A ball suspended by a thread of length $l$ at the point $O$ on the wall, forming a small angle $\alpha$ with the vertical (Fig.). Then the thread with the ball deviated through a small angle $\beta$ $(\beta>\alpha)$ and set free. Assuming the collision of the ball with the wall to be perfectly elastic, find the oscillation period of such a pendulum.

From fig $2\theta$ part of SHM will be skipped over, $\cos\theta=\dfrac{\alpha}{\beta}\implies\cos2\theta=\dfrac{2\alpha^2-\beta^2}{\beta^2}$. So time it would be skpping over will be $t=\dfrac{1}{\omega}\cos^{-1}\bigg(\dfrac{2\alpha^2-\beta^2}{\beta^2}\bigg)$. On substrating from total time $T=2\pi\sqrt{\dfrac{l}{g}}$ gives me $T'=2\sqrt{\dfrac{l}{g}}\bigg[\dfrac{\pi}{2}+\sin^{-1}\bigg(\dfrac{2\alpha^2-\beta^2}{\beta^2}\bigg)\bigg]$. But in official answer argument of sin inverse is $\dfrac{\alpha}{\beta}$.

Please help.