化简lg[cos（2π-x）cot（3π/2-x)+1-2xin^2x/2]+lg[根号2cos(x-π/4)]-lg[1-sin_数学解答

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!

人气：459 ℃　时间：2020-03-28 14:22:21

解答

lg[(cosx*tanx)+cosx](cosx+sinx)/(1+sin2x)]
=lg[(sinx+cosx)(cosx+sinx)/(sinx+cosx)^2]
=lg1
=0