化简sin(2π-α)*tan(π+α)*cot(-α-π)/cos(π-α)*tan(3π-α)(急)
化简sin(2π-α)*tan(π+α)*cot(-α-π)/cos(π-α)*tan(3π-α)(急)
cot(-α-π)这部希望能说明下
cot(-α-π)这部希望能说明下
数学人气:253 ℃时间:2019-12-15 11:53:39
优质解答
sin(2π-α)=sin(-α)=-sinαtan(π+α)=tanαcot(-α-π)=-cot(α+π)=-cotαcos(π-α)=-cosαtan(3π-α)=tan(-α)=-tanα所以原式=sinαtanαcotα/cosαtanα=sinα/(cosα*sinα/cosα)=sinα/sinα=1
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