$ = a\sqrt {{{\sin }^2}\alpha + {{\cos }^2}\alpha + {{\cos }^2}\beta + {{\sin }^2}\beta - 2\cos \alpha \cos \beta - 2\sin \alpha \sin \beta } $
$ = a\sqrt {2\left\{ {1 - \cos \,(\alpha - \beta )} \right\}} $
$= 2a\,\sin \,\left( {\frac{{\alpha - \beta }}{2}} \right)$
Trick : Put $a = 1,\,\,\alpha = \frac{\pi }{2},\,\beta = \frac{\pi }{6},$ then the points will be $(0, 1)$ and $\left( {\frac{{\sqrt 3 }}{2},\,\,\frac{1}{2}} \right)$.
Obviously, the distance between these two points is 1 which is given by $(d)$.
$\left\{ {\,\,2a\,\sin \frac{{\alpha - \beta }}{2} = 2 \times 1 \times \sin \frac{{(\pi /2) - (\pi /6)}}{2} = 2 \times \frac{1}{2} = 1} \right\}$
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$\sin ^{-1}\left(\sum_{i=1}^{\infty} x^{i+1}-x \sum_{i=1}^{\infty}\left(\frac{x}{2}\right)^i\right)=\frac{\pi}{2}-\cos ^{-1}\left(\sum_{i=1}^{\infty}\left(-\frac{x}{2}\right)^i-\sum_{i=1}^{\infty}(-x)^i\right)$
lying in the interval $\left(-\frac{1}{2}, \frac{1}{2}\right)$ is. . . . .
(Here, the inverse trigonometric functions $\sin ^{-1} x$ and $\cos ^{-1} x$ assume values in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ and $[0, \pi]$, respectively.)