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Question:

Sin α + Sin β = $\frac{\sqrt{2}}{\sqrt{3}}$ and Cos α + Cos β = $\frac{1}{\sqrt{3}}$, then the value of ${\left(20 \cos \left(\frac{\alpha -\beta }{2}\right)\right)}^2$ is _______.

Explanation:

Sin α + Sin β = $\frac{\sqrt{2}}{\sqrt{3}}$   ...(1) and
Cos α + Cos β = $\frac{1}{\sqrt{3}}$   ...(2),

Squaring both the equations and adding them we get.

⇒ Sin² α + Sin² β + 2 × Sin α × Sin β + Cos² α + Cos² β + 2 × Cos α × Cos β = $\frac{2}{3}$ + $\frac{1}{3}$

⇒ Sin² α + Cos² α + Sin² β + Cos² β + 2 × Sin α × Sin β + 2 × Cos α × Cos β = 1

⇒ 1 + 1 + 2 × cos (α - β) = 1

⇒ cos (α - β) = - 1/2

⇒ 2 × ${\left(\cos \left(\frac{\alpha -\beta }{2}\right)\right)}^2$ - 1 = - 1/2

⇒ ${\left(\cos \left(\frac{\alpha -\beta }{2}\right)\right)}^2$ = 1/4

∴ 20² × ${\left(\cos \left(\frac{\alpha -\beta }{2}\right)\right)}^2$ = 400

⇒  ${\left(20 \cos \left(\frac{\alpha -\beta }{2}\right)\right)}^2$ = 100.

Hence, 100.