Question

Fie a,b €R ,astfel încât a+b= pi/3. Demonstrați ca sin2a-2sin2b-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) *cos(pi/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)=0Este fara 2 in fata la sin2b, nu da exercitiul daca e cu 2 in fata.
Am folosit formula sin2a-sin2b=2cos[(2a+2b)/2]*sin[(2a-2b)/2]