Question

1)sin(a-b)-sin(a+b) 2)sin(30°-a)+cos(60°-a) 3)

Answer 1

$frac{x}{y} 1); ; sin(alpha -beta )-sin(alpha +beta)=2cdot sindfrac{alpha-beta-(alpha +beta )}{2}cdot cosdfrac{alpha -beta+(alpha +beta )}{2}=\\\=2cdot sin(-beta)cdot cosalpha =-2, sinbetacdot cosalpha \\\2); ; sin(30-a)+cos(60-a)=\\=sin30cdot cosa-cos30cdot sina+cos60cdot cosa+sin60cdot sina=\\=dfrac{1}{2}cdot cosa-dfrac{sqrt3}{2}cdot sina+dfrac{1}{2}cdot cosa+dfrac{sqrt3}{2}cdot sina=cosa$