Question

Prove: cos(theta) = 1/2[e^i(theta) + e^(-i(theta))

I know we use Eulers relation: e^i(theta) = cos(theta) + jsin(theta)

so far this is what I get:
cos(theta) = e^i(theta) - isin(theta)

Answer 1

so the Q is \[\cos \theta =\frac{ e^{i \theta}+e^{-i \theta}}{ 2 }\]

Answer 2

yes

Answer 3

i dont know how to write it like that...

Answer 4

and you have to know that cos is an even function and sin is an odd function

Answer 5

which means \[\cos(-\theta) = \cos \theta, \sin(-\theta)=-\sin(\theta)\]

Answer 6

so \[e ^{-i \theta}=\cos(\theta)- isin(\theta)\]

Answer 7

while\[e^{i \theta}=\cos(\theta)+isin(\theta)\]

Answer 8

oh I think I know where your going with this.

Answer 9

if you add this two together you will get 2 cos theta, and a half of this is the LHS

Answer 10

yes ok thats what I thought Im just doing the math now. Thanks!

Answer 11

BTW, can you look at my problem? I know the method but get stack with the calculation

Answer 12

is it posted? sure