Question

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

Answer 1

$\frac{x}{y} 1)\; \; sin(\alpha -\beta )-sin(\alpha +\beta)=2\cdot sin\dfrac{\alpha-\beta-(\alpha +\beta )}{2}\cdot cos\dfrac{\alpha -\beta+(\alpha +\beta )}{2}=\\\\\\=2\cdot sin(-\beta)\cdot cos\alpha =-2\, sin\beta\cdot cos\alpha \\\\\\2)\; \; sin(30-a)+cos(60-a)=\\\\=sin30\cdot cosa-cos30\cdot sina+cos60\cdot cosa+sin60\cdot sina=\\\\=\dfrac{1}{2}\cdot cosa-\dfrac{\sqrt3}{2}\cdot sina+\dfrac{1}{2}\cdot cosa+\dfrac{\sqrt3}{2}\cdot sina=cosa$