已知α是第二象限角,且cos[α-(π/2)=1/5求{sin(π-α)cos(π-α)tan[(-3π/2)-α]}/{tan[(π/2)+α]cos[(3π/2)α]}式子错了应为{sin(π-α)cos(π-α)tan[(-3π/2)-α]}/{tan[(π/2)+α]cos[(3π/2)+α]}

Question

已知α是第二象限角,且cos[α-(π/2)=1/5 求{sin(π-α)cos(π-α)tan[(-3π/2)-α]}/{tan[(π/2)+α]cos[(3π/2)α]}
式子错了 应为{sin(π-α)cos(π-α)tan[(-3π/2)-α]}/{tan[(π/2)+α]cos[(3π/2)+α]}

Answer 1

cos[α-(π/2)]=cos(π/2-α)=sinα
cos[α-(π/2)]=1/5
sinα=1/5
α是第二象限角,
cosα=-根号[1-1/5)^2]=-(2根号6)/5
{sin(π-α) cos(π-α) tan[(-3π/2)-α]} / {tan[(π/2)+α] cos[(3π/2)+α]}
={sinα (-cosα) [-tan(π+π/2+α)]} / {tan[(π/2)+α] cos(2π-(π/2-α)]}
={sinα cosα tan(π/2+α)]} / {tan[(π/2)+α] [cos(π/2-α)}
=sinα cosα / [cos(π/2-α)}
=sinα cosα / sinα
=cos α
=-(2根号6)/5