Proving sin(x) & cos(x) with Euler's Formula

[poobar]

Homework Statement

Show that cos(x) = (e^ix + e^-ix)/2. Show that sin(x) = (e^ix - e^-ix)/2i.

Homework Equations

sin(x) = $$\sum (x^{2n+1}(-1)^{n})/(2n+1)!$$ for n = zero to infinity

cos(x) = $$\sum (x^{2n}(-1)^{n})/(2n)!$$ for n = zero to infinity

The Attempt at a Solution

I found the first five or six terms for (e^ix + e^-ix)/2 and (e^ix - e^-ix)/2i
I just don't know how I am can prove they equal the trigonometric functions because there are terms that have "i" in them and there are not any i's in the series for sine or cosine.

 

[Mark44]

Homework Statement

Show that cos(x) = (e^ix + e^-ix)/2. Show that sin(x) = (e^ix - e^-ix)/2i.

Homework Equations

sin(x) = $$\sum (x^{2n+1}(-1)^{n})/(2n+1)!$$ for n = zero to infinity

cos(x) = $$\sum (x^{2n}(-1)^{n})/(2n)!$$ for n = zero to infinity

The Attempt at a Solution

I found the first five or six terms for (e^ix + e^-ix)/2 and (e^ix - e^-ix)/2i
I just don't know how I am can prove they equal the trigonometric functions because there are terms that have "i" in them and there are not any i's in the series for sine or cosine.

Do you have to do this using series? If not, you can use the facts that e^ix = cos(x) + isin(x), and e^-ix = cos(x) - isin(x), and some algebra.

 

[poobar]

ah, that makes this problem much simpler. I was on the wrong track because I wrongly assumed I had to use series. Thank you so much!