1
$\begingroup$

I have the following code:

S0 = {{a, 0}, {0, b}};
R[θ_] := {{Cos[θ], Sin[θ]}, {-Sin[θ], 
    Cos[θ]}};
S[θ_] := Dot[R[θ], S0, R[-θ]] // FullSimplify
S[θ] = S[θ] /. {(Cos[θ])^2 -> (1 + Cos[2 θ])/2} // FullSimplify
(*
{{a Cos[θ]^2 + 
   b Sin[θ]^2, (-a + b) Cos[θ] Sin[θ]}, {(-a + 
     b) Cos[θ] Sin[θ], 
  b Cos[θ]^2 + a Sin[θ]^2}}
*)
S[θ] = S[θ] /. {(Sin[θ])^2 -> (1 - Cos[2 θ])/2} // FullSimplify
(*
{{1/2 (a + b + (a - b) Cos[2 θ]), (-a + 
     b) Cos[θ] Sin[θ]}, {(-a + 
     b) Cos[θ] Sin[θ], 
  1/2 (a + b + (-a + b) Cos[2 θ])}}
*)
S[θ] = S[θ] /. {Cos[θ] Sin[θ] -> 1/2 Sin[2 θ]} //FullSimplify
(*
{{1/2 (a + b + (a - b) Cos[2 θ]), (-a + 
     b) Cos[θ] Sin[θ]}, {(-a + 
     b) Cos[θ] Sin[θ], 
  1/2 (a + b + (-a + b) Cos[2 θ])}}
*)

I wonder why the replacement rule in the last line of code doesn't work?

$\endgroup$
  • $\begingroup$ Check 1/2 Sin[2 θ] // FullSimplify $\endgroup$ – Simon Woods Oct 19 '15 at 18:04
2
$\begingroup$

The reason is that FullSimplify
Mathematica regards $\cos\theta\sin\theta$ simpler than $\frac{1}{2}\sin 2\theta$, so when you run the code, FullSimplify right after the replacement rule, $\frac{1}{2}\sin 2\theta$ is converted to $\cos\theta\sin\theta$ again.
The solution is to write the replacement rule after FullSimplify
So the right order for the code can be as follows:

S0 = {{a, 0}, {0, b}};
R[θ_] := {{Cos[θ], Sin[θ]}, {-Sin[θ], 
    Cos[θ]}};
S[θ_] := Dot[R[θ], S0, R[-θ]] // FullSimplify;
S[θ] = 
  S[θ] /. {(Cos[θ])^2 -> (1 + Cos[2 θ])/2};
S[θ] = 
  S[θ] /. {(Sin[θ])^2 -> (1 - Cos[2 θ])/2};
S[θ] = S[θ] // FullSimplify;
S[θ] = 
  S[θ] /. {Cos[θ] Sin[θ] -> Sin[2 θ]/2};
S[θ] = S[θ] /. {a + b -> C};
S[θ] = S[θ] /. {a - b -> D };
S[θ] = S[θ] /. {b - a -> -D };
S[θ] // TraditionalForm

$\left( \begin{array}{cc} \frac{1}{2} (C+D (\cos (2 \theta ))) & -\frac{1}{2} D (\sin (2 \theta )) \\ -\frac{1}{2} D (\sin (2 \theta )) & \frac{1}{2} (C-D (\cos (2 \theta ))) \\ \end{array} \right)$

$\endgroup$

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.