# how can tangent half angle substitution be done in Mathematica?

Is it possible to replace sine and cosine in terms of tangent half-angles and again the tangent half-angles by a variable say t?

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The answer to this question is based on tangent half-angle substitutions. – Simon May 19 at 4:39

rule = {Cos[_] :> (1 - t^2)/(1 + t^2), Sin[_] :> 2*t/(1 + t^2)};
Sin[x] /. rule
Sin[x] /. rule /. t -> Tan[x/2] // Simplify


$\dfrac{2 t}{t^2+1}$

$\sin(x)$

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I'd suggest something like this

 rule = {
Cos[x_] :> (1 - tan[x/2]^2)/(1 + tan[x/2]^2),
Sin[x_] :> 2 tan[x/2]/(1 + tan[x/2]^2)};

Sin[x]/.rule -> (2 tan[x/2])/(1 + tan[x/2]^2)


Note the lowercase tan. This will not simplify to other trig functions on you unless you replace with the real Tan:

  Sin[x]/.rule /. tan[x_]:>Tan[x] //fullSimplify -> Sin[x]


This way you carry around the original argument:

otherwise you get behavior like this:

   Sin[x]-Sin[y] /. Sin[_] :> 2*t/(1 + t^2)


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you'd get the same result with Öska's answer. Try his Sin[x] - Sin[y] /. rule. – eldo Jun 20 '14 at 20:19
that was my point, that answer is potentially dangerous unless obviously you take care all your Sin,Cos's have the same argument. (in which case you may as well do `rule=Sin(x)-> ... ). – george2079 Jun 22 '14 at 16:30