If ((23π)/(24)) = 2 p - q - 1 4 r , then the value of (p^ 2 + q^ 2...

\sin \left(\dfrac{23\pi}{4}\right) =\sin \left(\pi - \dfrac{\pi}{24}\right) =\sin \left(\dfrac{\pi}{24}\right)
Now, \cos \left(\dfrac{\pi}{12}\right) =$cos {15}^{o} =$\dfrac{\sqrt{3} + 1}{2\sqrt{2}}
Using $1 -$cos\theta =2sin^{2}\dfrac{\theta}{2}, we get
2sin^{2}\dfrac{\pi}{24} =$1 -$cos\dfrac{\pi}{12} =$1 - \left(\dfrac{\sqrt{3} + 1}{2\sqrt{2}}\right)$
$⇒ sin^{2}\dfrac{\pi}{24} = \dfrac{2\sqrt{2} - \sqrt{3} - 1}{4\sqrt{2}}$
$⇒ sin\dfrac{\pi}{24} =\sin \left(\dfrac{23\pi}{24}\right) = \sqrt{\dfrac{2\sqrt{2} - \sqrt{3} - 1}{4\sqrt{2}}}$
$⇒ p = 2, q = 3, r = 2$
$⇒ p^{2} + q^{2} - r^{2} = 9$