题目内容
在△ABC中,已知
,
,求
的值.
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823120252935433.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823120252966520.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823120253013364.gif)
sinB=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823120253029255.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=
.………14分
(注:不讨论扣2分)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823120253060213.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823120253278213.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823120253278213.gif)
∴cos(B-A)=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823120253060213.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823120253341225.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823120253060213.gif)
=sinAcos(B-A)+cosAsin(B-A)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823120253419561.gif)
若cos(B-A)=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823120253341225.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823120253434553.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823120253029255.gif)
(注:不讨论扣2分)
![](http://thumb.zyjl.cn/images/loading.gif)
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