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I have the expression

ss = 
  As μ 
    (2 I k η1 Sinh[y η1] - 
      (I k (2 k^2 - k2^2) Cosh[H η1] Sinh[y η2])/(η2 Cosh[H η2]) + 
      (I k2^2 (2 k^2 - k2^2) Cosh[H η1] Sinh[y η2])/(2 k η2 Cosh[H η2]));

I want to simplify it using the relation

-4 k^2 η1 η2 Cosh[H η2] Sinh[H η1] + (-2 k^2 + k2^2)^2 Cosh[H η1] Sinh[H η2] == 0

The answer is:

the answer

However, I couldn't do get that using Mathematica.

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2 Answers 2

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You can include your condition in Solve to eliminate k2 in the equation ss:

Solve[{ss == As \[Mu]
(2 I k \[Eta]1 Sinh[
    y \[Eta]1] - (I k (2 k^2 - k2^2) Cosh[H \[Eta]1] Sinh[
      y \[Eta]2])/(\[Eta]2 Cosh[
      H \[Eta]2]) + (I k2^2 (2 k^2 - k2^2) Cosh[H \[Eta]1] Sinh[
      y \[Eta]2])/(2 k \[Eta]2 Cosh[
      H \[Eta]2])), -4 k^2 \[Eta]1 \[Eta]2 Cosh[H \[Eta]2] Sinh[
  H \[Eta]1] + (-2 k^2 + k2^2)^2 Cosh[H \[Eta]1] Sinh[
  H \[Eta]2] == 0}, ss, k2, Complexes]

(* {{ss -> 2 I As k \[Eta]1 \[Mu] Csch[H \[Eta]2] (Sinh[y \[Eta]1] Sinh[H \[Eta]2] - 
  Sinh[H \[Eta]1] Sinh[y \[Eta]2])}} *)

which is what you are looking for.

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    $\begingroup$ Putting List brackets around k2 will clarify that it is a variable to be eliminated rather than a domain specification and eliminate the need to explicitly state the domain as Complexes to avoid the warning message. Solve[eqns, ss, {k2}] $\endgroup$
    – Bob Hanlon
    Dec 1, 2017 at 13:33
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If I interpret your answer(!) right you can solve your problem as follows:

ergk2=Solve[-4 k^2 η1 η2 Cosh[H η2] Sinh[H η1] + (-2 k^2 + k2^2)^2 Cosh[H η1] Sinh[H η2] == 0,k2]

Simplify[ss /. ergk2]

gives four solutions similar to the answer you expect!

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