If $$\sin \alpha + \sin \beta = \frac{\sqrt{2}}{\sqrt{3}}$$ and $$\cos \alpha + \cos \beta = \frac{1}{\sqrt{3}}$$, then the value of $$(20 \cos (\frac{\alpha - \beta}{2}))^{2}$$ is_____

$$\sin \alpha + \sin \beta = \dfrac{\sqrt{2}}{\sqrt{3}}$$

Squaring this equation we get, $$\sin^2\ \alpha\ +\sin^2\beta\ +2\sin\ \alpha\ \sin\ \beta\ =\dfrac{2}{3}$$ ------>(1)

$$\cos \alpha + \cos \beta = \dfrac{1}{\sqrt{3}}$$

Squaring this equation we get, $$\cos^2\alpha\ +\cos^2\beta\ +2\cos\ \alpha\ \cos\ \beta\ =\dfrac{1}{3}$$ -------->(2)

Adding equation (1) and (2),

$$\left(\sin^2\alpha\ +\cos^2\alpha\right)\ +\left(\sin^2\beta\ +\cos^2\beta\right)\ +2\left(\sin\alpha\ \sin\beta\ +\cos\alpha\ \cos\beta\ \right)=\dfrac{2}{3}+\dfrac{1}{3}=1$$

or, $$1+1+2\cos\left(\alpha\ -\beta\ \right)=1$$

or, $$2\cos\left(\alpha\ -\beta\ \right)=-1$$

or, $$\cos\left(\alpha\ -\beta\ \right)=-\dfrac{1}{2}$$

Now, $$\cos\left(\alpha\ -\beta\ \right)=2\cos^2\left(\dfrac{\alpha\ -\beta\ }{2}\right)-1$$

or, $$-\dfrac{1}{2}=2\cos^2\left(\dfrac{\alpha\ -\beta\ }{2}\right)-1$$

or, $$2\cos^2\dfrac{\alpha\ -\beta\ }{2}=1-\dfrac{1}{2}=\dfrac{1}{2}$$

or, $$\cos^2\dfrac{\alpha\ -\beta\ }{2}=\dfrac{1}{4}$$

So, $$\left(20\cos\dfrac{\alpha\ -\beta\ }{2}\right)^2=400\cdot\cos^2\left(\dfrac{\alpha\ -\beta\ }{2}\right)=400\cdot\dfrac{1}{4}=100$$