求证[tan(2π-α)sin(-2π-α)cos(6π-α)]/[sin(α+3π/2)cos(α+3π/2)]=-tanα
人气:126 ℃ 时间:2020-04-20 01:59:13
解答
这都是诱导公式tan(2π-α)=tan(-α)=-tanαsin(-2π-α)=sin(-α)=-sinαcos(6π-α)=cos(-α)=cosαsin(α+3π/2)=-cosαcos(α+3π/2)=sinα所以原式=(-tanα)(-sinα)cosα/[(-cosα)sinα]=-tanα
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