Question

已知cos2α/sin(α-π/4)=-√2/5

(1)求cosα+sinα的值;
(2)若-π<α<0,求sinα-cosα的值.

Answer 1

cos2α/sin(α-π/4)=-√2/5
(cos²a-sin²a)/[(√2/2)(sina-cosa)]=-√2/5
-(cosa-sina)(cosa+sina)/[(√2/2)(cosa-sina)]=-√2/5
-√2(cosa+sina)=-√2/5
得
cosa+sina=1/5

sinα-cosα
=√2sin(a-π/4)
-π<α<0
所以 -5π/4<a-π/4<-π/4
所以 √2sin(a-π/4)的取值范围为 [-√2,1]
sina-cosa
=±√[2-(sina+cosa)²]
=±√(2-(1/25))
=±√(49/25)
=±7/5
因为 sina-cosa∈【-√2,1】
所以
sina-cosa=-7/5

Answer 2

cos2α/sin(α-π/4)=-√2/5
cos2α=-√2/5sin(α-π/4)
(cos²α-sin²α)=-√2/5[sinαcos(π/4)-cosαsin(π/4)]=(cosα-sinα)/5
(cosα+sinα)(cosα-sinα)=(cosα-sinα)/5
所以 cosα+sinα=1/5 (1)
平方
1+2sinαcosα=1/25
2sinαcosα=-24/25<0
-π<α<0, 所以 -π/2<α<0,
(cosα-sinα)²=1-2sinαcosα=49/25
因为 -π/2<α<0,
cosα-sinα>0
cosα-sinα=7/5
sinα-cosα=-7/5

Answer 3

sα的