Dividing both sides of Pythagorean theorem \[a^2 + b^2 = c^2\] with \[c^2\] a \textbf{fundamental trigonometric identity} is obtained

,,\qquad \sin^2(\alpha) + \cos^2(\alpha) = 1

Solving equation invarious ways produces conversion formulas

,,\qquad |\sin(\alpha)| =  \sqrt{1 - \cos^2(\alpha)} = \frac{\tan(x)}{\sqrt{\tan^2(x) + 1}}

,,\qquad |\cos(\alpha)| =  \sqrt{1 - \sin^2(\alpha)} = \frac{1}{\sqrt{\tan^2(x) + 1}}

\textbf{Trigonometric function values}

\begin{tabular*}
\[\alpha=\] & \[0^\text{o}\]& \[30^\text{o}\]& \[45^\text{o}\]& \[60^\text{o}\]& \[90^\text{o}\]\\
\[\sin\alpha =\] & \[0\] & \[\frac{1}{2}\] & \[\frac{\sqrt{2}}{2}\]& \[\frac{\sqrt{3}}{2}\]&\[1\]\\
\[\cos\alpha =\] & \[1\] & \[\frac{\sqrt{3}}{2}\] & \[\frac{\sqrt{2}}{2}\]& \[\frac{1}{2}\]&\[1\]\\
\[\tan\alpha =\] & \[0\] & \[\frac{\sqrt{3}}{3}\] & \[1\]& \[\sqrt{3}\]&\[+\infty\]
\end{tabular*}
Easy way to remember the values 

\begin{tabular*}
\[\frac{\sqrt{0}}{2}\] &\[\frac{\sqrt{1}}{2}\] & \[\frac{\sqrt{2}}{2}\] & \[\frac{\sqrt{3}}{2}\] & \[\frac{\sqrt{4}}{2}\]
\end{tabular*}
Of course, this table is only a mnemonic and \[\sin(15^\text{o})\] is NOT \[\frac{\sqrt{0.5}}{2}\], although the value can be found through the half angle formula
,,\qquad \sin(15^\text{o}) = \frac{1 - \cos(30^\text{o})}{2} = \frac{2 - \sqrt{3}}{ 4}