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How to eliminate a trigonometric function between the equations?

Posted 11 years ago
Dear All,

I have an equation

Cos[2 x] == 0

and I want to get Mathematica to substitute the "a" for Cos[ x ]
the equation should get the form
2*a^2-1==0

 eqn = {Cos[2 x] == 0, Cos[ x ] == a}
 Eliminate[eqn, x]

doesn't work. Do anybody have any idea how to do it?

Thanks!
POSTED BY: Maxim Ilyn
5 Replies
Solve[2 ArcCos[a] == \[Pi]/2, a]
{{a -> 1/Sqrt[2]}}
POSTED BY: Frank Kampas
Posted 11 years ago
Dear Frank,

It works, but it solves the equation for x. What I wanted to do is to substitute the a for the Cos[ x ] and  get the equation. So I would like to convert

eqn={Cos[2 x] == 0}
into
eqn2={a^2-1==0}
using the function Eliminate.
But I'm getting
In[3]:= eqn = {Cos[2 x] == 0, Cos[x] == a}
Out[3]= {Cos[2 x] == 0, Cos[x] == a}
In[5]:= eqn2 = Eliminate[eqn, x]
During evaluation of In[5]:= Eliminate::ifun: Inverse functions are being used by Eliminate, so some solutions may not be found; use Reduce for complete solution information. >>
Out[5]= 2 ArcCos[a] == \[Pi]/2
In[6]:= eqn2
Out[6]= 2 ArcCos[a] == \[Pi]/2
So I wonder how to get the output
a^2-1==0
instead of
2 ArcCos[a] == \[Pi]/2

Thanks!
POSTED BY: Maxim Ilyn
Reduce[Cos[2 x] == 0 && a == Cos[x], {a, x}]
(C[1] \[Element] Integers &&
   a == -(1/Sqrt[2]) && (x == -2 ArcTan[1 + Sqrt[2]] + 2 \[Pi] C[1] ||
      x == 2 ArcTan[1 + Sqrt[2]] + 2 \[Pi] C[1])) || (C[1] \[Element]
    Integers &&
   a == 1/Sqrt[
    2] && (x == 2 ArcTan[1 - Sqrt[2]] + 2 \[Pi] C[1] ||
     x == -2 ArcTan[1 - Sqrt[2]] + 2 \[Pi] C[1]))
POSTED BY: Frank Kampas
Posted 11 years ago
Dear Frank,

Thanks for answering! I understand that I can do it manualy. It´s possible to transform the equation into the form when it contains only Cos[ x ] and substitute a for Cos[ x ] using \ . command. However the equations I'm working with are much more complex that one I used as an example and I wonder if I can get Mathematics do this job for me.  
POSTED BY: Maxim Ilyn
The first step would be
TrigExpand[Cos[2 x]]
to get
Cos[x]^2 - Sin[x]^2
POSTED BY: Frank Kampas
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