## anonymous 4 years ago Given Euler's formula, prove that cos(x)=(e(^x)+e(^-x))/2 and sin(x)=(e(^x)-e(^-x))/2i

1. anonymous

$\cos(x)=(e^x+e^{-x})/2$ and $\sin(x)=(e^x-e^{-x})/2i$ more neatly

2. EarthCitizen

$e ^{j \theta} =\cos( \theta)+jsin(\theta)$

3. EarthCitizen

$\cos(x)=(e ^{x}/2)+(e ^{-x}/2)$

4. anonymous

Ah, yes! It's easier when you express e^-x, e^x and cos/sin(x) as an infinite series. I think I get it now.

5. y2o2

I think this can't be true bec. $iSin(x) = {(e^x - e^{-x} ) \over 2}$ $\cos(x) + isin(x) = {{e^x + e^{-x}}\over 2} + {{e^x - e^{-x}}\over 2} = {2e^x \over 2} = e^x$ which is wrong bec cos(x) + isin(x) = e^(ix) Not e^x

6. anonymous

Bah! Typo- sorry for utterly wasting your time!

7. EarthCitizen

$e ^{jx}$ not e^x