Question

Hey - need some help deriving trig. identities from Euler's formula.

Answer 1

what was eulers formula?

Answer 2

e^(ix) = cosx + isinx

Answer 3

So would it then be true that e^(i2x)= (cosx + isinx)^2 as well as e^(i2x) = cos(2x) + isin(2x) ?

Answer 4

Yes.

Answer 5

So e^(-i2x)=cos(2x)-isin(2x)=(cos(x)-isin(x))^2?

Answer 6

The problem I have is that it seems that when I add e^(i2x)+e^(-i2x), I get 2cos2x=2cos^2(x) ! Obviously I'm going wrong somewhere.

Answer 7

cos(2x) = the real part of (cos(x)-isin(x)^2

Answer 8

Nevermind, I think I figured it out.

Answer 9

Err the real part of (cos(x)+isin(x))^2