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已知Cos(α%β)=%5/13,Cosβ=4/5,α∈(π/2,π),β∈(0,π/2)...

sinα=3/5,α∈(π/2,π),则cosα=-4/5 cosβ=-5/13,β∈(π,3/2π),则sinβ=-12/13 cos(α-β)=cosαcosβ+sinαsinβ=4/5×5/13-3/5×12/13=-16/65 麻烦采纳,谢谢!

a∈(0,π),b∈(0,π),a+b∈(0,2π),a/2∈(0,π/2), sin(a+b)=5/13>0,所以 a+b∈(0,π) sina=2tan(a/2)/(1+tan²(a/2))=2*(1/2)/(1+1/4)=4/5 cosa=(1-tan²(a/2))/(1+tan²(a/2))=(1-1/4)/(1+1/4)=-3/5 所以a∈(π/2,π),a+b∈(π/2,π) sin²x+c...

∵α∈(0,π) α/2∈(0,π/2), ∵sinα/2=√5/5,,cosα =1-2sin²α/2=3/5,sinα=4/5 ∴ sin2α=2sinαcosα=24/25 ∵cos(α+β)=5/13,∴sin(α+β)=12/13 sinβ=sin[(α+β)-α]=sin(α+β)cosα-cos(α+β)sinα =12/13×3/5-5/13×4/5=16/65

解如图。

α-π/4∈(0,π/2), cos(α-π/4)=3/5, sin(α-π/4)=4/5 3π/4+β∈(3π/4,π) sin(3π/4+β)=5/13 cos(3π/4+β)=-12/13 sin(α+β)=-cos(α+β+π/2)=-cos(α-π/4+3π/4+β) =-[cos(α-π/4)cos(3π/4+β)-sin(α-π/4)sin(3π/4+β)] =-[3/5*(-12/13)-4/5*5/13] 剩下就是...

sin(2a)=sin(a+b+a-b)=sin(a+b)cos(a-b)+sin(a-b)cos(a+b) 因为cos(a-b)=12/13,且π/2

cos[(π/4-a)-(3π/4+β)] =cos(-π/2-a-β) =cos(π/2+a+β) =sin(a+β) sin(a+β)=(3/5)(12/13)+(5/13)(4/5) =56/65 2. cos(π/4-α)=3/5 π/4

sinα =sin[(α+β)-β]=sin(α+β)cosβ - cos(α+β)sinβ =12/13 *4/5 - 5/13*3/5 =33/65 注意:以上把α、β当做锐角,若给出角的范围,应视具体情况作答。

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