Coterminal angles are angles that, when drawn in the standard position, share a terminal side. You can find these angles by adding or subtracting 360 to the given angle. Thus, the only angle measurement that works from the answers given is \(\displaystyle 305^{\circ}\). 

For an angle to be coterminal with \(\displaystyle \theta\), that angle must be of the form \(\displaystyle \theta + 2\pi N\) for some integer \(\displaystyle N\) - or, equivalently, the difference of the angle measures multiplied by \(\displaystyle \frac{1}{2\pi}\)must be an integer. We apply this test to all four choices.

\(\displaystyle \frac{41\pi}{9}\):

\(\displaystyle \left ( \frac{1}{2\pi } \right ) \left (\frac{41\pi}{9}- \frac{5\pi}{9}\right )\)

\(\displaystyle =\left ( \frac{1}{2\pi } \right )\left ( \frac{36\pi}{9} \right ) =\left ( \frac{1}{2\pi } \right )4 \pi =2\)

\(\displaystyle \frac{23\pi}{9}\):

\(\displaystyle \left ( \frac{1}{2\pi } \right ) \left (\frac{23\pi}{9}- \frac{5\pi}{9}\right )\)

\(\displaystyle =\left ( \frac{1}{2\pi } \right )\left ( \frac{18\pi}{9} \right ) =\left ( \frac{1}{2\pi } \right )2 \pi =1\)

\(\displaystyle -\frac{13\pi}{9}\):

\(\displaystyle \left ( \frac{1}{2\pi } \right ) \left (-\frac{13\pi}{9}- \frac{5\pi}{9}\right )\)

\(\displaystyle =\left ( \frac{1}{2\pi } \right )\left ( \frac{-18\pi}{9} \right ) =\left ( \frac{1}{2\pi } \right )\left (- 2 \pi \right )=-1\)

\(\displaystyle -\frac{31\pi}{9}\):

\(\displaystyle \left ( \frac{1}{2\pi } \right ) \left (-\frac{31\pi}{9}- \frac{5\pi}{9}\right )\)

\(\displaystyle =\left ( \frac{1}{2\pi } \right )\left ( \frac{-36\pi}{9} \right ) =\left ( \frac{1}{2\pi } \right )\left (- 4 \pi \right )=-2\)

All four choices pass the test, so all four angles are coterminal with \(\displaystyle \frac{5\pi}{9}\).

\(\displaystyle 90\textup{ degrees }=\frac{\pi }{2}\:\: \textup{ Coterminal angles share the same position on the unit circle.}\)

\(\displaystyle \textup{Therefore, coterminal angles can be found by adding or subtracting multiples of }2\pi\).

\(\displaystyle \frac{\pi }{2}+2\pi=\frac{5\pi }{2}\)

For two angles to be coterminal, they must differ by \(\displaystyle 2\pi N\) for some integer \(\displaystyle N\) - or, equivalently, the difference of the angle measures multiplied by \(\displaystyle \frac{1}{2\pi}\)must be an integer. We apply this test to all five choices.

\(\displaystyle \frac{17\pi }{2} \textrm{ and } \frac{23\pi }{2}\):

\(\displaystyle \frac{1}{2\pi } \cdot \left ( \frac{23\pi }{2} - \frac{17\pi }{2} \right ) = \frac{1}{2\pi } \cdot \frac{6\pi }{2} = \frac{3 }{2}\)

\(\displaystyle \frac{17\pi }{3} \textrm{ and } \frac{23\pi }{3}\):

\(\displaystyle \frac{1}{2\pi } \cdot \left ( \frac{23\pi }{3} - \frac{17\pi }{3} \right ) = \frac{1}{2\pi } \cdot \frac{6\pi }{3} = 1\)

\(\displaystyle \frac{17\pi }{4} \textrm{ and } \frac{23\pi }{4}\):

\(\displaystyle \frac{1}{2\pi } \cdot \left ( \frac{23\pi }{4} - \frac{17\pi }{4} \right ) = \frac{1}{2\pi } \cdot \frac{6\pi }{4} = \frac{3 }{4}\)

\(\displaystyle \frac{17\pi }{5} \textrm{ and } \frac{23\pi }{5}\):

\(\displaystyle \frac{1}{2\pi } \cdot \left ( \frac{23\pi }{5} - \frac{17\pi }{5} \right ) = \frac{1}{2\pi } \cdot \frac{6\pi }{5} = \frac{3 }{5}\)

\(\displaystyle \frac{17\pi }{6} \textrm{ and } \frac{23\pi }{6}\):

\(\displaystyle \frac{1}{2\pi } \cdot \left ( \frac{23\pi }{6} - \frac{17\pi }{6} \right ) = \frac{1}{2\pi } \cdot \frac{6\pi }{6} = \frac{1 }{2}\)

The only angles that pass the test - and are therefore coterminal - are \(\displaystyle \frac{17\pi }{3} \textrm{ and } \frac{23\pi }{3}\).

The pattern for positive functions is All Student Take Calculus. In quandrant I, all trigonometric functions are positive. In quadrant II, sine is positive. In qudrant III, tangent is positive. In quadrant IV, cosine is positive.

If you examine the unit circle, you'll see that the the \(\displaystyle \sin(\frac{\pi}{2})=1\). If you were to graph a sine function, you would also see that it crosses through the point \(\displaystyle (\frac{\pi}{2},1)\).

If you look at the unit circle, you'll see that \(\displaystyle \cos(\frac{2\pi}{3})=\frac{1}{2}\). You can also think of this as the cosine of \(\displaystyle 60^\circ\), which is also \(\displaystyle \frac{1}{2}\).

If you look at the unit circle, you'll see that \(\displaystyle \sin(\frac{2\pi}{3})=\frac{\sqrt3}{2}\). You can also think of this as the sine of \(\displaystyle 60^\circ\), which is also \(\displaystyle \frac{\sqrt3}{2}\).

Using the unit circle, \(\displaystyle \cos(\frac{\pi}{4})=\frac{\sqrt2}{2 }\). You can also think of this as the cosine of \(\displaystyle 45^\circ\), which would also be \(\displaystyle \frac{\sqrt2}{2}\).

Using the unit circle, you can see that \(\displaystyle \cos(\frac{\pi}{2})=0\). If you were to graph a cosine function, you would also see that it crosses through the point \(\displaystyle (\frac{\pi}{2},0)\).