Sin (pi/6-x)-cos(pi/3+x)
Как решить ?
по какой формуле ... ?
спасибо

Sinπ/6*cosx-cosπ/6*sinx-cosπ/3*cosx+sinπ/3*sinx=
=1/2*cosx-√3/2*sinx-1/2*cosx+√3/2*sinx=0

Или
sin(π/6 -x) - cos(π/3 +x) =sin( π/2 - ( π/3 +x) )  - cos(π/3 +x) =
 cos(π/3 +x) - cos(π/3 +x) =0.
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sin( π/2 - α)  =cosα .