If cosα + cosβ = 3/2 and sinα + sinβ = 1/2 and θ is the arithmetic mean of α and β, then sin2θ + cos2θ is equal to?

Given θ is the arithmetic mean of α and β ⇒ θ = (α + β)/2

We have cosα + cosβ = 3/2 and sinα + sinβ = 1/2
Squaring and adding these two equations, we get
(cosα + cosβ)² + (sinα + sinβ)² = (3/2)² + (1/2)²
cos²α + cos²β + 2cosαcosβ + sin²α + sin²β + 2sinαsinβ = 9/4 + 1/4
(cos²α + sin²α) + (cos²β + sin²β) + 2(cosαcosβ + sinαsinβ) = 10/4 = 5/2
1 + 1 + 2cos(α - β) = 5/2
2 + 2cos(α - β) = 5/2
2cos(α - β) = 5/2 - 4/2 = 1/2
cos(α - β) = 1/4

Now, sin2θ + cos2θ = sin(α + β) + cos(α + β)
= sinαcosβ + cosαsinβ + cosαcosβ - sinαsinβ
= cosαcosβ + sinαsinβ + sinαcosβ + cosαsinβ
= cos(α - β) + sin(α + β)
We know that cos(α - β) = 1/4
Let's find sin(α + β)
(sinα + sinβ)² = (1/2)²
sin²α + sin²β + 2sinαsinβ = 1/4
(cosα + cosβ)² = (3/2)²
cos²α + cos²β + 2cosαcosβ = 9/4
Adding these two equations,
cos²α + sin²α + cos²β + sin²β + 2(cosαcosβ + sinαsinβ) = 9/4 + 1/4 = 10/4 = 5/2
2 + 2(cosαcosβ + sinαsinβ) = 5/2
2(cosαcosβ + sinαsinβ) = 1/2
cosαcosβ + sinαsinβ = 1/4
cos(α - β) = 1/4

We have sin2θ + cos2θ = 1
However, sin2θ + cos2θ = √2 sin(2θ + π/4)
The maximum value is √2, and the minimum value is -√2. Since sin2θ + cos2θ must be between -√2 and √2, none of the options are correct.