If $0<A<B<\mathrm{\pi }, \mathrm{sinA}-\mathrm{sinB}=\frac{1}{\sqrt{2}}$, $\cos A-\cos B=\sqrt{\frac{3}{2}}$, then $A+B=$

Solution:

${(sinA-sinB)}^2+{(cosA-cosB)}^2=\frac{1}{2}+\frac{3}{2}=2$

$\Rightarrow 2(sinAsinB+cosAcosB)=0$

$\Rightarrow cos(B-A)=0\Rightarrow B=A+\frac{\mathrm{\pi }}{2}$

Now, $\sin A - \sin B =\frac{1}{\sqrt{2}}\Rightarrow \sin A-\cos A=\frac{1}{\sqrt{2}}$

$$\begin{gathered} \Rightarrow \frac{\sin A}{\sqrt{2}}-\frac{\cos A}{\sqrt{2}}=\frac{1}{2} \\ \Rightarrow \sin \left(A-\frac{\mathrm{\pi }}{4}\right)=\sin \frac{\mathrm{\pi }}{6}\Rightarrow A-\frac{\mathrm{\pi }}{4}=\frac{\mathrm{\pi }}{6}\Rightarrow A=\frac{5\mathrm{\pi }}{12} \\ \because A+ B =A+ A+\frac{\mathrm{\pi }}{2}=\frac{5\mathrm{\pi }}{6}+\frac{\mathrm{\pi }}{2}=\frac{4\mathrm{\pi }}{3} \end{gathered}$$