Problem 30 - Precalculus: A Concise Course

Question

Determine two coterminal angles (one positive and one negative) for each angle. Give your answers in radians. (a) \(\theta=-\frac{9 \pi}{4}\) (b) \(\theta=-\frac{2 \pi}{15}\)

Step-by-Step Solution

Answer

The positive and negative coterminal angles for \(\theta=-\frac{9\pi}{4}\) are \(\frac{3\pi}{4}\) and \(-\frac{17\pi}{4}\) respectively. For \(\theta=-\frac{2\pi}{15}\), the positive and negative coterminal angles are \(\frac{28\pi}{15}\) and \(-\frac{2\pi}{15}\) respectively.

Step 1 Determine Positive Coterminal Angle for \(\theta=-\frac{9\pi}{4}\)

The positive coterminal angle for any theta can be found by repeatedly adding \(2\pi\) until within the range \(0 \leq \theta < 2\pi \). Thus, \(\theta_{1} = -\frac{9\pi}{4}+2\pi \times n\), where n is an integer such that \(0 \leq \theta_{1} < 2\pi \). Here, it's found by trial that n=3 works. So, \(\theta_{1}= -\frac{9\pi}{4}+2\pi \times 3 = \frac{3\pi}{4}\)

Step 2 Determine Negative Coterminal Angle for \(\theta=-\frac{9\pi}{4}\)

The negative coterminal angle for any theta can be found by repeatedly subtracting \(2\pi\) until within the range \(-2\pi < \theta \leq 0 \). Thus, \(\theta_{2} = -\frac{9\pi}{4}-2\pi \times m\), where m is an integer such that \(-2\pi < \theta_{2} \leq 0 \). Here, it's found by trial that m=1 works. So, \(\theta_{2}= -\frac{9\pi}{4}-2\pi \times 1 = -\frac{17\pi}{4}\)

Step 3 Determine Positive Coterminal Angle for \(\theta=-\frac{2\pi}{15}\)

Using similar steps as Step 1, \(\theta_{1} = -\frac{2\pi}{15}+2\pi \times n\). Here, it's found by trial that n=1 works. So, \(\theta_{1}= -\frac{2\pi}{15}+2\pi \times 1 = \frac{28\pi}{15}\)

Step 4 Determine Negative Coterminal Angle for \(\theta=-\frac{2\pi}{15}\)

Using similar steps as Step 2, \(\theta_{2} = -\frac{2\pi}{15}-2\pi \times m\). Since, \(\theta = -\frac{2\pi}{15}\) is already in the negative range \(-2\pi < \theta \leq 0 \), \(\theta_{2} = -\frac{2\pi}{15}\)