-(cos^2π/12-sin^2π/12)为什么等于-(cosα+π/12)
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(1)sin(-5π/12) (2)cos(-61π/12)
由cos(a+b)=cos a cos b-sin a sin b cos(a-b)=cos a cos b+sin a sin b解题设a为锐角,证:1、2分之根3乘cos a + 2分之1乘sin a=cos(6分之π-a)2、cos a-sin a=根号2cos(4分之π+a)
cosαcos[(2k+1)π]-sinαsin[(2k+1)π]为什么等于-cosα
(1+sinα+cosα)*[sin(α/2)-cos(α/2)]/√(2+2cosα)化简 (3/2*π
设角a=-35π/6,则2sin(π+α)cos(π-α)-cos(π+α)/ 1+sin^2α+sin(π-α)-cos^2(π+α)sina=sin(-35π/6)=sin(-6π+π/6)=1/2 cosα=根号3/22sin(π+α)cos(π-α)-cos(π+α)/ [1+sin^2α+sin(π-α)-cos^2(π+α)]=2(-sinα)(-cosα)+cosα/[1+sin²
θ∈(0,π/2),比较cosθ、sin(cosθ)、cos(sinθ)的大小
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求证:sin^2/(sin-cos) - (sin+cos)/(tan^2 -1) =sin+cos
sin(α-β)sin(β-r)-cos(α-β)cos(r-β) 1.sin(α-β)sin(β-r)-cos(α-β)cos(r-β)2.( tan4分之5π+tan12分之5π)/(1-tan12分之5π)3.[ sin(α+β)-2sinαcosβ]/2sinαsinβ+cos(α+β)
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求值:已知sinθ+cosθ=1/5,已知θ∈(0,π).求(1)sinθ*cosθ(2)sinθ-cosθ(3)tanθ
化简:sin(-a)cos(2π+a)sin(-a-π)
cos(a+π/2)=sin?
若cosα>sinα,(-π/2
cos^6(π/8)-sin^6(π/8)=求值,[cos^2(π/8)-sin^2(π/8)][cos^4(π/8)+sin^4(π/8)+cos^2(π/8)sin^2(π/8)]=cos(π/4)[1-cos^2(π/8)sin^2(π/8)]=cos(π/4)[1-[sin^2(π/4)]/4]==7√2/16