Staredit Network > Forums > Null > Topic: Simple math question
Simple math question
Apr 17 2008, 3:52 am
By: MillenniumArmy  

Apr 17 2008, 3:52 am MillenniumArmy Post #1



I think this is something ridiculously simple but it just slipped out of my mind.

Say you are given an equation:
0.0036 = p(1-cos(0.6/p))

how would you solve for p (note: cos is measured in radians, not degrees)? The answer given is p = 50, but can someone show me the steps?

Post has been edited 2 time(s), last time on Apr 17 2008, 6:09 am by MillenniumArmy.



None.

Apr 17 2008, 5:45 am Rantent Post #2



The book is wrong, by a factor of ten.



None.

Apr 17 2008, 5:54 am A_of-s_t Post #3

aka idmontie

This is all I could get:

0.036 = p(1-cos(0.6/p))
0 = p(1-cos(0.6/p)) - 0.036
0 = p - pcos(0.6/p) - 0.036

cos(nx) = 2 cos([n-1]x)cos(x) - cos([n-2]x)

0.6/p = 3/5/p = 3/5p = 6/10p = 1.2/2p = 3/5 * 1/p

3/5/p = 3/p/5



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Apr 17 2008, 6:00 am MillenniumArmy Post #4



Quote from Rantent
The book is wrong, by a factor of ten.
LOL oops, i meant 0.0036, not 0.036 mb



None.

Apr 17 2008, 6:18 am DT_Battlekruser Post #5



I hate algebra..



None.

Apr 17 2008, 6:20 am cheeze Post #6



Expand to Taylor series and approximate. :D



None.

Apr 17 2008, 2:34 pm DT_Battlekruser Post #7



Win :P



None.

Apr 17 2008, 3:26 pm Clokr_ Post #8



Quote from MillenniumArmy
I think this is something ridiculously simple but it just slipped out of my mind.

Say you are given an equation:
0.0036 = p(1-cos(0.6/p))

how would you solve for p (note: cos is measured in radians, not degrees)? The answer given is p = 50, but can someone show me the steps?

You cannot really solve that, the solution is probably irrational. But you can approximate it by several methods, including Taylor series.



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