作业帮化简lg(cosx tanx+cosx)+lg[根号2cos(x-π/4)]-lg(1+2sinxcosx)已知0
来源：学生作业帮助网 编辑：作业帮 时间：2024/05/13 08:10:44
![化简lg(cosx tanx+cosx)+lg[根号2cos(x-π/4)]-lg(1+2sinxcosx)已知0](/uploads/image/z/2626973-53-3.jpg?t=%E5%8C%96%E7%AE%80lg%28cosx+tanx%2Bcosx%29%2Blg%5B%E6%A0%B9%E5%8F%B72cos%28x-%CF%80%2F4%29%5D-lg%281%2B2sinxcosx%29%E5%B7%B2%E7%9F%A50)
化简lg(cosx tanx+cosx)+lg[根号2cos(x-π/4)]-lg(1+2sinxcosx)已知0
化简lg(cosx tanx+cosx)+lg[根号2cos(x-π/4)]-lg(1+2sinxcosx)
已知0
化简lg(cosx tanx+cosx)+lg[根号2cos(x-π/4)]-lg(1+2sinxcosx)已知0
lg(cosx tanx+cosx)+lg[根号2cos(x-π/4)]-lg(1+2sinxcosx)
=lg(cosx ×sinx/cos+cosx)+lg[根号2cos(x-π/4)]-lg(sin²x+cos²x+2sinxcosx)
=lg(sinx+cosx)+lg[根号2cos(x-π/4)]-lg(sinx+cosx)²
=lg(sinx+cosx)+lg[根号2cosxcosπ/4+sinxsinπ/4)]-lg(sinx+cosx)²
=lg(sinx+cosx)+lg[cosx+sinx]-lg(sinx+cosx)²
=lg(sinx+cosx)(cosx+sinx)-lg(sinx+cosx)²
=lg[(sinx+cos)²/(sinx+cosx)²]
=lg1
=0