Question

在三角形ABC中,cosA=1/3,求tan^2[(B+C)/2]+sin^2(A/2)的值?

Answer 1

因为是A,B,C是三角形的内角,故

tan^2[(B+C)/2]+sin^2(A/2)

= tan^2[(180-A)/2]+sin^2(A/2) 【A+B+C=180】

= tan^2(90-A/2)+sin^2(A/2)

= ctg^2(A/2)+sin^2(A/2) 【诱导公式】

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

= [(1+cosA)/2] / [(1-cosA)/2] + [(1-cosA)/2] 【倍角公式】

= {[1+(1/3)]/2} / {[1-(1/3)]/2} + [1-(1/3)]/2 【代入cosA=1/3】

= 2 + (1/3)

= 7/3