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]

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