Question

Se considera a+b=pi/3. Demonstrati ca sin 2a - sin 2b - sin (a-b) = 0.

Answer 1

sin2a-sin2b-sin(a-b)=2sin ((2a-2b)/2)*cos((2a+2b)/2)-sin (a-b)=
=2sin(a-b)*cos(a+b)-sin(a-b)=

2sin (a-b) *cosπ/3-sin (a-b)=(2sin(a-b))*1/2-sin(a-b) =
(1/2)*2*sin(a-b)-sin(a-b)=sin(a-b)-sin(a-b)=0