If θ is an angle in standard position and its terminal side passes through point (−12,3√2)(−12,32)on the unit circle, what is a possible value of θ?

Break Down the Problem

1. We need to find the angle $\theta$ whose terminal side passes through the point $(-12, 3\sqrt{2})$.
2. Since the point is not on the unit circle as it does not satisfy the equation $x^2 + y^2 = 1$, we will first normalize the coordinates to find the corresponding point on the unit circle.

Relevant Concepts

1. The radius $r$ can be determined by the distance formula from the origin to the given point.
2. The angle $\theta$ can then be found using the inverse tangent function: $\theta = \tan^{-1}\left(\frac{y}{x}\right)$

Analysis and Detail

1. Calculate the radius $r$: $r = \sqrt{(-12)^2 + (3\sqrt{2})^2} = \sqrt{144 + 18} = \sqrt{162} = 9\sqrt{2}$

2. Find the normalized coordinates: The coordinates of the point on the unit circle can be calculated as: $\left( \frac{-12}{r}, \frac{3\sqrt{2}}{r} \right) = \left( \frac{-12}{9\sqrt{2}}, \frac{3\sqrt{2}}{9\sqrt{2}} \right) = \left( -\frac{4}{3\sqrt{2}}, \frac{1}{3} \right)$

3. Calculate $\theta$: Using the tangent: $\tan(\theta) = \frac{y}{x} = \frac{\frac{1}{3}}{-\frac{4}{3\sqrt{2}}} = -\frac{\sqrt{2}}{4}$ To find $\theta$, we take the arctangent: $\theta = \tan^{-1}\left(-\frac{\sqrt{2}}{4}\right)$

Verify and Summarize

Since the tangent function is negative, $\theta$ is in the second quadrant or the fourth quadrant.

Final Answer

A possible value of $\theta$ (in radians) is approximately: $\theta \approx \tan^{-1}\left(-\frac{\sqrt{2}}{4}\right) + \pi \quad (\text{2nd quadrant}) \approx 2.1588 \text{ radians} \text{ (or)} \quad 6.108 \text{ radians} \text{ (if in 4th quadrant)}$

So, one possible value is $\theta \approx 2.1588 \text{ radians}$.