化简[tan(π+α)cos(π+α)sin²(3π+α)]/[tan²α*cos³(-π-α)]
来源:学生作业帮助网 编辑:作业帮 时间:2024/09/16 08:40:31
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cosαtan(2π+α)化简
sin3(-α)cos(2π+α)tan(- 2- π)
化简:sin²(α+π)cos(-α+π)/tan(α+π)tan(α+2π)cos²(-α-π)1.cos(27°+α)cos(33°-α)-sin(27°+α)sin(33°-α) 2.sin²(α+π)cos(-α+π)/tan(α+π)tan(α+2π)cos²(-α-π)
求证:1/cosα-tanα=1/tan(π/4+α/2)
已知(1+tanα)(1+tanβ)=4cos(π/3)0
sin(2π-α)tan(α+π)cos(α-π)/cos(α+π)tan(3π-α)化简 快
化简:tan(π-α)·sin²(α+π/2)·cos(2π-α)/cos³(-α-π)·tan(α-2π)
化简:(1)sin(-α+180°)-tan(-α)+tan(-α+360°)/tan(α+180°)+cos(-α)+cos(α+180°)(2)sin^2(α+π)cos(-α+π)/tan(α+π)+tan(α+2π)cos^3(-α-π)
化简[tan(π+α)cos(π+α)sin²(3π+α)]/[tan²α*cos³(-π-α)]
化简sin(9π+α)tan(3/2π+α)/cos(-4π-α)tan(15π+α)tan(7/2π-α)
[sin(2π+α)·tan(-α)·cos(-α)]÷[sin(-α)·cos(2π+α)·tan(2π+α)化简
数学同角三角比的化简1.(sin(π+α)-tan(-α)-tan(π+α))/(tan(α-π)+cos(-α)+cos(π+α))
化简:4.化简:(2cos²α-1)/{[2tan((π/4)- α)]}*{cos³[(π/4)- α]}
化简:(2cos²α-1)/[2tan((π/4)- α)]*{cos³[(π/4)- α],
求证:[tan(2π-α)cos(3π/2-α)cos(6π-α)]/[sin(α+3π/2)cos(α+3π/2)]=-tanα
化简sin(π+α)/cos(-α)tan(2π-α)等于什么?
①化简1+cos(π/2 +α)*sin(π/2 -α)*tan(π+α)②计算(1)tan(π/5)+tan(2π/5)+tan(3π/5)+tan(4π/5) (2)sin(-60°)+cos(225°)+tan135°(3)cos(π/5)+cos(2π/5)+cos(3π/5)+cos(4π/5)(4)tan10°+tan170°+sin1866°-sin(-606°)
化简sin³(﹣α)cos(2π+α)tan(2π-α)/sin(α-2π)cos(α-3π/2)-tan(π-α)tan(3π/2-α)