practice another\nconsider the following.\n$t = \\frac{35\\pi}{6}$\n(a) find the reference number…

Explanation:

Step1: Find the coterminal angle

To find the coterminal angle of (t = \frac{35\pi}{6}), we subtract (2\pi) (a full - circle rotation) multiple times. [ \begin{align*} \frac{35\pi}{6}- 5\times2\pi&=\frac{35\pi}{6}-10\pi\ &=\frac{35\pi - 60\pi}{6}\ &=-\frac{25\pi}{6}\ \frac{35\pi}{6}- 4\times2\pi&=\frac{35\pi}{6}-8\pi\ &=\frac{35\pi - 48\pi}{6}\ &=-\frac{13\pi}{6}\ \frac{35\pi}{6}- 3\times2\pi&=\frac{35\pi}{6}-6\pi\ &=\frac{35\pi - 36\pi}{6}\ &=-\frac{\pi}{6}\ \frac{35\pi}{6}- 2\times2\pi&=\frac{35\pi}{6}-4\pi\ &=\frac{35\pi - 24\pi}{6}\ &=\frac{11\pi}{6} \end{align*} ] The coterminal angle of (t=\frac{35\pi}{6}) in the range ([0, 2\pi)) is (\frac{11\pi}{6})

Step2: Determine the terminal point

We know that for a real number (t), if the reference number is (\overline{t}) and the terminal point of (\overline{t}) is ((x',y')), then:

• If (t) is in the fourth quadrant ((\frac{3\pi}{2}<t<2\pi)), and the reference number (\overline{t}=2\pi - t) (or vice - versa for coterminal angles). For (t = \frac{11\pi}{6}), the reference number (\overline{t}=\frac{\pi}{6})
• The terminal point of (\overline{t}=\frac{\pi}{6}) is ((\frac{\sqrt{3}}{2},\frac{1}{2}))
• Since (t=\frac{11\pi}{6}) (or its coterminal angle (\frac{35\pi}{6})) is in the fourth quadrant ((x>0,y < 0))
• Using the formula ((x,y)=(x',-y')) where ((x',y')) is the terminal point of the reference number. For (\overline{t}=\frac{\pi}{6}) with ((x',y')=(\frac{\sqrt{3}}{2},\frac{1}{2})), we get ((x,y)=(\frac{\sqrt{3}}{2},-\frac{1}{2}))

Answer:

((x,y)=\left(\frac{\sqrt{3}}{2},-\frac{1}{2}\right))