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  • $\begingroup$ Thanks, in fact my expression is quite simple: ds2=dr^2 + 1/2 (d\[Theta]^2 + 2 d\[Phi]^2) r^2 - 1/2 d\[Theta]^2 r^2 1 - 2 Sin[\[Phi]]^2, and I get what I want by your method, ds2/. {Cos[2 t_] :> 1 - 2 Sin[t]^2} // Simplify $\endgroup$ Commented Jul 26, 2015 at 3:21
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    $\begingroup$ @vanabel You're welcome. Sometimes Simplify will convert the expression back to terms of Cos[2 x] and sometimes it will combine the Sin[x] with other terms and not convert it. Trig. is tricky to deal with. BTW, I would not redefine a basic function like Cos in the way RunnyKine does. Since there are, I believe, many rules built into Mathematica about sine and cosine, I would be afraid the redefinition would break something. Just what that is would be difficult to predict. $\endgroup$ Commented Jul 26, 2015 at 3:28