প্রদত্ত, $\sin \left(\frac{\pi}{2} \cos \alpha\right) = \cos \left(\frac{\pi}{2} \sin \alpha\right) = \sin \left(\frac{\pi}{2} - \frac{\pi}{2} \sin \alpha\right)$। সুতরাং, $\frac{\pi}{2} \cos \alpha = \frac{\pi}{2} - \frac{\pi}{2} \sin \alpha$ $$ \begin{aligned} \Rightarrow \cos \alpha + \sin \alpha &= 1 \\ \Rightarrow \frac{1}{\sqrt{2}} \cos \alpha + \frac{1}{\sqrt{2}} \sin \alpha &= \frac{1}{\sqrt{2}} \\ \Rightarrow \sin \left(\alpha + \frac{\pi}{4}\right) &= \sin \frac{\pi}{4} \end{aligned} $$ এখান থেকে সমাধান করলে পাওয়া যায় $\alpha = 0$ অথবা $\alpha = \frac{\pi}{2}$।