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Section 2.1 #44: Convert to radians: $-225^{\circ}$
Solution: Recall that $180^{\circ} = \pi \hspace{2pt} \mathrm{rad}$, so $\dfrac{\pi \hspace{2pt} \mathrm{rad}}{180^{\circ}}=1$. Therefore compute $$-225^{\circ} = -225^{\circ} \dfrac{\pi \hspace{2pt} \mathrm{rad}}{180^{\circ}}= - \dfrac{225 \pi}{180} \mathrm{rad} = - \dfrac{5\pi}{4} \mathrm{rad}.$$

Section 2.1 #48: Convert to degrees: $\dfrac{5\pi}{6}$
Solution: Recall that $180^{\circ} = \pi \hspace{2pt} \mathrm{rad}$, so $\dfrac{180^{\circ}}{\pi \hspace{2pt} \mathrm{rad}}=1$. Therefore compute $$\dfrac{5\pi}{6} \mathrm{rad} = \left( \dfrac{5\pi}{6} \mathrm{rad} \right) \left( \dfrac{180^{\circ}}{\pi \hspace{2pt} \mathrm{rad}} \right)=150^{\circ}.$$ Section 2.1 #73: Find the radius of a circle where an angle of $\theta = \dfrac{1}{3}$ radians subtends an arc length of $s=2$.
Solution: Recall the relationship between the radius $r$ of a circle, the angle $\theta$, and the arc subtended (of length $s$) is $s=r\theta$. Since we are seeing the radius, solve for $r$ to get $r=\dfrac{s}{\theta}$. Now plug in the information we know and get $$r=\dfrac{2}{\frac{1}{3}} = 6.$$ Section 2.2 #19: The point $\left( \dfrac{2 \sqrt{2}}{3}, -\dfrac{1}{3} \right)$ corresponds to some angle $\theta$ on the unit circle. Find the exact values of the six trigonometric functions of $\theta$.
Solution: Compute from definition: $$\sin(\theta)= -\dfrac{1}{3},$$ $$\cos(\theta) = \dfrac{2\sqrt{2}}{3},$$ $$\tan(\theta) = \dfrac{-\frac{1}{3}}{\frac{2\sqrt{2}}{3}}=- \dfrac{1}{2\sqrt{2}},$$ $$\csc(\theta)=\dfrac{1}{\sin(\theta)}=-3,$$ $$\sec(\theta)=\dfrac{1}{\cos(\theta)}=\dfrac{3}{2\sqrt{2}},$$ and $$\cot(\theta)=-2\sqrt{2}.$$