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化简:sin(π/8)Cos(π/8)=?

如图所示

已知sinαcosα=1/8,且π/4<α<π/2 (1)求cosα-sinα的值 (2)求co

sin105°-cos105° =√2(sin105°cos45°-cos105°sin45°)

奇变偶不变,符号看象限

-cos(π/16)sin(π/16)cos(π/8) =(-1/2)2cos(π/16)sin

cos(nπ/4)的周期是2π/(π/4)=8 sin(nπ/8)的周期是2π/(π/8)=16

cos(-5π/6)是-√3/2,sin(-5π/6)是-1/2。 解:cos(-5π/6)=cos

cos(nπ/4)的周期是2π/(π/4)=8 sin(nπ/8)的周期是2π/(π/8)=16 c

解: ∵0<θ<π ∴0<θ/2<π/2 ∴cosθ/2>0 于是 原式=[2sin(θ/2

[sin(π/8)]^2-[cos(π/8)]^2 =-cos(π/4) =-√2/2

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