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Re: Re: How to solve a simple Trig cofunction?
*To*: mathgroup at smc.vnet.net
*Subject*: [mg50561] Re: [mg50519] Re: [mg50500] How to solve a simple Trig cofunction?
*From*: George Woodrow III <georgevw3 at mac.com>
*Date*: Fri, 10 Sep 2004 04:06:02 -0400 (EDT)
*References*: <200409080914.FAA08518@smc.vnet.net> <913EB31F-018F-11D9-8C24-000A95B4967A@akikoz.net> <200409090917.FAA19326@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
In the equation, 4 and 5 are in radians. If you want to solve the
equation with everything in degrees, you need to enter 4 as 4 Degree,
etc.
george
On 9 Sep, 2004, at 5:17 am, 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/
>>
>
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