To determine whether the terminal side of the angle [tex]\(\theta = \frac{13\pi}{6}\)[/tex] lies in Quadrant IV, we first need to normalize [tex]\(\theta\)[/tex] to an equivalent angle between 0 and [tex]\(2\pi\)[/tex].
1. Normalize the angle [tex]\(\theta\)[/tex]:
[tex]\[
\theta = \frac{13\pi}{6}
\][/tex]
To normalize this angle, we find the equivalent angle within the range [tex]\(0 \leq \theta < 2\pi\)[/tex].
We do this by subtracting [tex]\(2\pi\)[/tex] (which is the same as subtracting a full rotation in radians) until the angle falls within the desired range:
[tex]\[
\frac{13\pi}{6} = \frac{13\pi}{6} \mod (2\pi)
\][/tex]
Converting [tex]\(2\pi\)[/tex] into a fraction with denominator 6, we have:
[tex]\[
2\pi = \frac{12\pi}{6}
\][/tex]
Now,
[tex]\[
\frac{13\pi}{6} \mod \left( \frac{12\pi}{6} \right) = \frac{13\pi}{6} - \frac{12\pi}{6} = \frac{\pi}{6}
\][/tex]
So, the equivalent angle is:
[tex]\[
\theta_{\text{normalized}} = \frac{\pi}{6}
\][/tex]
2. Determine the quadrant:
The angle [tex]\(\frac{\pi}{6}\)[/tex] is within the range [tex]\(0 \leq \theta < \frac{\pi}{2}\)[/tex]. This range corresponds to Quadrant I.
To summarize, [tex]\(\frac{13\pi}{6}\)[/tex] is equivalent to [tex]\(\frac{\pi}{6}\)[/tex] when normalized to the interval [tex]\([0, 2\pi)\)[/tex], and [tex]\(\frac{\pi}{6}\)[/tex] lies in Quadrant I, not Quadrant IV.
Therefore, the statement "The terminal side of [tex]\(\theta=\frac{13 \pi}{6}\)[/tex] is in Quadrant IV" is:
False
