若sin(π/6-α)=1/3,则cos(2π/3+2α)的值为?好像要运用巧妙换算的,希望写明解题步骤
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若sin(π/6-α)=1/3,则cos(2π/3+2α)的值为?好像要运用巧妙换算的,希望写明解题步骤若sin(π/6-α)=1/3,则cos(2π/3+2α)的值为?好像要运用巧妙换算的,希望写明
若sin(π/6-α)=1/3,则cos(2π/3+2α)的值为?好像要运用巧妙换算的,希望写明解题步骤
若sin(π/6-α)=1/3,则cos(2π/3+2α)的值为?
好像要运用巧妙换算的,希望写明解题步骤
若sin(π/6-α)=1/3,则cos(2π/3+2α)的值为?好像要运用巧妙换算的,希望写明解题步骤
cos(2π/3+2α)
=-cos[π-(2π/3+2α)]
=-cos(π/3-2α)
=-cos[2(π/6-α)]
=-{1-2[sin(π/6-α)]^2}
=-(1-2/9)
=-7/9
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