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For the following problems I used "FullSimplify" and "TrigReduce" but I manage to reduce more by hand.

FullSimplify[(Cos[2*a + b + c]*Tan[a + 2*b + 2*c])/Sin[(b + c - a)/2],
a + b + c == \[Pi]] // TrigReduce

Sin[2 a]/(1 + Cos[2 b + 2 c] (* Mathematica's solution *)

I am looking for expression containing:

a) -Tan[a] b) -Cot[a] c) Cot[a] d) Tan[a] e) Sin[a]

A similar situation is the following:

FullSimplify[(Sin[(6*Pi)/5 + x])^2 + (Sin[(4*Pi)/5 - x])^2, 
Cos[Pi/5 + x] == m^-1] // TrigReduce

1 - Sin[\[Pi]/10 - 2 x] (* Mathematica's solution *)

Here I am looking for expression containing:

a) 2(1-m^2) b) 1-1/m^2 c) 2(m^2-1)/m2 d) (m^2-1)/2m^2 e) (m^2+1)/(m^2-1)

Any suggestions?

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Simplify[(Cos[2*a + b + c]*Tan[a + 2*b + 2*c])/
    Sin[(b + c - a)/2]] /. (b -> Pi - a - c) // TrigExpand

Tan[a]

Simplify[(Sin[(6*Pi)/5 + x])^2 + (Sin[(4*Pi)/5 - x])^2 // Simplify, 
  Pi/5 + x == ArcCos[m^-1]] // Together

(2*(-1 + m^2))/m^2

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  • $\begingroup$ thanks...Why the "together" at the end of the second? $\endgroup$ – Fernando Silva May 24 '15 at 18:44
  • $\begingroup$ To put in the form that you requested (item c). If you want the simplest form then modify the question and leave off the Together. $\endgroup$ – Bob Hanlon May 24 '15 at 18:54

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