锐角三角形ABC 已知sinA=2√2/3

题目

锐角三角形ABC 已知sinA=2√2/3
求tan²（（B+C)/2 ）+sin² (A/2)

答案

sinA=2√2/3,因为是锐角三角形,所以cosA=1/3
tan^2[(B+C)/2]+sin^2 (A/2)=tan^2(π-A)/2+sin^2(A/2)
=cot^2(A/2)+sin^2(A/2)=(cos^2(A/2)/sin^2(A/2)+sin^2(A/2)
=[cos^2(A/2)+sin^2(A/2)*sin^2(A/2)]/sin^2(A/2)
=[cos^2(A/2)+(1-cos^2(A/2)*sin^2(A/2)]/sin^2(A/2)
=[cos^2(A/2)+sin^2(A/2)-cos^2(A/2)*sin^2(A/2)]/sin^2(A/2)
=[1-(1/4)sin^2A]/[(1-cosA)/2]
=[1-(1/4)2√2/3]/[(1-1/3)/2]
=(6-√2)/2
(2)
S=(1/2)bcsinA=(1/2)bc*2√2/3=√2
所以bc=3
根据余弦定理
a^2-b^2-c^2+2bccosA=0,即4-b^2-c^2+2bc*1/3=0
解得:b=c=√3