作业帮化简lg[cos（2π-x）cot（3π/2-x)+1-2xin^2x/2]+lg[根号2cos(x-π/4)]-lg[1-sin(π+2x）]请详细过程,谢谢!
来源：学生作业帮助网 编辑：作业帮 时间：2024/05/08 19:31:28
![化简lg[cos（2π-x）cot（3π/2-x)+1-2xin^2x/2]+lg[根号2cos(x-π/4)]-lg[1-sin(π+2x）]请详细过程,谢谢!](/uploads/image/z/2407724-44-4.jpg?t=%E5%8C%96%E7%AE%80lg%5Bcos%EF%BC%882%CF%80-x%EF%BC%89cot%EF%BC%883%CF%80%2F2-x%29%2B1-2xin%5E2x%2F2%5D%2Blg%5B%E6%A0%B9%E5%8F%B72cos%28x-%CF%80%2F4%29%5D-lg%5B1-sin%28%CF%80%2B2x%EF%BC%89%5D%E8%AF%B7%E8%AF%A6%E7%BB%86%E8%BF%87%E7%A8%8B%2C%E8%B0%A2%E8%B0%A2%21)
化简lg[cos（2π-x）cot（3π/2-x)+1-2xin^2x/2]+lg[根号2cos(x-π/4)]-lg[1-sin(π+2x）]请详细过程,谢谢!
化简lg[cos（2π-x）cot（3π/2-x)+1-2xin^2x/2]+lg[根号2cos(x-π/4)]-lg[1-sin(π+2x）]
请详细过程,谢谢!
化简lg[cos（2π-x）cot（3π/2-x)+1-2xin^2x/2]+lg[根号2cos(x-π/4)]-lg[1-sin(π+2x）]请详细过程,谢谢!
lg[(cosx*tanx)+cosx](cosx+sinx)/(1+sin2x)]
=lg[(sinx+cosx)(cosx+sinx)/(sinx+cosx)^2]
=lg1
=0