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化简[sin(π-α)cos(3π-α)tan(-α-π)]/[tan(4π-α)sin(5π+α)]
人气:439 ℃ 时间:2020-04-04 22:49:16
解答
sin(π-a)cos(3π-a)tan(-π-a)tan(a-2π)/tan(4π-a)sin(5π+a)
=sinacosatana/tanasina
=cosa
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