cosX-sinX>0
根据
sinX=[2tan（X/2)]/{1+[tan(X/2)]^2}
cosX={1-[tan(X/2)]^2}/{1+[tan(X/2)]^2}得
cosX-sinX
={1-[tan(X/2)]^2}/{1+[tan(X/2)]^2}-[2tan（X/2)]/{1+[tan(X/2)]^2}
={1-[tan(X/2)]^2-2tan（X/2)}/{1+[tan(X/2)]^2}
所以1-[tan(X/2)]^2-2tan（X/2)>0
又1-[tan(X/2)]^2-2tan（X/2)=-[1+tan(X/2)]^2+2>0
-1-√2＜tan(X/2)＜-1+√2
2kπ+arctan(-1-√2)＜X/2＜2kπ+arctan(-1+√2)
4kπ-2arctan(1+√2)＜X＜4kπ+2arctan(-1+√2)
2kπ-3π/4＜X＜2kπ+π/4