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anonymous

  • 5 years ago

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

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  1. amistre64
    • 5 years ago
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    what was eulers formula?

  2. anonymous
    • 5 years ago
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    e^(ix) = cosx + isinx

  3. anonymous
    • 5 years ago
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    So would it then be true that e^(i2x)= (cosx + isinx)^2 as well as e^(i2x) = cos(2x) + isin(2x) ?

  4. anonymous
    • 5 years ago
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    Yes.

  5. anonymous
    • 5 years ago
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    So e^(-i2x)=cos(2x)-isin(2x)=(cos(x)-isin(x))^2?

  6. anonymous
    • 5 years ago
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    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.

  7. anonymous
    • 5 years ago
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    cos(2x) = the real part of (cos(x)-isin(x)^2

  8. anonymous
    • 5 years ago
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    Nevermind, I think I figured it out.

  9. anonymous
    • 5 years ago
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    Err the real part of (cos(x)+isin(x))^2

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