Two substitutions fail, so do I forget some earlier steps?
For
Cos[x] == Sin[(1/2)*Pi + x] /. x -> (1/2)*x
= true
I don't get the final answer, but only true.
Cos[(1/2)*x] == Sin[(1/2)*Pi + (1/2)*x]
(final answer)
A second derivation case :
Sin[x] == 2*Cos[x/2]*Sin[x/2] /. Cos[(1/2)*x] -> Sin[(1/2)*Pi + (1/2)*x]
(*no substition possible *)
Trace[Cos[x] == Sin[(1/2)*Pi + x] /. x -> (1/2)*x]
. Note that the initial equation evaluates toTrue
before the replacement ever comes into play. $\endgroup$Trace
; highlight it in Mathematica and pressF1
for help. $\endgroup$