若sin(π/6-α)=1/3,则cos(2π/3+2α)的值?请针对每一步做说明,清楚一点.

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若sin(π/6-α)=1/3,则cos(2π/3+2α)的值?请针对每一步做说明,清楚一点.

若sin(π/6-α)=1/3,则cos(2π/3+2α)的值?请针对每一步做说明,清楚一点.
若sin(π/6-α)=1/3,则cos(2π/3+2α)的值?请针对每一步做说明,清楚一点.

若sin(π/6-α)=1/3,则cos(2π/3+2α)的值?请针对每一步做说明,清楚一点.
cos(2π/3+2α)= - cos(π-(2π/3+2α))
=- cos(π/3-2α)
=- cos(2α-π/3) (公式cos2α=1-2(sinα)^2)
=- (1-2(sin(α-π/6))^2)
因为 sin(π/6-α)=1/3 所以(sin(α-π/6))^2=1/9
故cos(2π/3+2α)= -(1-2*1/9)
=-7/9

π=180°,然后你可以用计算器算

sin(π/6-α)=1/3,
cos(π/3-2α)=1-2sin(π/6-α)sin(π/6-α)=1-2/9=7/9
cos(2π/3+2α)=-cos(π-π/3+2α)=-7/9