Question

Fie a si b numere reale, astfel incat a+b=pi/3. Sa se arate ca sin2a-sin2b-sin(a-b)=0

Answer 1

sin2a-sin2b=2sin[(2a-2b)/2]*cos[(2a+2b)/2]=2sin(a-b)cos(a+b)     /scadem sin(a-b)
2sin(a-b)cos(a+b)-sin(a-b)=sin(a-b)(2cos(a+b)-1)=sin(a-b)(2*cos(pi/3)-1)=
sin(a-b)(2*1/2-1)=0