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

题目

在锐角三角形ABC中..sinA=2根号2/3 求(sinB+C/2)^2+cos(3TT-2A)
RT...TT=帕而 既=3.1415926

答案

(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