△ABC中,A为锐角,f(A)=[cos(π-2A)-1]sin(π+A/2)sin(π/2-A/2)/sin²(π/2-A/2)-sin²（π-A/2)

题目

△ABC中,A为锐角,f(A)=[cos(π-2A)-1]sin(π+A/2)sin(π/2-A/2)/sin²(π/2-A/2)-sin²（π-A/2)
+cos²A,若A+B=7π/12，f(A）=1，BC=2，求三个内角及AC长

答案

f(A)=[cos(π-2A)-1]sin(π+A/2)sin(π/2-A/2)/sin²(π/2-A/2)-sin²（π-A/2)+cos²A
=[﹣cos(2A)-1][﹣sin(A/2)]cos(A/2)/[cos²(A/2)-sin²（A/2)]+cos²A
=cos(2A)+1]·½·2sin(A/2)cos(A/2)]／cosA+cos²A
=cos²AsinA／cosA＋cos²A
=sinAcosA＋cos²A
=½sin2A＋½﹙1＋cos2A﹚
=√2／2·sin﹙2A＋π/4﹚＋½=1
∴sin﹙2A＋π/4﹚=√2／2
∵0＜2A＜2×7π/12=7π/6
∴π/4＜2A＋π/4＜17π/12
∴2A＋π/4=3π/4
∴A=π/4
∴B=π/3,C=5π/12
AC=BCSINB／SINA
=2SINπ/3／SINπ/4
=√6