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.
来源:学生作业帮 编辑:搜狗做题网作业帮 分类:数学作业 时间:2024/07/22 19:11:51
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.](/uploads/image/z/8387985-57-5.jpg?t=%E8%8B%A5sin%28%CF%80%2F6-%CE%B1%29%3D1%2F3%2C%E5%88%99cos%282%CF%80%2F3%2B2%CE%B1%29%E7%9A%84%E5%80%BC%3F%E8%AF%B7%E9%92%88%E5%AF%B9%E6%AF%8F%E4%B8%80%E6%AD%A5%E5%81%9A%E8%AF%B4%E6%98%8E%2C%E6%B8%85%E6%A5%9A%E4%B8%80%E7%82%B9.)
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.
已知(sinα-cosα)/(sinα+cosα)=1/3,则cos^4(π/3+α)-cos^4(π/6-α)的值为
解方程:X+1/2X+1/3X=55,最好每一步都说清楚
已知tan(π-α)=2,求sinα-2sinαcosα-cosα/4cosα-3sinα的值
若α满足sinα-2cosαsinα+3cosα=2,则sinα•cosα的值为 ___ .
已知(sinα+3cosα)/(3cosα-sinα)=5 则sin^2α-sinαcosα的值是
已知6sin^2α+sinαcosα-cos^2α=0,求sin(2α+π/3)的值
已知(3sinα+cosα)/(3cosα-sinα)=2,则2-3sin(α-3π)sin(1.5π-α)-[cos(
已知sinα+3cosα=2,求(sinα-cosα)/(sinα+cosα)的值
若sinα-sinβ=1-根号3/2,cosα-cosβ=-1/2,则cos(α-β)的值为
若sinα-sinβ=根号3/2,cosα-cosβ=1/2,则cos(α-β)的值为
若(sinα+cosα)/(sinα-cosα)=2 则sin(α-5π)·sin(3π/2-α)=?