在△ABC中.
(1)已知sinA=cosBcosC,求證:tanC+tanB=1;
(2)求證:a2-2ab cos(60°+C)=b2-2bc cos(60°+A).
【答案】
分析:(1)根據(jù)A=B+C把sinA轉(zhuǎn)換成sin(A+B),進(jìn)而利用兩角和公式化簡整理,等式兩邊同時(shí)除以cosBcosC,即可證明原式.
(2)先利用兩角和公式對(duì)要證的結(jié)論化簡整理可得a
2-abcosC+ab
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214258654070204/SYS201310232142586540702018_DA/0.png)
sinC=c
2-bccosA+bc
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214258654070204/SYS201310232142586540702018_DA/1.png)
sinA 再利用余弦定理分別把cosC,cosA代入整理asinC=csinA,根據(jù)正弦定理可知在三角形中此等式恒成立,進(jìn)而使原式得證.
解答:解:(1)因?yàn)樵谌切蜛BC中,sinA=cosBcosC
∴sin(B+C)=cosBcosC
即sinBcosC+cosBsinC=cosBcosC
等式兩邊同時(shí)除以cosBcosC,得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214258654070204/SYS201310232142586540702018_DA/2.png)
+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214258654070204/SYS201310232142586540702018_DA/3.png)
=1
即tanB+tanC=1,原式得證.
(2)證明:要使a
2-2ab cos(60°+C)=b
2-2bc cos(60°+A).
需a
2-2ab(
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214258654070204/SYS201310232142586540702018_DA/4.png)
cosC-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214258654070204/SYS201310232142586540702018_DA/5.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214258654070204/SYS201310232142586540702018_DA/6.png)
sinC)=c
2-2bc(
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214258654070204/SYS201310232142586540702018_DA/7.png)
cosA-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214258654070204/SYS201310232142586540702018_DA/8.png)
sinA)
需a
2-abcosC+ab
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214258654070204/SYS201310232142586540702018_DA/9.png)
sinC=c
2-bccosA+bc
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214258654070204/SYS201310232142586540702018_DA/10.png)
sinA
需a
2-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214258654070204/SYS201310232142586540702018_DA/11.png)
(a
2+b
2-c
2)+ab
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214258654070204/SYS201310232142586540702018_DA/12.png)
sinC=c
2-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214258654070204/SYS201310232142586540702018_DA/13.png)
(b
2+c
2-a
2)+bc
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214258654070204/SYS201310232142586540702018_DA/14.png)
sinA
需a
2-b
2+c
2+2ab
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214258654070204/SYS201310232142586540702018_DA/15.png)
sinC=c
2-b
2+a
2+2bc
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214258654070204/SYS201310232142586540702018_DA/16.png)
sinA
需asinC=csinA
在三角形ABC中,根據(jù)正弦定理可知
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214258654070204/SYS201310232142586540702018_DA/17.png)
即asinC=csinA恒成立,
所以等式得證
點(diǎn)評(píng):本題主要考查了三角函數(shù)恒等式的證明,涉及了正弦定理,余弦定理,同角三角函數(shù)基本關(guān)系的應(yīng)用等.考查了學(xué)生綜合分析問題和演繹推理的能力.