化简sin²(α+π)·cos(π+α)·cot(-α-2π)/tan(π+α)·cos²(-α-π)求答案
人气:445 ℃ 时间:2019-10-25 09:56:41
解答
[sin²(α+π)·cos(π+α)·cot(-α-2π)]/[tan(π+α)·cos²(-α-π)]=[sin²α(-cosα)(-cotα)]/[tanαcos²α]=(sin²αcosα*cosα/sinα)/(sinα/cosα*cos²α) =cosα您好,答案上的得数是-1,您确定是对的吗?对于你输入的题目,我的是正解
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