\(cos\left ( \frac{9\pi}{12}\right )=\frac{\sqrt{2-\sqrt{3}}}{2}\)
Step-by-step explanation:
We know that
cos 2x = 2cos²x - 1
So we have
\(cosx=\sqrt{\frac{1+cos2x}{2}}\)
Substituting \(x=\frac{19\pi}{12}\)
\(cos\left ( \frac{19\pi}{12}\right )=\sqrt{\frac{1+cos2\left ( \frac{19\pi}{12}\right )}{2}}\\\\cos\left ( \frac{19\pi}{12}\right )=\sqrt{\frac{1+cos\left ( \frac{19\pi}{6}\right )}{2}}\\\\cos\left ( \frac{9\pi}{12}\right )=\sqrt{\frac{1+cos570}{2}}\\\\cos\left ( \frac{9\pi}{12}\right )=\sqrt{\frac{1-\frac{\sqrt{3}}{2}}{2}}\\\\cos\left ( \frac{9\pi}{12}\right )=\sqrt{\frac{2-\sqrt{3}}{4}}\\\\cos\left ( \frac{9\pi}{12}\right )=\frac{\sqrt{2-\sqrt{3}}}{2}\)
\(cos\left ( \frac{9\pi}{12}\right )=\frac{\sqrt{2-\sqrt{3}}}{2}\)
