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
问题描述：
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
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)/5 sin(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