how to simplify a*cos(x)+b*sin(x)

[Matth32]

Hello,
I would like to simplify a trigonometric function like acos(x)+bsin(x) into Rcos(x-c).
My expression is:

I have tried many functions, I thought that TR10i function of fu library could work but maybe it only works with numbers.

Does anybody have a suggestion?
Thanks

[Matth32]

Hello,

Nobody have an idea?

Thanks

[oscarbenjamin]

Do you know that a simpler form exists?

[Matth32]

Yes, I would like to simplify in this form (made by hand):

[Matth32]

I think you are right, atan returns an angle between –pi/2 and pi/2 so it only describes the right side of the trigonometric circle which means positive cosine.
So if the cosine was initially negative (alpha < 0), using atan (b/a) introduces an error of pi which leads to incorrect sign of the result.

For example:
-cos(x)+sin(x)

Applying the formula would give:
√2cos⁡(x-atan⁡(-1) )=√2cos⁡(x+π/4)

Expending back using cosine properties:
√2*(cos⁡(x)*cos⁡(π/4)-sin⁡(x)sin⁡(π/4))=√2(cos⁡(x)*√2/2-sin⁡(x)*√2/2)=cos(x)-sin(x)

There is probably a better way to find the angle knowing its sine and cosine, but I don’t have it.
The case where alpha = 0 is also problematic for this factorisation, but there is no need to do it in that case.

[oscarbenjamin]

There is probably a better way to find the angle knowing its sine and cosine, but I don’t have it.

That's atan2 which SymPy has:

In [11]: [atan2(y, x) for x, y in [(1, 1), (-1, 1), (-1, -1), (1, -1)]]
Out[11]: 
⎡π  3⋅π  -3⋅π   -π ⎤
⎢─, ───, ─────, ───⎥
⎣4   4     4     4 ⎦

The case where alpha = 0 is also problematic for this factorisation, but there is no need to do it in that case.

The atan2 function handles all of the special cases:

In [12]: [atan2(y, x) for x, y in [(0, 1), (0, -1), (1, 0), (-1, 0)]]
Out[12]: 
⎡π  -π       ⎤
⎢─, ───, 0, π⎥
⎣2   2       ⎦

[Matth32]

Yes atan2 seems to make the job for that.
So, if my understanding is right, in the actual code sympy is not able to make the complete simplification and it will be difficult to code a simplification which take all case into account?

[oscarbenjamin]

I think it can be done. There just is not a function for this particular rearrangement in SymPy yet (someone needs to make that function).

[Matth32]

Ok so is there a kind of pool of requests or something like this? Because I don't have the capability to code this function myself and I think could be an enhancement for sympy.

[ThePauliPrinciple]

To answer your question on how to rewrite asin(x)+bcos(x) into R*cos(x-c), we can do this as follows:

import sympy as sym

a,b,c,R,x= sym.symbols('a b c R x')

lhs=a*sym.cos(x)+b*sym.sin(x) # form we have
rhs=R*sym.cos(x-c) # form we want
eq=lhs-rhs #solve expects an equation of the type lhs-rhs=0 instead of lhs=rhs

#solve for the unknowns in the form we want
s1,s2=sym.solve(eq,c)  #as it turns out, two solutions are found for c (makes sense to me in a cos)
s3=sym.solve(eq,R)[0]  #solve returns an array, even if one solution is found```

I do realise this does not obtain the form of R and c you were looking for, but perhaps it is still useful?

[oscarbenjamin]

The formula needed is straight-forward:

In [48]: A, B, t = symbols('A, B, t')                                                                

In [49]: expr = sqrt(A**2 + B**2)*cos(t - atan2(B, A))                                               

In [50]: expr                                                                                        
Out[50]: 
   _________                     
  ╱  2    2                      
╲╱  A  + B  ⋅cos(t - atan2(B, A))

In [51]: expr.expand(trig=True)                                                                      
Out[51]: A⋅cos(t) + B⋅sin(t)

There just needs to be a function that will:

1. Identify the cos and sin terms that have the same argument.
2. Extract the coefficients A and B.