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(求步骤)化简:[sin(-α-3/2π)cos(3/2π-α)·tan2(π-α)]/[cos(π/2-α)·sin(π/2+α)]
人气:346 ℃ 时间:2020-05-09 18:19:48
解答
[sin(-α-3/2π)cos(3/2π-α)·tan2(π-α)]/[cos(π/2-α)·sin(π/2+α)]
=[cosa(-sina)(-tan2a)]/[sina(-cosa)]
=-tan2a
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