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