Option $2: \cos \beta / 2=2 \cos \alpha / 2$
Calculation:
According to the questions, resultant of two equal forces inclined at an angle $\alpha$ is half of resultant when they are inclined at the angle $\beta$.
Let $F$ be the force.
When angle between them is $\alpha$.
$\therefore$ Resultant $\mathrm{R}_1=\sqrt{\mathrm{F}^2+\mathrm{F}^2+2 \mathrm{~F}^2 \cos \alpha}$
When angle between them is $\beta$
$\therefore$ Resultant $\mathrm{R}_2=\sqrt{\mathrm{F}^2+\mathrm{F}^2+2 \mathrm{~F}^2 \cos \beta}$
According to the question:
$R_2=2 R_1$
$\Rightarrow R_2^2=4 R_1^2$
$\Rightarrow 2 \mathrm{~F}^2+2 \mathrm{~F}^2 \cos \beta=8 \mathrm{~F}^2+8 \mathrm{~F}^2 \cos \alpha$
$\Rightarrow 2 \mathrm{~F}^2(1+\cos \beta)=4 \times 2 \mathrm{~F}^2(1+\cos \alpha)$
$\Rightarrow(1+\cos \beta)=4(1+\cos \alpha)$
$\Rightarrow 2 \cos ^2(\beta / 2)=4 \times 2 \cos ^2(\alpha / 2)$
$\Rightarrow \cos (\beta / 2)=2 \cos (\alpha / 2)$