consider the following.\n\n$t = - \\frac { 61 \\pi } { 6 }$\n\n(a) find the reference number $\\overline { t…

Explanation:

Step1: Simplify ( t = -\frac{61\pi}{6} )

First, find a positive coterminal angle. Add ( 10\pi=\frac{60\pi}{6} ) to ( t ). ( t + 10\pi=-\frac{61\pi}{6}+\frac{60\pi}{6}=-\frac{\pi}{6} ). Then add ( 2\pi ) (since the period of the unit - circle trigonometric functions is ( 2\pi )) to get a positive angle. ( -\frac{\pi}{6}+2\pi=\frac{- \pi + 12\pi}{6}=\frac{11\pi}{6} ). The reference number formula for ( t=\frac{11\pi}{6} ) (which is in the fourth quadrant, ( \frac{3\pi}{2}<\frac{11\pi}{6}<2\pi )) is ( \overline{t}=2\pi - t ). ( \overline{t}=2\pi-\frac{11\pi}{6}=\frac{12\pi - 11\pi}{6}=\frac{\pi}{6} ).

Step2: Find the terminal point

The terminal point associated with ( \overline{t}=\frac{\pi}{6} ) is ( \left(\cos\frac{\pi}{6},\sin\frac{\pi}{6}\right)=\left(\frac{\sqrt{3}}{2},\frac{1}{2}\right) ). Since ( t = \frac{11\pi}{6} ) is in the fourth quadrant (( x>0,y < 0 )), the terminal point ( (x,y)=\left(\cos t,\sin t\right) ). ( \cos\frac{11\pi}{6}=\cos\left(2\pi-\frac{\pi}{6}\right)=\cos\frac{\pi}{6}=\frac{\sqrt{3}}{2} ), ( \sin\frac{11\pi}{6}=\sin\left(2\pi-\frac{\pi}{6}\right)=-\sin\frac{\pi}{6}=-\frac{1}{2} ).

Answer:

(a) ( \frac{\pi}{6} ) (b) ( \left(\frac{\sqrt{3}}{2},-\frac{1}{2}\right) )