Re: How to solve a simple Trig cofunction?

*To*: mathgroup at smc.vnet.net*Subject*: [mg50552] Re: [mg50500] How to solve a simple Trig cofunction?*From*: Andrzej Kozlowski <andrzej at akikoz.net>*Date*: Thu, 9 Sep 2004 05:19:46 -0400 (EDT)*References*: <45208C0E-0192-11D9-8737-000A95BB3776@mac.com> <DA976E8C-0193-11D9-8C24-000A95B4967A@akikoz.net> <560CD99A-0211-11D9-B9C5-000A95BB3776@mac.com>*Sender*: owner-wri-mathgroup at wolfram.com

You are quite right. I should have been more careful. What I should have done was: N[(x /. {ToRules[Reduce[{Cos[x + 4*Pi/180] == Sin[3*x + 2*Pi/180], 0 < x < Pi}, x]]})*(180/Pi)] But this is now taking very much longer, I have been waiting for quite a long time and Mathematica has not managed it yet. As I have not enough patience (and it may never be able to do it exactly) I have decided to switch to a numerical approach: << NumericalMath`IntervalRoots` (Mean /@ (List @@ (IntervalBisection[Cos[x + 4*Pi/ 180] - Sin[3*x + 2*Pi/180], x, Interval[{0., N[Pi]}], .01, MaxRecursion -> 10])))*180/Pi {20.918, 46.0547, 110.918} This seems to agree with what you expected. Andrzej On 9 Sep 2004, at 12:35, Brian Feeny wrote: > > Andrzej, > > I think the differences I am seeing in the solutions (21 vs. 26) is > because doing something like: > > N[Cos[x + 4] /.x -> 21*Degree] > > is not the same thing as: > > N[Cos[x Degree + 4 Degree] /. x -> 21] > > > My originally problem: > > Cos[x+4] = Sin[3x+2] > > The problem I think, is that with the way your examples and some > others are being done, the part being added, > such as "4", is not in Degree's. Probably because I didn't explain my > problem well. I do appreciate you and others > taking the time to show me excellent ways to solve these problems in > Mathematica. > > Brian > > > > On Sep 8, 2004, at 7:37 AM, Andrzej Kozlowski wrote: > >> *This message was transferred with a trial version of CommuniGate(tm) >> Pro* >> Well, Mathematica does not agree with you: >> >> >> N[Cos[x + 4] /. x -> 21*Degree] >> >> >> -0.3390151318887568 >> >> >> N[Sin[3*x + 2] /. x -> 21*Degree] >> >> >> 0.042022846832621065 >> >> which is way off, while >> >> In[58]:= >> N[Cos[x + 4] /. x -> 26.56*Degree] >> >> >> -0.24626970814204074 >> >> >> N[Sin[3*x + 2] /. x -> 26.56*Degree] >> >> >> -0.24651797391493188 >> >> which is much closer. Perhaps you had a different equation in mind? >> >> Andrzej >> >> >> >> On 8 Sep 2004, at 21:26, Brian Feeny wrote: >> >>> >>> On Sep 8, 2004, at 7:06 AM, Andrzej Kozlowski wrote: >>> >>>> I have no idea what a "Trig cofunction" is ;-) >>>> >>>> Mathematica can solve this exactly (so there is no need to use >>>> numerical methods, and in any case NSolve would be the wrong solver >>>> to use since it does not deal with trig equations) >>>> >>>> This gives the exact answer in radians: >>>> >>>> >>> >>> >>> Thanks for replying. The correct numerical answer is 21degrees >>> however, I wonder why the numerical >>> answers below are 5 degrees too much? >>> >>> You have given me alot to think about in ways to use Mathematica and >>> I will read in the manual further to try >>> to make sense of it, thanks. >>> >>> Brian >>> >>> >>> >>> >>> >>>> Reduce[{Cos[x + 4] == Sin[3*x + 2], 0 < x < Pi}, x] >>>> >>>> >>>> x == 1 - 2*ArcTan[1 - Sqrt[2]] || >>>> x == (1/2)*(-3 + 2*Pi - 2*ArcTan[1 - Sqrt[2]]) || >>>> x == (1/2)*(-3 + 2*Pi - 2*ArcTan[1 + Sqrt[2]]) >>>> >>>> If you prefer an answer in degrees you get do this: >>>> >>>> >>>> (x /. {ToRules[Reduce[{Cos[x + 4] == Sin[3*x + 2], >>>> 0 < x < Pi}, x]]})*(180/Pi) >>>> >>>> {(180*(1 - 2*ArcTan[1 - Sqrt[2]]))/Pi, >>>> (90*(-3 + 2*Pi - 2*ArcTan[1 - Sqrt[2]]))/Pi, >>>> (90*(-3 + 2*Pi - 2*ArcTan[1 + Sqrt[2]]))/Pi} >>>> >>>> And if you prefer a numerical answer you can just apply N: >>>> >>>> >>>> N[%] >>>> >>>> >>>> {102.296,116.556,26.5563} >>>> >>>> >>>> Andrzej Kozlowski >>>> >>>> Chiba, Japan >>>> http://www.akikoz.net/~andrzej/ >>>> http://www.mimuw.edu.pl/~akoz/ >>>> >>