若cosα=1/5α属于(0.π/2)则sin(α-π/3)求答案

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若cosα=1/5α属于(0.π/2)则sin(α-π/3)求答案
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若cosα=1/5α属于(0.π/2)则sin(α-π/3)求答案
若cosα=1/5α属于(0.π/2)则sin(α-π/3)
求答案

若cosα=1/5α属于(0.π/2)则sin(α-π/3)求答案
cosα=1/5α属于(0.π/2)
sinα=√1-cos平方α=2√6/5
所以
sin(α-π/3)
=sinαcosπ/3-cosαsinπ/3
=2√6/5·1/2-1/5·√3/2
=(2√6-√3)/10

由cosα=1/5和α属于(0.π/2)得:sinα=2√ 6/5,
所以:sin(α-π/3)=(sinα)/2-(√3cosα)/2=√ 6/5-3√2/5

cosα=1/5
那么sinα=(2*根号6)/5
sin(α-π/3)=sinα*cosπ/3-cosα*sinπ/3
=[(2*根号6)/5]*1/2+(1/5)*(根号3)/2
=(根号6)/5+(根号3)/10
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