Re: How to solve a simple Trig cofunction?

• To: mathgroup at smc.vnet.net
• Subject: [mg50572] Re: How to solve a simple Trig cofunction?
• From: "David W. Cantrell" <DWCantrell at sigmaxi.org>
• Date: Fri, 10 Sep 2004 04:06:39 -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>
• Sender: owner-wri-mathgroup at wolfram.com

```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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