Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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.

Answer 5

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

Answer 6

Bah! Typo- sorry for utterly wasting your time!

Answer 7

\[e ^{jx}\] not e^x