∵cos(π/8-α)=√3/3 ∴π/8-α=±π/6+2kπ,k∈Z α=7π/24+2kπ或-π/24+2kπ,k∈Z 当α=7π/24+2kπ时,原式=sin^2(7π/24+2kπ-π/6)-cos(5π/6+7π/24+2kπ) =sin^2π/8-cos9π/8 =sin^2π/8+cosπ/8 =1-√2/2+√(2+√2)/2 当α=-π/24+2kπ时,原式=sin^2(-π/24+2kπ-π/6)-cos(5π/6-π/24+2kπ) =sin^25π/24-cos19π/24 =sin^2π/24+cos5π/24 (PS:不懂追问)
cos(八分之π-α)=三分之根号三,则sin(α-六分之π)的平方-cos(六分之5π+α)的值 帮我下
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