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Re: How to solve a simple Trig cofunction?
*To*: mathgroup at smc.vnet.net
*Subject*: [mg50519] Re: [mg50500] How to solve a simple Trig cofunction?
*From*: Brian Feeny <bfeeny at mac.com>
*Date*: Thu, 9 Sep 2004 05:17:31 -0400 (EDT)
*References*: <200409080914.FAA08518@smc.vnet.net> <913EB31F-018F-11D9-8C24-000A95B4967A@akikoz.net>
*Sender*: owner-wri-mathgroup at wolfram.com
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/
>
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