Question

高一数学化简。

已知0<x<π/2 化简lg[cosxtanx+1-2sin^2(x/2)]+lg[√2cos(x-π/4)]-lg(1+sin2x)

Answer 1

因为，tanx=sinx/cosx,
cosx=1-2sin^2(x/2),
√2cos(x-π/4)=√2cosxcos(π/4)+√2sinxsin(π/4)=sinx+cosx
lgM+lgN=lgMN
(sinx+cosx)^2=1+sin2x
所以
lg[cosxtanx+1-2sin^2(x/2)]+lg[√2cos(x-π/4)]-lg(1+sin2x)
=lg(sinx+cosx)+lg(sinx+cosx)-lg(sinx+cosx)^2
=lg(sinx+cosx)^2-lg(sinx+cosx)^2=0

Answer 2

原式=lg[sinx+cosx]+lg{（cosx+sinx）}-lg{（sinx+cosx）^2} ,x属于0—π/2
=lg{(sinx+cosx)^2}-lg{(sinx+cosx)^2}
=lg1
=0

Answer 3

lg[cosxtanx+1-2sin^2(x/2)]+lg[√2cos(x-π/4)]-lg(1+sin2x)
=lg[sinx+cosx]+lg[√2cosxcosπ/4+√2sinxsinπ/4)]
-lg(sin²x+cos²x+2sinxcosx)
=lg[sinx+cosx]+lg[cosx+sinx]-lg(sinx+cosx)²
=lg[sinx+cosx]²-lg(sinx+cosx)²=0