首先sin(x)+cos(x) = √2·sin(x+π/4) ≤ √2 < π/2, 故sin(x) < π/2-cos(x).同理可得sin(x) < π/2+cos(x), 于是-π/2 < sin(x) < π/2-|cos(x)| ≤ π/2.由sin(x)在[-π/2,π/2]上严格单调递增, 有sin(sin(x)) ...那在[π/2,3π/2]上呢？这里是用到正弦函数在[-π/2,π/2]上严格递增, 不是对x分段讨论.由-π/2 < sin(x) < π/2-|cos(x)| ≤ π/2,取正弦得sin(sin(x)) < sin(π/2-|cos(x)|) = cos(|cos(x)|) = cos(cos(x)).证明过程中并没有限制x的取值范围.