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!

问题描述:

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