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#1 2007-02-24 07:47:26

John E. Franklin
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Registered: 2005-08-29
Posts: 3,588

Trig Deriv - is this right?

Are these correct derivatives?
Just add pi over 2 in radians?

Seems to be right to me, even though we usually don't see the derivatives with a phase shift, but usually flip between sine and plus or minus cosine.

Last edited by John E. Franklin (2007-02-24 07:48:27)


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#2 2007-02-24 12:28:18

John E. Franklin
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Registered: 2005-08-29
Posts: 3,588

Re: Trig Deriv - is this right?

Yeah!!  I was thinking about the unit circle, how I was taught trig in high school,
and trying to visualize stuff.  It's pretty exciting!!


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#3 2007-02-24 14:39:45

George,Y
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Registered: 2006-03-12
Posts: 1,379

Re: Trig Deriv - is this right?

Also
Cos[sup](n)[/sup](x)=Cos(x+nπ/2)
Sin[sup](n)[/sup](x)=Sin(x+nπ/2)
where (n) stands for the nth derivative.


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#4 2007-02-25 03:48:27

John E. Franklin
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Registered: 2005-08-29
Posts: 3,588

Re: Trig Deriv - is this right?

That's neat! George.  So the 4th and 8th derivatives just come back to where you started.


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#5 2007-02-25 03:55:58

luca-deltodesco
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Registered: 2006-05-05
Posts: 1,470

Re: Trig Deriv - is this right?

yes, you can see the same thing when just dealing with the derivitaves normally.


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#6 2007-02-25 04:03:39

John E. Franklin
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Registered: 2005-08-29
Posts: 3,588

Re: Trig Deriv - is this right?

Interesting lucadd.  It's funny how I memorize them.
It's hard to explain, but when they all stay cosine, then you go
counter-clockwise a quarter turn.  But if you change from
sin to cos to - sin to - cos to sin and around, you can remember
vectors pointing up, right, down, left, up, which goes clockwise
instead.


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#7 2007-02-25 15:21:04

George,Y
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Registered: 2006-03-12
Posts: 1,379

Re: Trig Deriv - is this right?

Yes I once tried to solve the reason behind it but failed. So I guess it's probably beautiful coincidence. smile


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