如何证明:当2kπ-π/4 0

题目

如何证明:当2kπ-π/4 <= x <= π/4+2kπ时,cos x>=sin x,sin x +cos x >0

答案

和差化积：
sinx-cosx=√2[(√2/2)sinx-(√2/2)cosx]
=√2[sin(x-π/4)]
sinx+cosx=√2[(√2/2)sinx+(√2/2)cosx]
=√2[sin(x+π/4)]
所以
2kπ-π/4 <= x <= π/4+2kπ时,
2kπ-π/2<=x-π/4<=2kπ
√2[sin(x-π/4)]<=0
2kπ<=x+π/4<=2kπ+π/2,
√2[sin(x+π/4)]>=0
所以sinx-cosx<=0,sinx+cosx>=0
得证