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Let the magnitude of each of the two equal forces be $P$.

When the forces act at an angle $2\alpha$, let their resultant be $R_1$.
Using the formula for the resultant of two forces:
$R_1 = \sqrt{P^2 + P^2 + 2P \cdot P \cos(2\alpha)}$
$R_1 = \sqrt{2P^2(1 + \cos 2\alpha)} = \sqrt{2P^2 \cdot 2\cos^2\alpha} = \sqrt{4P^2\cos^2\alpha} = 2P\cos\alpha$

When the forces act at an angle $2\beta$, let their resultant be $R_2$.
Similarly:
$R_2 = \sqrt{P^2 + P^2 + 2P \cdot P \cos(2\beta)}$
$R_2 = \sqrt{2P^2(1 + \cos 2\beta)} = \sqrt{4P^2\cos^2\beta} = 2P\cos\beta$

According to the problem:
$R_1 = 2 R_2$

Substituting the values of $R_1$ and $R_2$:
$2P \cos\alpha = 2(2P \cos\beta)$
$\implies 2P \cos\alpha = 4P \cos\beta$

Dividing both sides by $2P$ (since $P \neq 0$):
$\cos\alpha = 2\cos\beta$

Hence proved.