Question

已知锐角三角形ABC中,角A,B,C所对的边分别为a.b.c.已知sinA=（2根号2）/3 求tan^2(B+C)/2

Answer 1

sinA=2√2/3

锐角△ABC

所以cosA=√[1-(sinA)^2]=1/3

cosA=1-2[sin(A/2)]^2

所以[sin(A/2)]^2=1/3

[cos(A/2)]^2=1-[sin(A/2)]^2=2/3

tan^2(B+C)/2

=tan^2(90-A/2)

=cot^2A/2

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

=2

所以原式=2+1/3=7/3