问题
解答题
在△ABC中,A、B、C的对边分别为a、b、c, (Ⅰ)化简:bcosC+ccosB; (Ⅱ)求证:
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答案
(Ⅰ)bcosC+ccosB=b•
+c•a2+b2-c2 2ab a2+c2-b2 2ac
=
+a2+b2-c2 2a a2+c2-b2 2a
=a
(II)证明:∵
-1 a2
=cos2A a2
=1-cos2A a2
=4R2(R为三角形外接圆的半径)sin2A a2
-1 b2
=cos2B b2
=1-cos2B b2
=4R2sin2B b2
∴
-1 a2
=cos2A a2
-1 b2 cos2B b2
∴
-cos2A a2
=cos2B b2
-1 a2
.1 b2