mathematica, Simplify trig function

2
votes

if say i have a function given :

               singlepattern = Cosh[theta] + Cosh[3theta]

How do i get a rational expression in terms of x of the function if i want to substitute Cosh[theta] by 

               "Cosh[theta] = ( x )/ 2 "

expression?

3
wolfram-mathematica
Hi i'm trying to get a simplified rational expression for singlepattern in terms of x - Sunday
Aren't rational expressions ratio's of polynomials? I don't directly see the relationship with your Cosh functions. - Sjoerd C. de Vries
Well when you substitute the x/2 expression for cosh(theta) after doin simplepattern // ExpandTrig ..... u will find that it will now have expressions in terms of x - Sunday
You can also use a direct substitution of theta->ArcCosh[x/2], coupled with TrigExpand. Like so: Cosh[theta] + Cosh[3theta] /.theta->ArcCosh[x/2] // TrigExpand // Together. This gives (x^3-2*x)/2. This, of course, is similar in spirit to Sjoerd's answer below. - Sasha

3 Answers

2
votes

I retagged the question as a homework. You should look into ChebyshevT polynomials. It has the property that ChebyshevT[3, Cos[th] ]==Cos[3*th]. So for your problem the answer is 

In[236]:= x/2 + ChebyshevT[3, x/2]

Out[236]= -x + x^3/2

Alternatively, you could use TrigExpand:

In[237]:= Cos[th] + Cos[3*th] // TrigExpand

Out[237]= Cos[th] + Cos[th]^3 - 3 Cos[th] Sin[th]^2

In[238]:= % /. Sin[th]^2 -> 1 - Cos[th]^2 // Expand

Out[238]= -2 Cos[th] + 4 Cos[th]^3

In[239]:= % /. Cos[th] -> x/2

Out[239]= -x + x^3/2

EDITCosh[theta] == Cos[I*u]uutheta
2
votes

Use Solve to solve for theta, then substitute, Expand, and Simplify:

In[16]:= TrigExpand[Cosh[3 theta] + Cosh[theta]] /. 
  Solve[Cosh[theta] == (x)/2, theta] // FullSimplify

During evaluation of In[16]:= Solve::ifun: Inverse functions are being used by Solve,
so some solutions may not be found; use Reduce for complete solution information. >>

Out[16]= {1/2 x (-2 + x^2), 1/2 x (-2 + x^2)}