(1)若cos(75°+α)=35,(-180°<α<-90°),求sin(105°-α)+cos(375°-α)值;(2)在△ABC中,若sinA+cosA=-713,求sinA-cosA,tanA

题目简介

(1)若cos(75°+α)=35,(-180°<α<-90°),求sin(105°-α)+cos(375°-α)值;(2)在△ABC中,若sinA+cosA=-713,求sinA-cosA,tanA

题目详情

(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.
题型:解答题难度:中档来源:不详

答案

(1)sin(105°-α)=sin[180°-(75°+α)]=sin(75°+α)
∵-180°<α<-90°
-105°<75°+α<-15°又cos(75°+α)=class="stub"3
5
>0

∴-90°<75°+α<-15°
sin(7 +α)=-class="stub"4
5

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

∴原式=-class="stub"8
5

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

而0<A<π2sinAcosA=-class="stub"120
169
<0

class="stub"π
2
<A<π

1-2sinAcosA=class="stub"289
169

(sinA-cosA)2=(class="stub"17
13
)2

又sinA-cosA>0sinA-cosA=class="stub"17
13

sinA=class="stub"5
13
cosA=-class="stub"12
13

tanA=-class="stub"5
12

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