# How to eliminate a trigonometric function between the equations?

Posted 10 years ago
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 Dear All,I have an equation Cos[2 x] == 0and I want to get Mathematica to substitute the "a" for Cos[ x ]the equation should get the form2*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!
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Posted 10 years ago
 Solve[2 ArcCos[a] == \[Pi]/2, a]{{a -> 1/Sqrt}}
Posted 10 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 converteqn={Cos[2 x] == 0}into eqn2={a^2-1==0}using the function Eliminate.But I'm getting In:= eqn = {Cos[2 x] == 0, Cos[x] == a}Out= {Cos[2 x] == 0, Cos[x] == a}In:= eqn2 = Eliminate[eqn, x]During evaluation of In:= Eliminate::ifun: Inverse functions are being used by Eliminate, so some solutions may not be found; use Reduce for complete solution information. >>Out= 2 ArcCos[a] == \[Pi]/2In:= eqn2Out= 2 ArcCos[a] == \[Pi]/2So I wonder how to get the outputa^2-1==0instead of 2 ArcCos[a] == \[Pi]/2Thanks!
Posted 10 years ago
 Reduce[Cos[2 x] == 0 && a == Cos[x], {a, x}](C \[Element] Integers &&    a == -(1/Sqrt) && (x == -2 ArcTan[1 + Sqrt] + 2 \[Pi] C ||      x == 2 ArcTan[1 + Sqrt] + 2 \[Pi] C)) || (C \[Element]     Integers &&    a == 1/Sqrt[    2] && (x == 2 ArcTan[1 - Sqrt] + 2 \[Pi] C ||      x == -2 ArcTan[1 - Sqrt] + 2 \[Pi] C))
Posted 10 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 10 years ago
 The first step would beTrigExpand[Cos[2 x]]to getCos[x]^2 - Sin[x]^2