问题 解答题
(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.
答案

(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

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