cntkillme
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| Joined: 07 Apr 2008 |
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| 03 Nov 2013 01:19 PM |
I can't seem to derive sinh from Euler's formula, I can some reason get cosh though.
e^ix + e^-ix = cosx + isinx + cosx - isinx
cosh: 2cosx = e^ix + e^-ix cosx = (e^ix + e^-ix)/2 coshx = (e^x + e^-x)/2
sinh: ...
I some reason can't think of a way to get sinh |
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cntkillme
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| Joined: 07 Apr 2008 |
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| 03 Nov 2013 01:21 PM |
I know sinh is: (e^x - e^-x)/2 But I want to be able to "get it" |
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cntkillme
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| Joined: 07 Apr 2008 |
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| 03 Nov 2013 01:22 PM |
| I couldn't find the actual way over google, only the answer which I alreayd knew... |
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cntkillme
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| Joined: 07 Apr 2008 |
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| 03 Nov 2013 01:23 PM |
| OT is dumb, I just started algebra :P |
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| 03 Nov 2013 01:23 PM |
4
furga000's alt | banned for 2 days |
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cntkillme
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| Joined: 07 Apr 2008 |
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| 03 Nov 2013 01:23 PM |
| I'm pretty sure someone can help, I don't want to go into another forum where it is probably considered spam. |
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| 03 Nov 2013 01:26 PM |
| if the math is "simple" then why do you need help? |
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cntkillme
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| Joined: 07 Apr 2008 |
| Total Posts: 44956 |
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| 03 Nov 2013 01:27 PM |
| Because I feel like I'm not seeing a simple step, if cosh was that easy, why isn't sinh... |
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| 03 Nov 2013 01:27 PM |
| No one can help ok? Ask a family member instead nine year olds ._. |
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kitty234
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| Joined: 03 Feb 2009 |
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cntkillme
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| Joined: 07 Apr 2008 |
| Total Posts: 44956 |
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| 03 Nov 2013 01:28 PM |
Just kidding, I got it. It was simple.. |
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cntkillme
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| 03 Nov 2013 01:31 PM |
if e^ix = cosx + isinx and e^-ix = cosx - isinx then -(e^-ix) = -cosx + isinx therefor e^ix - e^-ix = cosx + isinx - cosx + isinx leads to e^ix - e^-ix = 2isinx which finally gives sinh = (e^x - e^-x)/2 |
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| 03 Nov 2013 01:32 PM |
oops, i posted late
damn floodcheck |
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cntkillme
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| Joined: 07 Apr 2008 |
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| 03 Nov 2013 01:33 PM |
I already knew that, I just wanted the "proof."
If you know how to read you would of seen that I had known that |
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| 03 Nov 2013 01:33 PM |
"If you know how to read you would of seen that I had known that"
*would have seen
Oh, the irony. |
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