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Trig simplification

Albert Rich, Independent Research
Posted 10 years ago

Mathematica 10's Simplify function has no problem simplifying 1-Cos[1/7]^2 to Sin[1/7]^2 and more importantly Sqrt[1-Cos[1/7]^2] to Sin[1/7].

However Simplify, and even FullSimplify, refuse to simplify 1-Cos[Pi/7]^2 to Sin[Pi/7]^2 and Sqrt[1-Cos[Pi/7]^2] to Sin[Pi/7].

Is there a good reason for this? The need to simplify these sort of trig constants come up a lot when integrating expressions of the form x^m/(a+b x^n).

Albert
POSTED BY: Albert Rich
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Thank you for your responses, but I know how to add the requisite knowledge to Mathematica.

Mathematica’s Simplify function simplifies expressions of the form Sqrt[1-Cos[n]^2] to Sin[n] for all constants n in the interval 0 to pi, except when n is of the form m Pi where m is a fraction whose denominator is greater than 6. So my question is why the exception?

Albert
POSTED BY: Albert Rich

Probably Sin[Pia] has precedence over Sin[ba] in all the transformations it tries, i guess?
POSTED BY: Sander Huisman

Hi Albert,

You can add things to try by Simplify:

ClearAll[t]
t[1-Sin[a_]^2]:=Cos[a]^2
t[1-Cos[a_]^2]:=Sin[a]^2
Simplify[1-Cos[\[Pi]/7]^2,TransformationFunctions->{Automatic,t}]

This will do it for all a, but you have to make sure yourself that those relations always hold.

POSTED BY: Sander Huisman
Jim Baldwin
Jim Baldwin, Retired
Posted 10 years ago

Simplification is sometimes in the eye of the beholder. Note that

Simplify[1 - Cos[a/7]^2]
(* Sin[a/7]^2 *)

works as you desire. Alternatively you can add extra rules for any special simplifications such as:

rule = {1 - Cos[x_]^2 -> Sin[x]^2};
1 - Cos[Pi/7]^2 /. rule
(* Sin[\[Pi]/7]^2 *)

But I don't know enough about Mathematica to know "why" the same simplification does not result from using a and Pi.

POSTED BY: Jim Baldwin
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