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1、化简 (tan2π-θ)sin(-2π-θ)cos(6π-θ)/cos(θ-π)sin(5π+θ)
2、化简 sin(15π/2+α)cos(α-π/2)/sin(9π/2-α)cos(3π/2+α)
3、求证 sin(π/2+θ)-cos(π-θ)/sin(π/2-θ)-sin(π-θ)=2/1-tanθ
人气:352 ℃ 时间:2020-05-09 18:18:14
解答
(tan2π-θ)sin(-2π-θ)cos(6π-θ)/cos(θ-π)sin(5π+θ)=(-tanθ)(-sinθ)cosθ/(-cosθ)(-sinθ)=tanθsinθcosθ/cosθsinθ=tanθsin(15π/2+α)cos(α-π/2)/sin(9π/2-α)cos(3π/2+...
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