Question

2sinb*cosb+sin(a-b) b-бэтта a-альфа

Answer 1

2sinB*cosB+sin(A-B)=2sinB*cosB+sinAcosB-cosAsinB=cosB(sinA+sinB)+sin B(cosB-cosA)=

=2cosBsin(A+B)/2*cos(A-B)/2-2sin Bsin (A+B)/2*sin(A-B)/2=

=2sin(A+B)/2*(cosBcos(A-B)/2-sinBsin(A-B)/2)=2sin(A+B)/2*cos(B+(A-B)/2)=

=2sin(A+B)/2*cos(A+B)/2=sin(A+B)

Отв sin(A+B)

Answer 2

2sinB*cosB+sin(A-B)=2sinB*cosB+sinA*cosB-sinB*cosA=cosB(sinA+sinB)-sinB(cosA+cosB)=2cosB*sin((A+B)/2)*cos((A-B)/2)-2sinB*cos((A+B)/2)*cos((A-B)/2)=2cos((A-B)/2)*( cosB*sin((A+B)/2)- sinB*cos((A+B)/2))=2cos((A-B)/2)*sin((A+B)/2-B)= 2cos((A-B)/2)*sin((A-B)/2)=sin(A-B)