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anonymous

  • 4 years ago

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

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  1. anonymous
    • 4 years ago
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    \[\cos(x)=(e^x+e^{-x})/2\] and \[\sin(x)=(e^x-e^{-x})/2i\] more neatly

  2. EarthCitizen
    • 4 years ago
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    \[e ^{j \theta} =\cos( \theta)+jsin(\theta)\]

  3. EarthCitizen
    • 4 years ago
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    \[\cos(x)=(e ^{x}/2)+(e ^{-x}/2)\]

  4. anonymous
    • 4 years ago
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    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.

  5. y2o2
    • 4 years ago
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    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

  6. anonymous
    • 4 years ago
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    Bah! Typo- sorry for utterly wasting your time!

  7. EarthCitizen
    • 4 years ago
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    \[e ^{jx}\] not e^x

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