1. **State the problem:** Given $t = \frac{8\pi}{3}$, evaluate $\sin t$, $\cos t$, $\tan t$, $\csc t$, $\sec t$, and $\cot t$. 2. **Recall the periodicity and reference angle:** The sine, cosine, and tangent functions are periodic with period $2\pi$. Since $t = \frac{8\pi}{3} = 2\pi + \frac{2\pi}{3}$, we can reduce $t$ by subtracting $2\pi$ to find the reference angle: $$t_{ref} = t - 2\pi = \frac{8\pi}{3} - 2\pi = \frac{8\pi}{3} - \frac{6\pi}{3} = \frac{2\pi}{3}$$ 3. **Evaluate sine and cosine at $\frac{2\pi}{3}$:** - $\sin \frac{2\pi}{3} = \sin \left(\pi - \frac{\pi}{3}\right) = \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$ - $\cos \frac{2\pi}{3} = \cos \left(\pi - \frac{\pi}{3}\right) = -\cos \frac{\pi}{3} = -\frac{1}{2}$ 4. **Evaluate tangent:** $$\tan t = \frac{\sin t}{\cos t} = \frac{\frac{\sqrt{3}}{2}}{-\frac{1}{2}} = -\sqrt{3}$$ 5. **Evaluate cosecant, secant, and cotangent:** - $\csc t = \frac{1}{\sin t} = \frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3}$ - $\sec t = \frac{1}{\cos t} = \frac{1}{-\frac{1}{2}} = -2$ - $\cot t = \frac{1}{\tan t} = \frac{1}{-\sqrt{3}} = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3}$ 6. **Summary of values:** - $\sin t = \frac{\sqrt{3}}{2}$ (Option 4) - $\cos t = -\frac{1}{2}$ - $\tan t = -\sqrt{3}$ - $\csc t = \frac{2\sqrt{3}}{3}$ - $\sec t = -2$ - $\cot t = -\frac{\sqrt{3}}{3}$ These are the exact trigonometric values for $t = \frac{8\pi}{3}$.