已知0<X<π/2，化简：lg(cosx•tanx+1-2sin²x/2)+lg[√2 cos(x

Question

已知0<X<π/2，化简：lg(cosx•tanx+1-2sin²x/2)+lg[√2 cos(x-π/4)]-lg(1+sin2x).

Answer 1

0<X<π/2 所以 1>=cosx>=0
1.cosx•tanx+1-2sin²x/2=cosx*sinx/cosx+cosx=sinx+cosx
2.[ √2 cos(x-π/4)]=√2 [1/√2 *cosx+1/√2*sinx ]=sinx+cosx
3.1+sin2x=sin^2x+2sinx*cosx+cos^2x=(sinx+cosx)^2
lg(cosx•tanx+1-2sin²x/2)+lg[√2 cos(x-π/4)]-lg(1+sin2x). 将1，2，3代入有
=lg(sinx+cosx)+lg(sinx+cosx)-2lg(sinx+cosx)=0