Section A.6 Trigonometry — Special Triangles
¶From the above pair of special triangles we have
\begin{align*}
\sin \frac{\pi}{4} &= \frac{1}{\sqrt{2}} &  \sin \frac{\pi}{6} &= \frac{1}{2} & \sin \frac{\pi}{3} &= \frac{\sqrt{3}}{2}\\
\cos \frac{\pi}{4} &= \frac{1}{\sqrt{2}} &  \cos \frac{\pi}{6} &= \frac{\sqrt{3}}{2} & \cos \frac{\pi}{3} &= \frac{1}{2}\\
\tan \frac{\pi}{4} &= 1 &  \tan \frac{\pi}{6} &= \frac{1}{\sqrt{3}} & \tan
\frac{\pi}{3} &= \sqrt{3}
\end{align*}