Adding 4pi to theta, we can find a coterminal angle between 0 and 2pi:
[tex]-\frac{11}{4}\pi+4\pi=-\frac{11}{4}\pi+\frac{16}{4}\pi=\frac{5}{4}\pi[/tex]
This new angle is in quadrant III, so the sine and cosine relations are negative.
Calculating all trigonometric functions of this angle, we have:
[tex]\begin{gathered} \sin (\frac{5}{4}\pi)=-\frac{\sqrt[]{2}}{2} \\ \cos (\frac{5}{4}\pi)=-\frac{\sqrt[]{2}}{2} \\ \tan (\frac{5}{4}\pi)=\frac{\sin (\frac{5}{4}\pi)}{\cos (\frac{5}{4}\pi)}=1 \\ \csc (\frac{5}{4}\pi)=\frac{1}{\sin(\frac{5}{4}\pi)}=-\sqrt[]{2} \\ \sec (\frac{5}{4}\pi)=\frac{1}{\cos(\frac{5}{4}\pi)}=-\sqrt[]{2} \\ \cot (\frac{5}{4}\pi)=\frac{1}{\tan (\frac{5}{4}\pi)}=1 \end{gathered}[/tex]
