Coefficient and Addition Formula

Question

If I type:

Coefficient[Cos[x + b], Cos[x]]

I get zero.

But I would like Mathematica to apply:

$$\cos(x+b)=\cos x\cos b-\sin x\sin b$$

Therefore the coefficient of $\cos x$ should be $\cos b$.

How do I fix this?

Edit:

The solution Coefficient[TrigExpand@f[x],Cos[x]] has been proposed, but this wouldn't work for expressions like $\sin(a+3x)$, in fact:

In[84]:= TrigExpand@Sin[a + 3 x]

Out[84]= Cos[x]^3 Sin[a] + 3 Cos[a] Cos[x]^2 Sin[x] - 
         3 Cos[x] Sin[a] Sin[x]^2 - Cos[a] Sin[x]^3

And therefore this would give me a wrong coefficient:

In[86]:= Coefficient[TrigExpand@Sin[a + 3 x], Cos[x]]

Out[86]= -3 Sin[a] Sin[x]^2

Because $\sin x$ is not a constant, and for what I am concerned with, it cannot be the coefficient of $\cos x$. Therefore, for all the expressions of the type $\cos(n x+ a)$ I am expecting to find zero, except when $n=1$.

Answer 1

Ok, here is another approach:

ClearAll[findCoeff];
findCoeff[expr_, termToIsolate_] := Module[{findCoeffIn},
  findCoeffIn[termToIsolate] = 1;
  findCoeffIn[exprIn_Plus, OptionsPattern[]] := findCoeffIn /@ exprIn;
  findCoeffIn[Times[lterms___, termToIsolate, rterms___]] := 
   If[FreeQ[#, x], #, 0] &@Times[lterms, rterms];
  findCoeffIn[exprIn : (Cos[_] | Sin[_] | Tan[_] | Cot[_]), 
    OptionsPattern[]] := findCoeffIn[TrigExpand[exprIn]];
  findCoeffIn[_] := 0;
  findCoeffIn[expr]
  ]

This seems to work on all the cases that I tried. Use it like this:

In[1]:= findCoeff[Cos[2 x] + Cos[x + a] + 2, Cos[x]]
Out[1]= Cos[a]

In[2]:= findCoeff[2 Cos[x], Cos[x]]
Out[2]= 2

In[3]:= findCoeff[Cos[x + a] + Sin[2 x] + Cos[x], Cos[x]]
Out[3]= 1 + Cos[a]

In[4]:= findCoeff[Cos[2 x], Cos[2 x]]
Out[4]= 1

Something like findCoeff[Sin[2 x+a],Cos[2 x]] will still not work as you want it to, though. One probably has to fiddle with FourierCosCoefficient to do that. In fact, I'm not even sure that the problem is well stated: consider for example Sin[2x]. How should it expand? FourierCosCoefficient[Sin[2 x],x ,1] gives the answer $8/3\pi$, while findCoeff[Sin[2 x],Cos[x]] gives 0. While the former result answers a well defined question, what exactly are the rules according to which you want the latter result?

Old

Here is a slightly improved version of Kuba's idea:

Options[findCoeff] = {termToIsolate -> Cos[x], 
   independentVariable -> x};
findCoeff[expr_, OptionsPattern[]] := Total@Cases[
    TrigExpand[expr],
    cl___ OptionValue@termToIsolate cr___ :> 
     cl cr /; FreeQ[cl cr, OptionValue@independentVariable]
    ] /. {} -> 0

Examples:

In[1]:= findCoeff[Sin[2 x] + 2 Cos[x] + Cos[x + a]]

Out[1]= 2 + Cos[a]

In[2]:= findCoeff[Cos[y + x] + Sin[x], termToIsolate -> Cos[y], 
 independentVariable -> y]

Out[2]= Cos[x]

I'm not sure it would work on more complicated cases though.

Answer 2

Does it fit your needs?

Cases[
  TrigExpand@Sin[a + x],
  Cos[x] coeff__ :> coeff /; FreeQ[coeff, x]
  ] /. {} -> 0

---

Old:

Not sure if I got your point in general but this should do: (no it shouldn't really)

FourierCosCoefficient[Cos[x + b], x, 1]

FourierCosCoefficient[Sin[a + 3 x], x, 1]

Cos[b]

0

Answer 3

EDIT: Ok, now this answer is almost the same as Kuba's...

ORIGINAL POST:

I think Kuba's answer is preferable, but you could also do this:

coeff[expr_, var_] := Boole[FreeQ[#,x]]*#&@
  Coefficient[TrigExpand@expr, var]

Then,

coeff[Sin[a + 3 x], Cos[x]]

0

and

coeff[Cos[a+x],Cos[x]]

Cos[a]