Question

在三角形ABC中,A为最小角C为最大角,已知cos(2A+C)=-4/5,sinB=4/5,cos2(B+C)的值.

Answer 1

cos(2a+c)=-4/5

cos[2(π-b-c)+c]=-4/5

cos(2π-2b-2c+c)=-4/5

cos(2π-2b-c)=-4/5

cos(-2b-c)=-4/5

cos(2b+c)=-4/5

cos2(b+c)=cos2[(2b+c)-b]

=2cos^2[(2b+c)-b]-1

cos[(2b+c)-b]

=cos(2b+c)cosb+sin(2b+c)sinb

=(-4/5)*(3/5)+(3/5)*(4/5)

=0

cos2(b+c)=2cos^2[(2b+c)-b]-1=-1.