Question

已知tana=2,则sin(π-a)cos(2π-a)sin(-a+3π/2)/tan(-a-π)sin(-π-a)=

Answer 1

sin(π-a)=sina ;
cos(2π-a)=cos(-a)=cosa;
sin(-a+3π/2)=(sin(-a)*cos(3π/2)+cos(-a)*sin3π/2)=-cosa
tan(-a-π)=-tana;sin(-π-a)=sina
即原式=sina*cosa*（-cosa）/(-tana)*sina
=sina*cosa/(tana)^2
=(sin2a)/8
用万能公式：
sin2α=(2tanα)/[1+(tanα)^2]=4/(1+4)=4/5
故，原式=（4/5）/8=1/10

Answer 2

已知tana=2,则sin(π-a)cos(2π-a)sin(-a+3π/2)/tan(-a-π)sin(-π-a)=

解：sin(π-α)cos(2π-α)sin(-α+3π/2)/tan(-α-π)sin(-π-α)
=sinαcosαsin(3π/2-α)/{[-tan(π+α)][-sin(π+α)]}
=sinαcosα(-cosα)/(-tanα)sinα=cos²α/tanα=1/(tanαsec²α)
=1/[tanα(1+tan²α)]=1/[2(1+4)]=1/10