题目内容
在△ABC中,已知
,
,求
的值.
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823133533471433.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823133533487520.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823133533502364.gif)
sinB=![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823133533518255.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823133533518255.gif)
在△ABC中,cosA=
,∴sinA=
.又sin(B-A)=
,∴ 0<B-A<π.
∴cos(B-A)=
,或cos(B-A)=
.……6分若cos(B-A)=
,则sinB=sin[A+(B-A)]
=sinAcos(B-A)+cosAsin(B-A)
.…12分若cos(B-A)=
,
则sinB=sin[A+(B-A)]=sinAcos(B-A)+cosAsin(B-A)
(舍去).
综上所述,得sinB=![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823133533518255.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823133533534213.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823133533534213.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823133533534213.gif)
∴cos(B-A)=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823133533534213.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823133533596225.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823133533534213.gif)
=sinAcos(B-A)+cosAsin(B-A)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823133533627561.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823133533596225.gif)
则sinB=sin[A+(B-A)]=sinAcos(B-A)+cosAsin(B-A)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823133533658553.gif)
综上所述,得sinB=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823133533518255.gif)
![](http://thumb.zyjl.cn/images/loading.gif)
练习册系列答案
相关题目