已知-π2<x<0,sinx+cosx=15.(1)求sinx-cosx的值;(2)求sin2x+2cos2x1+tanx的值.-数学

题目简介

已知-π2<x<0,sinx+cosx=15.(1)求sinx-cosx的值;(2)求sin2x+2cos2x1+tanx的值.-数学

题目详情

已知-
π
2
<x<0
sinx+cosx=
1
5

(1)求sinx-cosx的值;
(2)求
sin2x+2cos2x
1+tanx
的值.
题型:解答题难度:中档来源:福建

答案

(1)把sinx+cosx=class="stub"1
5
两边平方得1+2sinxcosx=class="stub"1
25
,有sin2x=-class="stub"24
25

(sinx-cosx)2=1-2sinxcosx=class="stub"49
25

-class="stub"π
2
<x<0
,得sinx<0,cosx>0,sinx-cosx<0,
sinx-cosx=-class="stub"7
5

(2)由sinx+cosx=class="stub"1
5
sinx-cosx=-class="stub"7
5
,得sin2x-cos2x=-class="stub"7
25

cos2x=cos2x-sin2x=class="stub"7
25

又由sinx+cosx=class="stub"1
5
sinx-cosx=-class="stub"7
5
解得sinx=-class="stub"3
5
,cosx=class="stub"4
5
,有tanx=-class="stub"3
4

class="stub"sin2x+2cos2x
1+tanx
=
-class="stub"24
25
+class="stub"14
25
1+(-class="stub"3
4
)
=-class="stub"8
5

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