find t and the terminal point determined by t for each point in the following figure, where t is increasing…
find t and the terminal point determined by t for each point in the following figure, where t is increasing in increments of π/6. (order your answers from smallest to largest t.)
Answer
Explanation:
Step1: Recall the unit - circle terminal - point formula
For a unit circle (x = \cos t) and (y=\sin t). When (t = 0), (\cos(0)=1) and (\sin(0) = 0).
Step2: Analyze the increment of (t)
Since (t) is increasing in increments of (\frac{\pi}{6}). When (t=\frac{\pi}{6}), (\cos(\frac{\pi}{6})=\frac{\sqrt{3}}{2}) and (\sin(\frac{\pi}{6})=\frac{1}{2}), the terminal point is ((\frac{\sqrt{3}}{2},\frac{1}{2})). When (t = \frac{\pi}{3}), (\cos(\frac{\pi}{3})=\frac{1}{2}) and (\sin(\frac{\pi}{3})=\frac{\sqrt{3}}{2}), the terminal point is ((\frac{1}{2},\frac{\sqrt{3}}{2})). When (t=\frac{\pi}{2}), (\cos(\frac{\pi}{2}) = 0) and (\sin(\frac{\pi}{2})=1), the terminal point is ((0,1)). When (t=\frac{2\pi}{3}), (\cos(\frac{2\pi}{3})=-\frac{1}{2}) and (\sin(\frac{2\pi}{3})=\frac{\sqrt{3}}{2}), the terminal point is ((-\frac{1}{2},\frac{\sqrt{3}}{2})). When (t=\frac{5\pi}{6}), (\cos(\frac{5\pi}{6})=-\frac{\sqrt{3}}{2}) and (\sin(\frac{5\pi}{6})=\frac{1}{2}), the terminal point is ((-\frac{\sqrt{3}}{2},\frac{1}{2})). When (t=\pi), (\cos(\pi)=- 1) and (\sin(\pi)=0), the terminal point is ((-1,0)). When (t=\frac{7\pi}{6}), (\cos(\frac{7\pi}{6})=-\frac{\sqrt{3}}{2}) and (\sin(\frac{7\pi}{6})=-\frac{1}{2}), the terminal point is ((-\frac{\sqrt{3}}{2},-\frac{1}{2})). When (t=\frac{4\pi}{3}), (\cos(\frac{4\pi}{3})=-\frac{1}{2}) and (\sin(\frac{4\pi}{3})=-\frac{\sqrt{3}}{2}), the terminal point is ((-\frac{1}{2},-\frac{\sqrt{3}}{2})). When (t=\frac{3\pi}{2}), (\cos(\frac{3\pi}{2}) = 0) and (\sin(\frac{3\pi}{2})=-1), the terminal point is ((0, - 1)). When (t=\frac{5\pi}{3}), (\cos(\frac{5\pi}{3})=\frac{1}{2}) and (\sin(\frac{5\pi}{3})=-\frac{\sqrt{3}}{2}), the terminal point is ((\frac{1}{2},-\frac{\sqrt{3}}{2})). When (t=\frac{11\pi}{6}), (\cos(\frac{11\pi}{6})=\frac{\sqrt{3}}{2}) and (\sin(\frac{11\pi}{6})=-\frac{1}{2}), the terminal point is ((\frac{\sqrt{3}}{2},-\frac{1}{2})).
Answer:
| (t) | Terminal Point |
|---|---|
| (0) | ((1,0)) |
| (\frac{\pi}{6}) | ((\frac{\sqrt{3}}{2},\frac{1}{2})) |
| (\frac{\pi}{3}) | ((\frac{1}{2},\frac{\sqrt{3}}{2})) |
| (\frac{\pi}{2}) | ((0,1)) |
| (\frac{2\pi}{3}) | ((-\frac{1}{2},\frac{\sqrt{3}}{2})) |
| (\frac{5\pi}{6}) | ((-\frac{\sqrt{3}}{2},\frac{1}{2})) |
| (\pi) | ((-1,0)) |
| (\frac{7\pi}{6}) | ((-\frac{\sqrt{3}}{2},-\frac{1}{2})) |
| (\frac{4\pi}{3}) | ((-\frac{1}{2},-\frac{\sqrt{3}}{2})) |
| (\frac{3\pi}{2}) | ((0,-1)) |
| (\frac{5\pi}{3}) | ((\frac{1}{2},-\frac{\sqrt{3}}{2})) |
| (\frac{11\pi}{6}) | ((\frac{\sqrt{3}}{2},-\frac{1}{2})) |