Question

求cos(π/2+α)sin(-π-α) / cos(11π/2 -α)sin(9π/2+α)的值

Answer 1

cos(π/2+α)=-sinα, 因为若α在第一象限，π/2+α在第二，cos是负值 (换成sin因为差的是π/2的倍数)
sin(-π-α)=sin(-π+2π-α) =sin(π-α)=sinα, α在第一象限,π-α在第二,sin是正值
cos(11π/2 -α)=cos(11π/2-4π -α)=cos(3π/2-α)=-sinα 3π/2-α 在第三象限，cos为负值
sin(9π/2+α)=sin(9π/2-4π+α)=sin(π/2+α)=cosα
所以cos(π/2+α)sin(-π-α) / cos(11π/2 -α)sin(9π/2+α)=[-sinα cosα]/[-sinα cosα]=1

Answer 2

cos(π/2+α)sin(-π-α) / cos(11π/2 -α)sin(9π/2+α)
=(-sina)sina/(-sina)cosa=tana

Answer 3

tan α