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.
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
再问: )= - cos(π-(2π/3+2α)) =- cos(π/3-2α) 这一步是如何变化过来的 =- cos(2α-π/3) 这一步是如何变化过来的
再答: cos(2π/3+2α)= - cos(π-(2π/3+2α)) //注:直接将括号内的展开就可以得到下面的 =- cos(π/3-2α) //注:根据公式cos(-β)=cosβ,这里有cos(π/3-2α)=cos(-(2α-π/3))=cos(2α-π/3) =- 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
=- 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
再问: )= - cos(π-(2π/3+2α)) =- cos(π/3-2α) 这一步是如何变化过来的 =- cos(2α-π/3) 这一步是如何变化过来的
再答: cos(2π/3+2α)= - cos(π-(2π/3+2α)) //注:直接将括号内的展开就可以得到下面的 =- cos(π/3-2α) //注:根据公式cos(-β)=cosβ,这里有cos(π/3-2α)=cos(-(2α-π/3))=cos(2α-π/3) =- 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
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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