化简lg(cosx tanx+cosx)+lg[根号2cos(x-π/4)]-lg(1+2sinxcosx)已知0
问题描述:
化简lg(cosx tanx+cosx)+lg[根号2cos(x-π/4)]-lg(1+2sinxcosx)
已知0
答
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...