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MathGroup Archive 2004

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Re: Re: How to solve a simple Trig cofunction?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg50580] Re: [mg50572] Re: How to solve a simple Trig cofunction?
  • From: Brian Feeny <bfeeny at mac.com>
  • Date: Sat, 11 Sep 2004 06:44:29 -0400 (EDT)
  • References: <45208C0E-0192-11D9-8737-000A95BB3776@mac.com> <DA976E8C-0193-11D9-8C24-000A95B4967A@akikoz.net> <560CD99A-0211-11D9-B9C5-000A95BB3776@mac.com> <chpan5$k3p$1@smc.vnet.net> <200409100806.EAA04857@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

I put all the ways people gave me to solve in a notebook, and this is 
by far the way I will
be doing these types of problems :)


On Sep 10, 2004, at 3:06 AM, David W. Cantrell wrote:

> I've only now looked at this thread. Having read all the replies, I'm
> rather surprised that no one has suggested the neat solution:
>
> In[1]:=
> FullSimplify[
>   Reduce[{Cos[(x + 4)Degree] == Sin[(3x + 2)Degree], 0 < x < 180}, x]]
> Out[1]=
> x == 21 || x == 46 || x == 111
>
> David Cantrell
>
>
> Andrzej Kozlowski <andrzej at akikoz.net> wrote:
>> 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
>



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