Question

已知-＜x＜0,sinx+cosx=.

(1)求sinx-cosx的值；

（2）求的值.

Answer 1

（1）sinx-cosx=-（2）

解析:

（1）方法一  聯(lián)立方程：

                                                                                     2分

由①得sinx=-cosx,將其代入②，整理得

25cos²x-5cosx-12=0.                                                                                                               4分

∵-＜x＜0,

∴,

所以sinx-cosx=-.                                                                                     7分

方法二  ∵sinx+cosx=,

∴(sinx+cosx)²=,

即1+2sinxcosx=,

∴2sinxcosx=-.                                                                                       2分

∵(sinx-cosx)²=sin²x-2sinxcosx+cos²x

=1-2sinxcosx=1+=                                                                                        ①              4分

又∵-＜x＜0,∴sinx＜0,cosx＞0,

∴sinx-cosx＜0                                                                                                      ②

由①②可知：sinx-cosx=-.                                                                       7分

(2)由已知條件及（1）可知

,解得,                                                                                             9分

∴tanx=-.                                                                                                                 11分

又∵

=

=                                                                                                                       13分

=.                                                                                         14分