a*sin(pi/4 - x) - b*cos(pi/4 - x) = a*sin(pi/4 + x) - b*cos(pi/4 + x)

问题描述:

a*sin(pi/4 - x) - b*cos(pi/4 - x) = a*sin(pi/4 + x) - b*cos(pi/4 + x)
怎么得到(a + b)* sinx = 0

asinπ/4cosx-acosπ/4sinx-bcosπ/4cosx-bsinπ/4sinx=asinπ/4cosx+acosπ/4sinx-bcosπ/4cosx+bsinπ/4sinx
acosπ/4sinx+bsinπ/4sinx=0
了吧除以√2/2
所以(a+b)sinx=0