sin23π24=2p−q−14rthen the value of p2−q2−r2=

sin23π24=sinπ-π24 =sinπ24We have 2sin2π24=1-cosπ12 =1-cos15°∴sin2π24=1-3+1222⇒sinπ24=22-3-142⇒p=2,q=3 and r=2⇒p2+q2-r2=9

sin23π24=2p−q−14rthen the value of p2−q2−r2=

sin23π24=sinπ-π24 =sinπ24We have 2sin2π24=1-cosπ12 =1-cos15°∴sin2π24=1-3+1222⇒sinπ24=22-3-142⇒p=2,q=3 and r=2⇒p2+q2-r2=9

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$\sin \frac{23 π}{24} = \sqrt{\frac{2 \sqrt{p} - \sqrt{q} - 1}{4 \sqrt{r}}}$ then the value of $p^{2} - q^{2} - r^{2} =$

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Detailed Solution

$\begin{array}{l} \sin (\frac{23 π}{24}) = \sin (π - \frac{π}{24}) \\ = \sin \frac{π}{24} \end{array}$

$\begin{array}{l} W e h a v e 2 \sin^{2} \frac{π}{24} = 1 - \cos \frac{π}{12} \\ = 1 - \cos 15 ° \\ ∴ \sin^{2} \frac{π}{24} = \frac{1 - \frac{\sqrt{3} + 1}{2 \sqrt{2}}}{2} \\ \Rightarrow \sin \frac{π}{24} = \sqrt{\frac{2 \sqrt{2} - \sqrt{3} - 1}{4 \sqrt{2}}} \\ \Rightarrow p = 2, q = 3 a n d r = 2 \\ \Rightarrow p^{2} + q^{2} - r^{2} = 9 \end{array}$