Question

Нужно до завтрашнего ! sina*sinb+cos(a+b)= корень из 3* на cosa-2*cos(a-п/6) (cos(a-b)-2*sina*sinb)/2*sina*sinb+sin(b-a)

Answer 1

1) $sin\alpha*sin\beta+cos(\alpha+\beta)=sin\alpha*sin\beta+cos\alpha*cos\beta-sin\alpha*sin\beta=cos\alpha*cos\beta$

2) $\sqrt{3}*cos\alpha-2*cos(\alpha-\pi/6)=\sqrt{3}*cos\alpha-2*(\frac{\sqrt{3}*cos\alpha}{2}+\frac{sin\alpha}{2})=\sqrt{3}*cos\alpha-\sqrt{3}*cos\alpha-sin\alpha=-sin\alpha$

 3) $\frac{(cos(\alpha-\beta)-2*sin\alpha*sin\beta)}{2*sin\alpha*sin\beta+sin(\beta-\alpha)}=\frac{cos\alpha*cos\beta+sin\alpha*\sin\beta-2*sin\alpha*sin\beta}{2*sin\alpha*sin\beta+sin\beta*cos\alpha-cos\beta*sin\alpha}=\frac{cos\alpha*cos\beta-sin\alpha*sin\beta}{cos(\alpha-\beta)-cos(\alpha+\beta)+sin\beta*cos\alpha-cos\beta*sin\alpha}$$\frac{cos\alpha*cos\beta-sin\alpha*sin\beta}{cos\alpha*cos\beta+sin\alpha*sin\beta-cos\alpha*cos\beta+sin\alpha*sin\beta+sin\beta*cos\alpha-cos\beta*sin\alpha}=\frac{cos\alpha*cos\beta-sin\alpha*sin\beta}{2*sin\alpha*sin\beta+sin\beta*cos\alpha-cos\beta*sin\alpha}$