若sin(π/6-α)=1/3,则cos(2π/3+2α)=
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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
已知6sin²α+sinαcosα-2cos²α=0,α∈(π/2,π),求值1).(sinα-3cosα)/(sinα-cosα);2).sinαcosα-sin²α;3).sin²α-3cosαsinα-2
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cos(π/3-α)=1/8,则cosα+(根号3)sinα的值为____已知cosα-cosβ=1/2,sinα-sinβ=1/3,求cos(α+β)的值
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若sin(π/6-α)=1/3,则cos(2π/3+2α)=
若sin(π/6-α)=1/3 则cos(2π/3+2α)等于多少
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若cos(α-β)=1/3,则(sinα+sinβ)^2+(cosα+cosβ)^2=?
若f(sinα+cosα)=sinαcosα,则f(cosπ/6)等于..麻烦讲清楚过程,辛苦各位了.
已知sin(π+α)=3cos(π+α),则sin(π-α)cos(2π-α)/2sin²α-1的值
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