions which extend the arguments to the complex case.
• Apr 5th 2006, 01:16 PM
ThePerfectHacker
Quote:

Originally Posted by TD!
I'd use the definitions which extend the arguments to the complex case.

You mean the power series for $\displaystyle e^z$ for complex numbers. From here we can use the fact,
$\displaystyle e^{zi}=\cos z+i\sin z$ then you can easily denomstarte that,
$\displaystyle \sin(z_1+z_2)=\sin z_1\cos z_2+\cos z_1\sin z_2$.

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 $\displaystyle \pi$ on some interval which is a zero of the sine function by the use of the indetermediate value theorem. Then, demonstate that is $\displaystyle \pi$ is a zero then so is $\displaystyle \pi k$ but that does not prove the "only if" part.
• Apr 5th 2006, 01:56 PM
TD!
You can use that identity of course, but I was referring to the definition:

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

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

$\displaystyle \sin z = 0 \Leftrightarrow \frac{{e^{iz} - e^{ - iz} }}{{2i}} = 0 \Leftrightarrow e^{iz} - e^{ - iz} = 0 \Leftrightarrow e^{iz} = e^{ - iz}$

Now we use the periodicy of e^z.

$\displaystyle e^{iz} = e^{ - iz} \Leftrightarrow iz = - iz + 2k\pi i \Leftrightarrow 2zi = 2k\pi i \Leftrightarrow z = k\pi$
• Apr 5th 2006, 02:29 PM
ThePerfectHacker
Quote:

Originally Posted by TD!
You can use that identity of course, but I was referring to the definition:

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

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

$\displaystyle \sin z = 0 \Leftrightarrow \frac{{e^{iz} - e^{ - iz} }}{{2i}} = 0 \Leftrightarrow e^{iz} - e^{ - iz} = 0 \Leftrightarrow e^{iz} = e^{ - iz}$

Now we use the periodicy of e^z.

$\displaystyle e^{iz} = e^{ - iz} \Leftrightarrow iz = - iz + 2k\pi i \Leftrightarrow 2zi = 2k\pi i \Leftrightarrow z = k\pi$

How do you show the exponential function is periodic. (You better not tell me that because sine and cosine are periodic :D )
• Apr 6th 2006, 08:14 AM
TD!
That is indeed easy to show by using sin & cos :o :D