已知sin(x-π/4）=1/3,则cos（π/4+x）=?这一步是怎么得来的cos[π/2-(π/4-x）]sin(x-π/4）=1/3sin(π/4-x）=-1/3sin(π/4-x）=cos[π/2-(π/4-x）]=cos（π/4+x）=-1/3

Question

已知sin(x-π/4）=1/3,则cos（π/4+x）=?这一步是怎么得来的cos[π/2-(π/4-x）]
sin(x-π/4）=1/3
sin(π/4-x）=-1/3
sin(π/4-x）=cos[π/2-(π/4-x）]
=cos（π/4+x）=-1/3

Answer 1

诱导公式
sin(π/2-a)=cosa 即两个角(π/2-a)+a=π/2 两角互余,则余函数值相等
（x-π/4)+(π/4+x)=π/2 两角互余,
所以
sin(π/4-x）=cos[π/2-(π/4-x）]
=cos（π/4+x）=-1/3