Question

Tính \(\sin \frac{\pi }{{12}}\) và \(\tan \frac{\pi }{{12}}\)

Answer 1

\(sin\left(\dfrac{\pi}{12}\right)=sin\left(\dfrac{\pi}{3}-\dfrac{\pi}{4}\right)=sin\dfrac{\pi}{3}cos\dfrac{\pi}{4}-cos\dfrac{\pi}{3}sin\dfrac{\pi}{4}=\dfrac{\sqrt{3}}{2}\cdot\dfrac{\sqrt{2}}{2}-\dfrac{1}{2}\cdot\dfrac{\sqrt{2}}{2}=\dfrac{\sqrt{6}-\sqrt{2}}{4}\\ cos\left(\dfrac{\pi}{12}\right)=\dfrac{\sqrt{6}+\sqrt{2}}{4}\\ tan\left(\dfrac{\pi}{12}\right)=\dfrac{sin\dfrac{\pi}{12}}{cos\dfrac{\pi}{12}}=\dfrac{\dfrac{\sqrt{6}-\sqrt{2}}{4}}{\dfrac{\sqrt{6}+\sqrt{2}}{4}}=2-\sqrt{3}\)