# Thread: Rigorous Sine and Cosine

1. ## 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.

2. I'd use the definitions which extend the arguments to the complex case.

3. 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.

4. 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.

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

5. 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 )

6. That is indeed easy to show by using sin & cos