若sin((π/6)-α)=1/3,则cos((2π/3)+2α)等于多少?
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若sin((π/6)-α)=1/3,则cos((2π/3)+2α)等于多少?
![若sin((π/6)-α)=1/3,则cos((2π/3)+2α)等于多少?](/uploads/image/z/17883763-43-3.jpg?t=%E8%8B%A5sin%EF%BC%88%EF%BC%88%CF%80%EF%BC%8F6%EF%BC%89%EF%BC%8D%CE%B1%EF%BC%89%3D1%EF%BC%8F3%2C%E5%88%99cos%EF%BC%88%EF%BC%882%CF%80%EF%BC%8F3%EF%BC%89%2B2%CE%B1%EF%BC%89%E7%AD%89%E4%BA%8E%E5%A4%9A%E5%B0%91%3F)
sin((π/6)-α)=1/3,
而sin((π/6)-α)=cos[π/2-((π/6)-α))]
= cos((π/3)+α),
∴cos((π/3)+α)=1/3.
所以cos((2π/3)+2α)= cos[2((π/3)+α)]
= 2cos²((π/3)+α)-1
=-7/9.
而sin((π/6)-α)=cos[π/2-((π/6)-α))]
= cos((π/3)+α),
∴cos((π/3)+α)=1/3.
所以cos((2π/3)+2α)= cos[2((π/3)+α)]
= 2cos²((π/3)+α)-1
=-7/9.
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