[cos(α-π)*cos(19/2*π-α)]/[sin(π/2-α)* tan[(2k+1)π+α]]
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/07/17 13:18:49
[cos(α-π)*cos(19/2*π-α)]/[sin(π/2-α)* tan[(2k+1)π+α]]
cos(α - π) = cos(π - α) = -cosα
cos(19π/2 - α) = cos[(20 - 1)π/2 - α] = cos(10π - π/2 - α) = cos(π/2 + α) = -sinα
sin(π/2-α) = cosα
tan[(2k+1)π+α] = tanα
原式 = -cosα(-sinα)/(cosαtanα) = sinαcosα/(cosα*sinα/cosα) = sinαcosα/sinα = cosα
cos(19π/2 - α) = cos[(20 - 1)π/2 - α] = cos(10π - π/2 - α) = cos(π/2 + α) = -sinα
sin(π/2-α) = cosα
tan[(2k+1)π+α] = tanα
原式 = -cosα(-sinα)/(cosαtanα) = sinαcosα/(cosα*sinα/cosα) = sinαcosα/sinα = cosα
[cos(α-π)*cos(19/2*π-α)]/[sin(π/2-α)* tan[(2k+1)π+α]]
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