已知O是锐角三角形ABC的外心,△BOC,△COA,△AOB的面积数依次成等差数列.
(1)推算tanAtanC是否为定值?说明理由;
(2)求证:tanA,tanB,tanC也成等差数列.
【答案】
分析:如图所示,设△ABC的外接圆半径为R,则
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024183636992327237/SYS201310241836369923272021_DA/0.png)
,
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024183636992327237/SYS201310241836369923272021_DA/1.png)
,由题意可得2S
△COA=S
△BOC+S
△AOB,整理可得2sin2B=sin2A+sin2C,结合三角形的内角和公式及和差角公式整理得 sinA•sinC=3cosA•cosC.
(1)因△ABC是锐角三角形,
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024183636992327237/SYS201310241836369923272021_DA/2.png)
,可知cosA≠0,cosC≠0,可求tanAtanC.
(2)要证tanA,tanB,tanC成等差数列.只要证明2tanB=tanA+tanC即可
解答:![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024183636992327237/SYS201310241836369923272021_DA/images3.png)
解:如图所示,设△ABC的外接圆半径为R,
则
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024183636992327237/SYS201310241836369923272021_DA/3.png)
,
同理:
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024183636992327237/SYS201310241836369923272021_DA/4.png)
.
∵S
△BOC,S
△COA,S
△AOB成等差数列,
∴2S
△COA=S
△BOC+S
△AOB,
即
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024183636992327237/SYS201310241836369923272021_DA/5.png)
.
∴2sin2B=sin2A+sin2C,∴2sin2B=sin[(A+C)+(A-C)]+sin[(A+C)-(A-C)],
∴4sinBcosB=2sin(A+C)cos(A-C).
又A+B+C=π,故sinB=sin(A+C)≠0.
∴2cosB=cos(A-C).
又A+B+C=π,∴-2cos(A+C)=cos(A-C).
整理得 sinA•sinC=3cosA•cosC.
(1)因△ABC是锐角三角形,
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024183636992327237/SYS201310241836369923272021_DA/6.png)
,可知cosA≠0,cosC≠0,∴tanAtanC=3,
故tanAtanC为定值.
(2)∵
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024183636992327237/SYS201310241836369923272021_DA/7.png)
=
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024183636992327237/SYS201310241836369923272021_DA/8.png)
.∴2tanB=tanA+tanC,
即tanA,tanB,tanC成等差数列.
点评:本题主要以等差数列的性质为切入点,主要考查了三角形中正弦定理、两角和与差的三角公式,三角形的内角和公式等知识的综合应用.