题目内容
已知θ是第二象限角,![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024185003028702625/SYS201310241850030287026005_ST/0.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024185003028702625/SYS201310241850030287026005_ST/1.png)
A.7
B.-7
C.
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024185003028702625/SYS201310241850030287026005_ST/2.png)
D.
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024185003028702625/SYS201310241850030287026005_ST/3.png)
【答案】分析:由
,θ是第二象限角,可求cosθ,从而可求sin(
)与cos(
),
可求.
解答:解:∵
,θ是第二象限角,∴cosθ=
,
∴sin(
)=
,
cos(
)=
,
∴
.
故选A.
点评:本题考查两角和与差的正切函数,着重考查学生掌握两角和与差的三角函数公式的能力,属于中档题.
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024185003028702625/SYS201310241850030287026005_DA/0.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024185003028702625/SYS201310241850030287026005_DA/1.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024185003028702625/SYS201310241850030287026005_DA/2.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024185003028702625/SYS201310241850030287026005_DA/3.png)
解答:解:∵
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024185003028702625/SYS201310241850030287026005_DA/4.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024185003028702625/SYS201310241850030287026005_DA/5.png)
∴sin(
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024185003028702625/SYS201310241850030287026005_DA/6.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024185003028702625/SYS201310241850030287026005_DA/7.png)
cos(
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024185003028702625/SYS201310241850030287026005_DA/8.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024185003028702625/SYS201310241850030287026005_DA/9.png)
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024185003028702625/SYS201310241850030287026005_DA/10.png)
故选A.
点评:本题考查两角和与差的正切函数,着重考查学生掌握两角和与差的三角函数公式的能力,属于中档题.
![](http://thumb.zyjl.cn/images/loading.gif)
练习册系列答案
相关题目