Let us find the reference angle:
[tex]\begin{gathered} \frac{41}{6}\pi\Rightarrow\frac{41}{6}\cdot\frac{2}{2}\pi=\frac{41}{12}\cdot2\pi=(3+\frac{5}{12})2\pi \\ \frac{41}{6}\pi\Rightarrow\frac{5}{6}\pi \end{gathered}[/tex]Now, let us to calculate the sine and cosine of this angle:
[tex]\begin{gathered} \sin (\frac{5}{6}\pi)=\sin (150\degree)=0.5 \\ \cos (\frac{5}{6}\pi)=\cos (150\degree)=\sqrt[]{1-\sin ^2(\frac{5}{6}\pi)}=\sqrt[]{1-0.5^2}=\sqrt[]{\frac{3}{4}}=\frac{\sqrt[]{3}}{2} \\ \end{gathered}[/tex]From this, we have:
[tex]\begin{gathered} \sin (\frac{41\pi}{6})=\frac{1}{2} \\ \cos (\frac{41\pi}{6})=\frac{\sqrt[]{3}}{2} \end{gathered}[/tex]And because:
[tex]\tan (\theta)=\frac{\sin (\theta)}{\cos (\theta)}[/tex]we can calculate:
[tex]\tan (\frac{41\pi}{6})=\frac{\frac{1}{2}}{\frac{\sqrt[]{3}}{2}}=\frac{1}{2}\cdot\frac{2}{\sqrt[]{3}}=\frac{\sqrt[]{3}}{3}[/tex]From the solution developed above, we are able to summarize the solution as:
[tex]\begin{gathered} \sin (\frac{41\pi}{6})=\frac{1}{2} \\ \\ \cos (\frac{41\pi}{6})=\frac{\sqrt[]{3}}{2} \\ \\ \tan (\frac{41\pi}{6})=\frac{\sqrt[]{3}}{3} \end{gathered}[/tex]