Очень нужно до завтрашнего утра!!!!!!!!!!!!!!!!!!! упростить 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)

1) [latex]sin\alpha*sin\beta+cos(\alpha+\beta)=sin\alpha*sin\beta+cos\alpha*cos\beta-sin\alpha*sin\beta=cos\alpha*cos\beta[/latex] 2) [latex]\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[/latex]    3) [latex]\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}[/latex] [latex]\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}[/latex]