题目内容
(1) 若cos(75°+α)=
,(-180°<α<-90°),求sin(105°-α)+cos(375°-α)值;
(2) 在△ABC中,若sinA+cosA=-
,求sinA-cosA,tanA的值.
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(2) 在△ABC中,若sinA+cosA=-
| 7 |
| 13 |
(1)sin(105°-α)=sin[180°-(75°+α)]=sin(75°+α)
∵-180°<α<-90°
∴-105°<75°+α<-15°又cos(75°+α)=
>0
∴-90°<75°+α<-15°
∴sin(75° +α)=-
cos(375°-α)=cos(15°-α)=cos[90° -(75°+α)]=sin(75°+α)=-
∴原式=-
(2)由sinA+cosA=-
两边平方得1+2sinAcosA=
而0<A<π2sinAcosA=-
<0
∴
<A<π
∴1-2sinAcosA=
即(sinA-cosA)2=(
)2
又sinA-cosA>0sinA-cosA=
∴
∴tanA=-
∵-180°<α<-90°
∴-105°<75°+α<-15°又cos(75°+α)=
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| 5 |
∴-90°<75°+α<-15°
∴sin(75° +α)=-
| 4 |
| 5 |
cos(375°-α)=cos(15°-α)=cos[90° -(75°+α)]=sin(75°+α)=-
| 4 |
| 5 |
∴原式=-
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(2)由sinA+cosA=-
| 7 |
| 13 |
| 49 |
| 169 |
而0<A<π2sinAcosA=-
| 120 |
| 169 |
∴
| π |
| 2 |
∴1-2sinAcosA=
| 289 |
| 169 |
即(sinA-cosA)2=(
| 17 |
| 13 |
又sinA-cosA>0sinA-cosA=
| 17 |
| 13 |
∴
|
∴tanA=-
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