Question

在锐角△ABC中，角A,B,C所对的边分别为a，b，c，已知sinA=2根号2/3.（1）求tan^2[（B+C）/2]+sin^2（A/2）

Answer 1

:sinA=2√2/3,因为是锐角三角形,所以cosA=1/3
tan²[(B+C)/2]+sin² (A/2)
=tan²(π-A)/2+sin²(A/2)
=cot²(A/2)+sin²(A/2)
=(cos²(A/2)/sin²(A/2)+sin²(A/2)
=[cos²(A/2)+sin²(A/2)×sin²(A/2)]/sin²(A/2)
=[cos²(A/2)+(1-cos²(A/2)×sin²(A/2)]/sin²(A/2)
=[cos²(A/2)+sin²(A/2)-cos²(A/2)×sin²(A/2)]/sin²(A/2)
=[1-(1/4)sin²A]/[(1-cosA)/2]
=[1-(1/4)2√2/3]/[(1-1/3)/2]
=(6-√2)/2

Answer 2

:sinA=2√2/3,因为是锐角三角形,所以cosA=1/3
tan²[(B+C)/2]+sin² (A/2)
=tan²(π-A)/2+sin²(A/2)
=cot²(A/2)+sin²(A/2)
=(cos²(A/2)/sin²(A/2)+sin²(A/2)
cos^2(A/2)=(1+cosA)/2=2/3
sin^2(A/2)=1/3
∴原式=2+1/3=7/3