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Mathematics Algebra Wolfram Language Symbolic Computations

Convert sum of Sin[] or Cos[] to product?

Cesar Lozada

Posted 8 years ago

Hi, I am trying to convert a sum/difference of Cos[] into a product with the well know set of identities: r0 = {Cos[U_] -> Cos[Expand[U]], Sin[U_] -> Sin[Expand[U]], Cos[U_] + Cos[V_] -> 2Cos[(U + V)/2]Cos[(U - V)/2], Cos[U_] - Cos[V_] -> -2Sin[(U + V)/2]Sin[(U - V)/2], Sin[U_] + Sin[V_] -> 2Sin[(U + V)/2]Cos[(U - V)/2], Sin[U_] - Sin[V_] -> -2Sin[(U - V)/2]Cos[(U + V)/2], k_(B + C) -> kPi - kA, B + C -> Pi - A} ( The expression to be converted is : ) zz = 13 Cos[A] - 8 Cos[3 A] + Cos[5 A] - Cos[B - 5 C] + 5 Cos[B - 3 C] - 9 Cos[B - C] - 4 Cos[4 A] Cos[B - C] - Cos[3 (B - C)] + 5 Cos[3 B - C] - Cos[5 B - C] + 8 Cos[3 B + C] - 2 Cos[5 B + C] + 8 Cos[B + 3 C] - 2 Cos[B + 5 C] ( For a better conversion, some terms should be paired previously, for examples: 8 Cos[3 B + C] + 8 Cos[B + 3 C] ( Note the form k_Cos[p_B + q_C] + k_Cos[q_B + p_C] ) and - Cos[B - 5 C] - Cos[5 B - C] ( Note the form k_Cos[p_B - q_C] + k_Cos[q_B - p_C] ) so, I created the following set of rules: ) r1 = {Cos[U_] -> Cos[Expand[U]], Sin[U_] -> Sin[Expand[U]], k_Cos[p_B + q_C] + k_Cos[q_B + p_C] -> 2kCos[(p + q)/2(B + C)]Cos[(p - q)/2(B - C)], k_Cos[p_B + q_C] - k_Cos[q_B + p_C] -> -2kSin[(p + q)/2(B + C)]Sin[(p - q)/2(B - C)], k_Sin[p_B + q_C] + k_Sin[q_B + p_C] -> 2kSin[(p + q)/2(B + C)]Cos[(p - q)/2(B - C)], k_Sin[p_B + q_C] - k_Sin[q_B + p_C] -> -2kSin[(p + q)/2(B + C)]Cos[(p - q)/2(B - C)], k_(B + C) -> kPi - kA, B + C -> Pi - A} r2 = {Cos[U_] -> Cos[Expand[U]], Sin[U_] -> Sin[Expand[U]], k_Cos[p_B - q_C] + k_Cos[q_B - p_C] -> 2kCos[(p + q)/2(B - C)]Cos[(p - q)/2(B + C)], k_Cos[p_B - q_C] - k_Cos[q_B - p_C] -> -2kSin[(p + q)/2(B - C)]Sin[(p - q)/2(B + C)], k_Sin[p_B - q_C] + k_Sin[q_B - p_C] -> 2kSin[(p + q)/2(B - C)]Cos[(p - q)/2(B + C)], k_Sin[p_B - q_C] - k_Sin[q_B - p_C] -> -2kSin[(p + q)/2(B - C)]Cos[(p - q)/2(B + C)], k_B + k_C -> kPi - kA, k_(B + C) -> kPi - k*A, B + C -> Pi - A} The problem is that none of the following commands does the desirable work; z1=zz //. r1; z1=Simplify[zz //. r1,Trig]; z2=z1 //. r2; z2=Simplify[z1 //. r2,Trig]; I also has changed Trig to False without success. Does any one have a suggestion in order to make these conversions ? Thanks in advance César Lozada

Hi,

I am trying to convert a sum/difference of Cos[] into a product with the well know set of identities:

r0 = {Cos[U_] -> Cos[Expand[U]], Sin[U_] -> Sin[Expand[U]],
        Cos[U_] + Cos[V_] -> 2*Cos[(U + V)/2]*Cos[(U - V)/2],
        Cos[U_] - Cos[V_] -> -2*Sin[(U + V)/2]*Sin[(U - V)/2],
        Sin[U_] + Sin[V_] -> 2*Sin[(U + V)/2]*Cos[(U - V)/2],
        Sin[U_] - Sin[V_] -> -2*Sin[(U - V)/2]*Cos[(U + V)/2],
         k_*(B + C) -> k*Pi - k*A, B + C -> Pi - A}
(* The expression to be converted is : *)
zz = 13 Cos[A] - 8 Cos[3 A] + Cos[5 A] - Cos[B - 5 C] + 5 Cos[B - 3 C] -  9 Cos[B - C] - 4 Cos[4 A] Cos[B - C] 
- Cos[3 (B - C)] + 5 Cos[3 B - C] - Cos[5 B - C] + 8 Cos[3 B + C] - 2 Cos[5 B + C] +  8 Cos[B + 3 C] - 2 Cos[B + 5 C]

(* For a better conversion, some terms should be paired previously, for examples:
    8 Cos[3 B + C] +  8 Cos[B + 3 C]       ( Note the form k_*Cos[p_*B + q_*C] + k_*Cos[q_*B + p_*C] )
  and
    - Cos[B - 5 C] - Cos[5 B - C]          ( Note the form k_*Cos[p_*B - q_*C] + k_*Cos[q_*B - p_*C] )
  so, I created the following set of rules: *)
r1 = {Cos[U_] -> Cos[Expand[U]], Sin[U_] -> Sin[Expand[U]], 
  k_*Cos[p_*B + q_*C] + k_*Cos[q_*B + p_*C] ->  2*k*Cos[(p + q)/2*(B + C)]*Cos[(p - q)/2*(B - C)], 
  k_*Cos[p_*B + q_*C] - k_*Cos[q_*B + p_*C] -> -2*k*Sin[(p + q)/2*(B + C)]*Sin[(p - q)/2*(B - C)], 
  k_*Sin[p_*B + q_*C] + k_*Sin[q_*B + p_*C] ->  2*k*Sin[(p + q)/2*(B + C)]*Cos[(p - q)/2*(B - C)], 
  k_*Sin[p_*B + q_*C] - k_*Sin[q_*B + p_*C] -> -2*k*Sin[(p + q)/2*(B + C)]*Cos[(p - q)/2*(B - C)],
  k_*(B + C) -> k*Pi - k*A, B + C -> Pi - A}

r2 = {Cos[U_] -> Cos[Expand[U]], Sin[U_] -> Sin[Expand[U]], 
  k_*Cos[p_*B - q_*C] + k_*Cos[q_*B - p_*C] ->  2*k*Cos[(p + q)/2*(B - C)]*Cos[(p - q)/2*(B + C)], 
  k_*Cos[p_*B - q_*C] - k_*Cos[q_*B - p_*C] -> -2*k*Sin[(p + q)/2*(B - C)]*Sin[(p - q)/2*(B + C)], 
  k_*Sin[p_*B - q_*C] + k_*Sin[q_*B - p_*C] ->  2*k*Sin[(p + q)/2*(B - C)]*Cos[(p - q)/2*(B + C)], 
  k_*Sin[p_*B - q_*C] - k_*Sin[q_*B - p_*C] -> -2*k*Sin[(p + q)/2*(B - C)]*Cos[(p - q)/2*(B + C)],
  k_*B + k_*C -> k*Pi - k*A, k_*(B + C) -> k*Pi - k*A, B + C -> Pi - A}

The problem is that none of the following commands does the desirable work;

z1=zz //. r1;
z1=Simplify[zz //. r1,Trig];
z2=z1 //. r2;
z2=Simplify[z1 //. r2,Trig];

I also has changed Trig to False without success.

Does any one have a suggestion in order to make these conversions ?

Thanks in advance

César Lozada

POSTED BY: Cesar Lozada

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Yehuda Ben-Shimol

Yehuda Ben-Shimol, BGU - Ben Gurion University

Posted 8 years ago

Seems that r0 need some minor (and rather important) corrections `Cos[U_]-> Cos[Expand[U]]` - Expand will not be used as this is an immediate rule and need to be replaced with a delayed rule `:>`, etc (see below) `Cos[U_]+Cos[V_]-> 2Cos[...` etc. you do not cover the case of each of the cosines be multiplied with the same numbers. so you need to change this to a. Cos[U] + a. Cos[V] :> ... which means that if there is no `a` it is equal to 1 and the pattern is matched. Although the delayed rule here is not crucial, I prefer to use it almost always. Only the cases where the right hand side of the rule needs heavy computation that can be performed once I use a rule of the form `->` Now r0 = {Cos[U_] :> Cos[Expand[U]], Sin[U_] :> Sin[Expand[U]], a_. Cos[U_] + a_. Cos[V_] :> 2a Cos[(U + V)/2]Cos[(U - V)/2], a_. Cos[U_] - a_. Cos[V_] :> -2a Sin[(U + V)/2]Sin[(U - V)/2], a_. Sin[U_] + a_. Sin[V_] :> 2 aSin[(U + V)/2]Cos[(U - V)/2], a_. Sin[U_] - a_. Sin[V_] :> -2 aSin[(U - V)/2]Cos[(U + V)/2], k_Integer(B + C) :> kPi - kA, (B + C) :> Pi - A} and zz = 13 Cos[A] - 8 Cos[3 A] + Cos[5 A] - Cos[B - 5 C] + 5 Cos[B - 3 C] - 9 Cos[B - C] - 4 Cos[4 A] Cos[B - C] - Cos[3 (B - C)] + 5 Cos[3 B - C] - Cos[5 B - C] + 8 Cos[3 B + C] - 2 Cos[5 B + C] + 8 Cos[B + 3 C] - 2 Cos[B + 5 C] and zz //. r0 /. {Cos[x_] :> Cos[Simplify[x]], Sin[x_] :> Sin[Simplify[x]]} returnes 13 Cos[A] - 8 Cos[3 A] - 9 Cos[B - C] - 4 Cos[4 A] Cos[B - C] - 2 Cos[2 (B - 2 C)] Cos[B + C] + 10 Cos[2 (B - C)] Cos[B + C] + 16 Cos[B - C] Cos[2 (B + C)] - 4 Cos[2 (B - C)] Cos[3 (B + C)] - 2 Sin[1/2 (5 A + 5 B - C)] Sin[1/2 (5 A - 5 B + C)] which I believe that is closer to what you are looking for

Seems that r0 need some minor (and rather important) corrections

Cos[U_]-> Cos[Expand[U]] - Expand will not be used as this is an immediate rule and need to be replaced with a delayed rule :>, etc (see below)
Cos[U_]+Cos[V_]-> 2*Cos[... etc. you do not cover the case of each of the cosines be multiplied with the same numbers. so you need to change this to a. Cos[U] + a. Cos[V] :> ... which means that if there is no a it is equal to 1 and the pattern is matched. Although the delayed rule here is not crucial, I prefer to use it almost always. Only the cases where the right hand side of the rule needs heavy computation that can be performed once I use a rule of the form -> Now
```
r0 = {Cos[U_] :> Cos[Expand[U]], Sin[U_] :> Sin[Expand[U]], 
  a_. Cos[U_] + a_. Cos[V_] :> 2*a Cos[(U + V)/2]*Cos[(U - V)/2], 
  a_. Cos[U_] - a_. Cos[V_] :> -2*a Sin[(U + V)/2]*Sin[(U - V)/2], 
  a_. Sin[U_] + a_. Sin[V_] :> 2 a*Sin[(U + V)/2]*Cos[(U - V)/2], 
  a_. Sin[U_] - a_. Sin[V_] :> -2 a*Sin[(U - V)/2]*Cos[(U + V)/2], 
  k_Integer*(B + C) :> k*Pi - k*A, (B + C) :> Pi - A}
```

and

zz = 13 Cos[A] - 8 Cos[3 A] + Cos[5 A] - Cos[B - 5 C] + 
  5 Cos[B - 3 C] - 9 Cos[B - C] - 4 Cos[4 A] Cos[B - C] - 
  Cos[3 (B - C)] + 5 Cos[3 B - C] - Cos[5 B - C] + 8 Cos[3 B + C] - 
  2 Cos[5 B + C] + 8 Cos[B + 3 C] - 2 Cos[B + 5 C]

and

zz //. r0 /. {Cos[x_] :> Cos[Simplify[x]], 
  Sin[x_] :> Sin[Simplify[x]]}

returnes

13 Cos[A] - 8 Cos[3 A] - 9 Cos[B - C] - 4 Cos[4 A] Cos[B - C] - 
 2 Cos[2 (B - 2 C)] Cos[B + C] + 10 Cos[2 (B - C)] Cos[B + C] + 
 16 Cos[B - C] Cos[2 (B + C)] - 4 Cos[2 (B - C)] Cos[3 (B + C)] - 
 2 Sin[1/2 (5 A + 5 B - C)] Sin[1/2 (5 A - 5 B + C)]

which I believe that is closer to what you are looking for

POSTED BY: Yehuda Ben-Shimol

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