在銳角△ABC中,角A、B、C的對(duì)邊分別為a、b、c,已知2acosA=ccosB+bcosC.
(1)求A的大小;
(2)求cosB+cosC的取值范圍.
【答案】
分析:(1)根據(jù)正弦定理與兩角和的正弦公式,將已知等式化簡(jiǎn),得2sinAcosA=sin(B+C),結(jié)合三角形內(nèi)角和定理與誘導(dǎo)公式,得2cosA-1=0,所以A=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025122933539726526/SYS201310251229335397265017_DA/0.png)
;
(2)因?yàn)锳=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025122933539726526/SYS201310251229335397265017_DA/1.png)
,結(jié)合B是銳角△ABC的內(nèi)角,可得B
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025122933539726526/SYS201310251229335397265017_DA/2.png)
.再將cosB+cosC化簡(jiǎn)整理為sin(B+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025122933539726526/SYS201310251229335397265017_DA/3.png)
),結(jié)合三角函數(shù)的圖象與性質(zhì),不難得到cosB+cosC的取值范圍.
解答:解:(1)∵2acosA=ccosB+bcosC
∴由正弦定理,得2sinAcosA=sinCcosB+sinBcosC=sin(B+C)
∵△ABC中,B+C=π-A,∴2sinAcosA=sinA,得sinA(2cosA-1)=0
∵A∈(0,π),得sinA>0,∴2cosA-1=0,得cosA=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025122933539726526/SYS201310251229335397265017_DA/4.png)
,得A=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025122933539726526/SYS201310251229335397265017_DA/5.png)
(2)∵B+C=π-A=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025122933539726526/SYS201310251229335397265017_DA/6.png)
,得C=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025122933539726526/SYS201310251229335397265017_DA/7.png)
-B,
∴cosB+cosC=cosB+cos(
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025122933539726526/SYS201310251229335397265017_DA/8.png)
-B)=cosB+cos
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025122933539726526/SYS201310251229335397265017_DA/9.png)
cosB+sin
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025122933539726526/SYS201310251229335397265017_DA/10.png)
sinB=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025122933539726526/SYS201310251229335397265017_DA/11.png)
cosB+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025122933539726526/SYS201310251229335397265017_DA/12.png)
sinB=sin(B+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025122933539726526/SYS201310251229335397265017_DA/13.png)
)
∵B是銳角△ABC的內(nèi)角,可得B
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025122933539726526/SYS201310251229335397265017_DA/14.png)
∴B+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025122933539726526/SYS201310251229335397265017_DA/15.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025122933539726526/SYS201310251229335397265017_DA/16.png)
,可得sin(B+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025122933539726526/SYS201310251229335397265017_DA/17.png)
)的最小值大于sin
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025122933539726526/SYS201310251229335397265017_DA/18.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025122933539726526/SYS201310251229335397265017_DA/19.png)
當(dāng)B=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025122933539726526/SYS201310251229335397265017_DA/20.png)
時(shí),sin(B+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025122933539726526/SYS201310251229335397265017_DA/21.png)
)有最大值為1
由此可得,cosB+cosC的取值范圍是(
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025122933539726526/SYS201310251229335397265017_DA/22.png)
,1].
點(diǎn)評(píng):本題在銳角△ABC中,求兩個(gè)角余弦和的取值范圍.著重考查了正弦定理、兩角和的正弦公式和三角函數(shù)的圖象與性質(zhì)等知識(shí),屬于基礎(chǔ)題.