作业帮已知0＜x＜pai/2,化简：lg[cos xtan x+1-2sin^2(pai/2)]+lg[跟号2cos(x-pai/4)]-lg(1+sin2x) ...已知0＜x＜pai/2,化简：lg[cos xtan x+1-2sin^2(pai/2)]+lg[跟号2cos(x-pai/4)]-lg(1+sin2x)
来源：学生作业帮助网 编辑：作业帮 时间：2024/04/29 21:39:07
![已知0＜x＜pai/2,化简：lg[cos xtan x+1-2sin^2(pai/2)]+lg[跟号2cos(x-pai/4)]-lg(1+sin2x) ...已知0＜x＜pai/2,化简：lg[cos xtan x+1-2sin^2(pai/2)]+lg[跟号2cos(x-pai/4)]-lg(1+sin2x)](/uploads/image/z/3995433-9-3.jpg?t=%E5%B7%B2%E7%9F%A50%EF%BC%9Cx%EF%BC%9Cpai%2F2%2C%E5%8C%96%E7%AE%80%EF%BC%9Alg%5Bcos+xtan+x%2B1-2sin%5E2%28pai%2F2%29%5D%2Blg%5B%E8%B7%9F%E5%8F%B72cos%28x-pai%2F4%29%5D-lg%281%2Bsin2x%29+...%E5%B7%B2%E7%9F%A50%EF%BC%9Cx%EF%BC%9Cpai%2F2%2C%E5%8C%96%E7%AE%80%EF%BC%9Alg%5Bcos+xtan+x%2B1-2sin%5E2%28pai%2F2%29%5D%2Blg%5B%E8%B7%9F%E5%8F%B72cos%28x-pai%2F4%29%5D-lg%281%2Bsin2x%29)
已知0＜x＜pai/2,化简：lg[cos xtan x+1-2sin^2(pai/2)]+lg[跟号2cos(x-pai/4)]-lg(1+sin2x) ...已知0＜x＜pai/2,化简：lg[cos xtan x+1-2sin^2(pai/2)]+lg[跟号2cos(x-pai/4)]-lg(1+sin2x)
已知0＜x＜pai/2,化简：lg[cos xtan x+1-2sin^2(pai/2)]+lg[跟号2cos(x-pai/4)]-lg(1+sin2x) ...
已知0＜x＜pai/2,化简：lg[cos xtan x+1-2sin^2(pai/2)]+lg[跟号2cos(x-pai/4)]-lg(1+sin2x)
已知0＜x＜pai/2,化简：lg[cos xtan x+1-2sin^2(pai/2)]+lg[跟号2cos(x-pai/4)]-lg(1+sin2x) ...已知0＜x＜pai/2,化简：lg[cos xtan x+1-2sin^2(pai/2)]+lg[跟号2cos(x-pai/4)]-lg(1+sin2x)
lg[cosxtanx+1-2sin²(x/2)]+lg[√2cos(x-π/4)]-lg(1+sin2x)
=lg(sinx+cosx)+lg(cosx+sinx)-lg[(sinx+cosx) ²]
=lg1=0