已知cos(α+4分之π)=5分之4,则sin2α等于?
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已知cos(α+4分之π)=5分之4,则sin2α等于?
已知cos(α+4分之π)=5分之4,则sin2α等于?
已知cos(α+4分之π)=5分之4,则sin2α等于?
cos(α+π/4)=sin[π/2-(α+π/4)]=sin(π/4-α)=4/5
sin(2α)=cos(π/2 -2α)
=cos[2(π/4 -α)]
=1-2sin²(π/4 -α)
=1-2×(4/5)²
=-7/25
cos(α+4分之π)=5分之4
cos[2(α+4分之π)]=2[cos(α+4分之π)]^2-1
cos(2a+2分之π)=2*(5分之4)^2-1
sin2a=25分之32-1
sin2a=25分之7
就是这样
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解1:∵(π/4-α)+(π/4+α)=π/2
∴sin(π/4-α)=cos(π/4+α)=cos(α+π/4) = 4/5
∴sin(π/4-α)=4/5
∵sin2α=cos(π/2-α)
=cos2(π/4-α)
=1-2sin...
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解1:∵(π/4-α)+(π/4+α)=π/2
∴sin(π/4-α)=cos(π/4+α)=cos(α+π/4) = 4/5
∴sin(π/4-α)=4/5
∵sin2α=cos(π/2-α)
=cos2(π/4-α)
=1-2sin ²(π/4-α)
=1-2×(4/5)²
=﹣7/25
∴sin2α =﹣7/25
解2:∵cos(α+π/4) = 4/5
∴cosα cos(π/4)-sinα sin(π/4)= 4/5
∴√2/2(cosα-sinα )= 4/5
∴1/2(cos²α-2cosα sinα+sin²α )= ( 4/5)² (上式两边平方)
∵cos²α+sin²α=1 , 2cosα sinα=sin2α
∴1-sin2α=32/25
∴sin2α=﹣7/25
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