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求证:[tan(2π-α)cos(3π/2-α)cos(6π-α)]/[sin(α+3π/2)cos(α+3π/2)]
求证:[tan(2π-α)cos(3π/2-α)cos(6π-α)]/[sin(α+3π/2)cos(α+3π/2)]=-tanα
人气:271 ℃ 时间:2020-05-13 11:59:31
解答
证明:左边=[-tanα*cos(-π/2-α)cos(-α)]/[sin(α-π/2)cos(α-π/2)]
=[-tanα*cos(π/2+α)*cosα]/[-sin(π/2-α)cos(π/2 -α)]
=[tanα*(-sinα)*cosα]/(cosα*sinα)
=-tanα
=右边
等式得证.
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