The problem is thus treated in steps:

We are given that:

cos(π/4 + θ) + cos(π/4 - θ) = √2 cosθ

We shall consider the left-hand side (LHS):

The sum and difference formulas for cosine are useful here:

cos(A + B) = cosA * cosB - sinA * sinB

cos(A - B) = cosA * cosB + sinA * sinB

Substituting into our equation we get:

LHS = [cos(π/4) * cosθ - sin(π/4) * sinθ] + [cos(π/4) * cosθ + sin(π/4) * sinθ]

Now we simplify:

LHS = 2 cos(π/4) cosθ

cos(π/4) = 1/√2. Thus:

LHS = 2(1/√2) cosθ

LHS = √2 cosθ.

Hence, LHS = RHS.

So we have proved that the given equation stands correct.