Question

Find the value of $\sin \frac{23\pi }{12}$

Answer 1

$\sin \left(\frac{23\pi }{12}\right)$

$=\sin (\frac{5\pi }{3}+\frac{\pi }{4})$

$=\sin \left(\frac{5\pi }{3}\right)\cos \left(\frac{\pi }{4}\right)+\cos \left(\frac{5\pi }{3}\right)\sin \left(\frac{\pi }{4}\right)$

$=\left(\frac{-\sqrt{3}}{2}\right)\cos \left(\frac{\pi }{4}\right)+\left(\frac{1}{2}\right)\sin \left(\frac{\pi }{4}\right)$

$=\frac{-\sqrt{6}+\sqrt{2}}{4}$

$\sin \left(\frac{23\pi }{12}\right)=\frac{-\sqrt{6}+\sqrt{2}}{4}$

$\cos \left(\frac{23\pi }{12}\right)=\sqrt{1-\frac{(-\sqrt{6}+\sqrt{2}{)}^2}{16}}=\sqrt{\frac{16-6-2+2\sqrt{12}}{16}}=\frac{4+\sqrt{12}}{8}$

$\cos \left(\frac{23\pi }{12}\right)=1-2{\sin }^2\left(\frac{23\pi }{24}\right)$

$\sin \left(\frac{23\pi }{12}\right)=\frac{(1-\frac{4+\sqrt{12}}{8})}{2}$

$\sin \left(\frac{23\pi }{12}\right)=\frac{4+\sqrt{12}}{16}=\frac{2+\sqrt{3}}{8}$

$\sin \left(\frac{23\pi }{12}\right)=\pm \frac{\sqrt{2+\sqrt{3}}}{2\sqrt{2}}$