α是第三象限角,cos(α-3π/2)=1/5,求[sin(α-π/2)]cos(3π/2+α)tan(π-α)] / tan(-a-π)sin(-π-α)的值

α是第三象限角,cos(α-3π/2)=1/5,求[sin(α-π/2)]cos(3π/2+α)tan(π-α)] / tan(-a-π)sin(-π-α)的值
数学人气:612 ℃时间:2020-04-20 00:50:43
优质解答
cos(α-3π/2)=-cos(α-3π/2+π)=-cos(α-π/2)=-cos(π/2-α)=-sinα=1/5,所以sinα=-1/5.α是第三象限角,所以cosα=-√[1-(sinα)^2]=-(2√6)/5sin(α-π/2)=-sin(π/2-α)=-cosα cos(3π/2+α)=cos(2π-(3π/2+α))=cos(π/2-α)=sinα tan(π-α)=tan(-α)=-tanα tan(-a-π)=-tan(π+α)=-tanα sin(-π-α)=-sin(π+α)=sinα 所以,[sin(α-π/2)]cos(3π/2+α)tan(π-α)] / [tan(-a-π)sin(-π-α)]=[-cosα×sinα×(-tanα)]/[-tanα×sinα]=-cosα=(2√6)/5
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