If I use TrigFactor, it gives the result in terms of Sin:
In[483]:=
expr := Cos[x] + Sin[x]
TrigFactor[expr]
Out[485]=Sqrt[2]*Sin[Pi/4 + x]
What I want to get is
Sqrt[2]*Cos[-Pi/4 + x]
Using ReplaceAll doesn't help, since it does the replacement, but then converts back to Sin.
In[489]:=
expr := Cos[x] + Sin[x]
TrigFactor[expr] /. Sin[t_] -> Cos[t - Pi/2]
Out[490]=Sqrt[2] Sin[Pi/4 + x]
Try this:
Simplify[TrigFactor[Cos[x] + Sin[x]],
ComplexityFunction -> (Count[{#1}, _Sin, Infinity] &)]
(* Sqrt[2] Cos[1/4 ([Pi] - 4 x)] *)
Have fun!
Use RuleDelayed (:>) instead of Rule (->) (see: tutorial/ImmediateAndDelayedDefinitions):
TrigFactor[expr] /. Sin[t_] :> (Cos[t - Pi/2])
Sqrt[2] Cos[π/4 - x]
Alternatively,
TrigFactor[expr] /. Sin -> (Cos[# - π/2] &)
Sqrt[2] Cos[π/4 - x]
TrigFactor[expr] /. Sin[Pi/4 + t_] -> HoldForm@Cos[t - Pi/4]– Daniel Huber Nov 28 '21 at 17:20(((expr // TrigFactor) /. Sin[x_] :> Inactive[Cos][x - Pi/2]) // Activate) /. Cos[x_] :> Cos[-x]The last replacement is not necessary if you are satisfied with the negative of the desired argument. – Bob Hanlon Nov 28 '21 at 17:57