已知cos2α/sin(α-π/4)=-√2/5(1)求cosα+sinα的值;(2)若-π

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已知cos2α/sin(α-π/4)=-√2/5(1)求cosα+sinα的值;(2)若-π已知cos2α/sin(α-π/4)=-√2/5(1)求cosα+sinα的值;(2)若-π已知cos2α/

已知cos2α/sin(α-π/4)=-√2/5(1)求cosα+sinα的值;(2)若-π
已知cos2α/sin(α-π/4)=-√2/5
(1)求cosα+sinα的值;
(2)若-π

已知cos2α/sin(α-π/4)=-√2/5(1)求cosα+sinα的值;(2)若-π
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

cosa+sina=1/5
sinα-cosα
=√2sin(a-π/4)

sα的

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

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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

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