Question

sin($pi$/4-α)-sin($pi$/4+α)=√2cos($pi$/2+α)

Answer 1

$1);sinleft(fracpi4-alpharight)-sinleft(fracpi4+alpharight)=sinfracpi4cosalpha-cosfracpi4sinalpha-sinfracpi4cosalpha-cosfracpi4sinalpha=\=-2cosfracpi4sinalpha==-2cdotfrac{sqrt2}2sinalpha=-sqrt2sinalpha\2);sqrt2cosleft(fracpi2+alpharight)=-sqrt2sinalpha\\\-sqrt2sinalpha=-sqrt2sinalphaRightarrowsinleft(fracpi4-alpharight)-sinleft(fracpi4+alpharight)=sqrt2cosleft(fracpi2+alpharight)$