若sin(π/6-a)=1/3 则cos(2π/3+2a)=?
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若sin(π/6-a)=1/3 则cos(2π/3+2a)=?
若sin(π/6-a)=1/3 则cos(2π/3+2a)=?
若sin(π/6-a)=1/3 则cos(2π/3+2a)=?
sin(π/6-a)=cos[π/2-(π/6-a)]=cos(π/3+a)=1/3
则原式=cos[2(π/3+a)]
=2cos²(π/3+a)-1
=-7/9
cos(2π/3+2a)
=-cos[π-(2π/3+2a)]
=-cos(π/3-2a)
=-cos[2(π/6-a)]
=-[1-2sin²(π/6-a)]
=-[1-2×(1/3)²]
=-(1-2/9)
=-7/9
sin(π/6-a)=cos[π/2-(π/6-a)]=cos(π/3+a)1/3
cos(2π/3+2a)=cos2(π/3+a)=2[cos(π/3+a)]^2-1=-7/9
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
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