Question

在锐角三角形ABC中..sinA=2根号2/3 求(sinB+C/2)^2+cos(3TT-2A)

Answer 1

(sinB+C/2)^2+cos(3TT-2A)

=[sin(pi/2-A/2)]^2+cos(pi-2A)

=[cos(A/2)]^2-cos(2A)

=[(cosA+1)/2]-[1-2(sinA)^2]

因为sinA=2根号2/3 ,所以在锐角三角形中cosA=1/3

(sinB+C/2)^2+cos(3TT-2A)

=[(cosA+1)/2]-[1-2(sinA)^2]

=[(1/3+1)/2]-[1-2(2根号2/3)^2]

=2/3+7/9

=13/9