anonymous
  • anonymous
Hey - need some help deriving trig. identities from Euler's formula.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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amistre64
  • amistre64
what was eulers formula?
anonymous
  • anonymous
e^(ix) = cosx + isinx
anonymous
  • anonymous
So would it then be true that e^(i2x)= (cosx + isinx)^2 as well as e^(i2x) = cos(2x) + isin(2x) ?

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anonymous
  • anonymous
Yes.
anonymous
  • anonymous
So e^(-i2x)=cos(2x)-isin(2x)=(cos(x)-isin(x))^2?
anonymous
  • anonymous
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.
anonymous
  • anonymous
cos(2x) = the real part of (cos(x)-isin(x)^2
anonymous
  • anonymous
Nevermind, I think I figured it out.
anonymous
  • anonymous
Err the real part of (cos(x)+isin(x))^2

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