ACOS logarithmic form
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04-28-2016, 03:07 PM
(This post was last modified: 04-28-2016 03:10 PM by Claudio L..)
Post: #1
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ACOS logarithmic form
Here's my dilemma:
I looked at Wikipedia here for the logarithmic forms of ACOS. The second row is the one that interests me. ACOS(Z)=-i*ln(Z+sqrt(Z^2-1)) Now let's make Z=2 (real, but outside range so result will be complex). 2^2-1 = 3 ln(2+sqrt(3))=1.3169... (all real numbers so far) Now -i*1.3169... = (0 -1.3169) Now go to your 50g and do 2 ACOS, you'll get (0 1.3169...) (positive imaginary part) Wolfram Alpha agrees with the 50g, so that imaginary part has to be positive. Then a quick check: ACOS(Z)=pi/2-ASIN(Z) Doing that on the 50g we also get positive imaginary part. Is the formula wrong in wikipedia? Perhaps it should be i*ln(...) instead of -i ? I just need some help proving it, so it's not just me against the world. EDIT: BTW, the ASIN() formula in Wikipedia works in agreement with the 50g. |
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