1
$\begingroup$

I have two trigonometric expressions. I would like to eliminate the fraction from my 2nd expression.

A = (-Cos[ψ2] Sin[η2] Sin[θ2] + Cos[η2] Sin[ψ2]) 

B = (-D + d Cos[η2] Cos[θ2] Cos[ϕ2] + 1/2 l Cos[θ2] Sin[η2] + Q2 Sin[θ2] - d Sin[θ2] Sin[ϕ2] + (Q1 Cos[θ1] Cos[θ2] Sin[η2] (Cos[η2] Cos[θ2] Cos[ϕ2] - Sin[θ2] Sin[ϕ2])) / (Cos[ψ2] Sin[η2] Sin[θ2] - Cos[η2] Sin[ψ2]))

I tried Simplify and FullSimplify to suppress the denominator but the denominator in the expression B (equal to A) is not suppressed.

Simplify[A*B]

Please help me to force Mathematica to make the simplifications so as to eliminate the denominator of the expression B which is equal to A ?

$\endgroup$
  • 3
    $\begingroup$ Try Simplify@Expand[A*B] $\endgroup$ – Lukas Lang May 30 '18 at 13:54
  • $\begingroup$ Or Cancel[A*B]. $\endgroup$ – John Doty May 30 '18 at 15:14
1
$\begingroup$

First let me explain what is going on.

 A // FullForm
Plus[Times[-1, Cos[ψ2], Sin[η2], Sin[θ2]], Times[Cos[η2], Sin[ψ2]]]

However, what you consider the denominator of the fractional part of b is

(Cos[ψ2] Sin[η2] Sin[θ2] - Cos[η2] Sin[ψ2]) // FullForm

Plus[Times[Cos[ψ2], Sin[η2], Sin[θ2]], Times[-1, Cos[η2], Sin[ψ2]]]

So the two forms don't look the same to Mathematica and don't get cancelled.

There are many ways to solve the problem. One has already been suggested in a comment. Here is another.

AA = -1 A;
Simplify[AA B]

trig_expression

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.