题目内容

(1) 若cos(75°+α)=
3
5
,(-180°<α<-90°)
,求sin(105°-α)+cos(375°-α)值;
(2) 在△ABC中,若sinA+cosA=-
7
13
,求sinA-cosA,tanA的值.
(1)sin(105°-α)=sin[180°-(75°+α)]=sin(75°+α)
∵-180°<α<-90°
-105°<75°+α<-15°又cos(75°+α)=
3
5
>0

∴-90°<75°+α<-15°
sin(7 +α)=-
4
5

cos(375°-α)=cos(15°-α)=cos[9 -(75°+α)]=sin(75°+α)=-
4
5

∴原式=-
8
5

(2)由sinA+cosA=-
7
13
两边平方得1+2sinAcosA=
49
169

而0<A<π2sinAcosA=-
120
169
<0

π
2
<A<π

1-2sinAcosA=
289
169

(sinA-cosA)2=(
17
13
)2

又sinA-cosA>0sinA-cosA=
17
13

sinA=
5
13
cosA=-
12
13

tanA=-
5
12
练习册系列答案
相关题目

违法和不良信息举报电话:027-86699610 举报邮箱:58377363@163.com

精英家教网