find the reference number for each value of t. (a) ( t=\frac{4 pi}{3} ) (b) ( t=\frac{5 pi}{3} ) (c) (…

Explanation:

Step1: Determine the quadrant for (t = \frac{4\pi}{3})

Since (\pi<\frac{4\pi}{3}<\frac{3\pi}{2}), (t=\frac{4\pi}{3}) is in the third - quadrant. The formula for the reference number (\overline{t}) in the third - quadrant is (\overline{t}=t - \pi). So, (\overline{t}=\frac{4\pi}{3}-\pi=\frac{4\pi - 3\pi}{3}=\frac{\pi}{3})

Step2: Determine the quadrant for (t=\frac{5\pi}{3})

Since (\frac{3\pi}{2}<\frac{5\pi}{3}<2\pi), (t = \frac{5\pi}{3}) is in the fourth - quadrant. The formula for the reference number (\overline{t}) in the fourth - quadrant is (\overline{t}=2\pi - t). So, (\overline{t}=2\pi-\frac{5\pi}{3}=\frac{6\pi - 5\pi}{3}=\frac{\pi}{3})

Step3: Determine the quadrant for (t =-\frac{7\pi}{6})

First, find the positive coterminal angle. Add (2\pi) to (t): (t'=-\frac{7\pi}{6}+2\pi=\frac{-7\pi + 12\pi}{6}=\frac{5\pi}{6}) Since (\frac{\pi}{2}<\frac{5\pi}{6}<\pi), the angle is in the second - quadrant. The formula for the reference number (\overline{t}) in the second - quadrant is (\overline{t}=\pi - t) (for the positive coterminal angle (t'=\frac{5\pi}{6})). So, (\overline{t}=\pi-\frac{5\pi}{6}=\frac{\pi}{6})

Step4: Determine the quadrant for (t = 3.7)

Since (\pi\approx3.14) and (2\pi\approx6.28), and (3.14<3.7<6.28). The formula for the reference number (\overline{t}) when (t) is in the third or fourth quadrant (here (t = 3.7) is in the third quadrant as (3.7-\pi\approx3.7 - 3.14=0.56) and (3.7<\pi + 1.57\approx4.71) is wrong, actually (t = 3.7) is in the third quadrant). The formula for the reference number (\overline{t}) is (\overline{t}=t-\pi) (if (t) is in the third quadrant, (t\in(\pi,\frac{3\pi}{2}))). (\overline{t}=3.7-\pi\approx3.7 - 3.14 = 0.56)

Answer:

(a) (\frac{\pi}{3}) (b) (\frac{\pi}{3}) (c) (\frac{\pi}{6}) (d) (3.7-\pi\approx0.56)