A searchlight located at the origin rotates counterclockwise and points at an angle theta (in radians) measured from the positive...

1. TRANSLATE the problem information

• Given information:
  • Searchlight at origin pointing at angle \(\theta = -\frac{70\pi}{3}\) radians
  • Need to find slope of the beam line
• What this tells us: We need the slope of a line making angle \(\theta\) with the positive x-axis

2. INFER the relationship and strategy

• The slope of a line making angle \(\theta\) with the positive x-axis is \(\tan(\theta)\)
• Since \(-\frac{70\pi}{3}\) is a very large negative angle, we should use the periodic property of tangent to find an equivalent angle that's easier to work with
• Strategy: Find \(\tan(-\frac{70\pi}{3})\) by reducing the angle first

3. SIMPLIFY the angle using periodicity

• Since \(\tan\) has period \(\pi\), we have \(\tan(\theta) = \tan(\theta + n\pi)\) for any integer n
• Add multiples of \(\pi\) to \(-\frac{70\pi}{3}\):
  \(-\frac{70\pi}{3} + n\pi = -\frac{70\pi}{3} + \frac{3n\pi}{3} = \frac{(-70 + 3n)\pi}{3}\)
• Choose \(n = 24\): \(\frac{(-70 + 72)\pi}{3} = \frac{2\pi}{3}\)
• Therefore: \(\tan(-\frac{70\pi}{3}) = \tan(\frac{2\pi}{3})\)

4. SIMPLIFY the trigonometric evaluation

• Since \(\frac{2\pi}{3} = \pi - \frac{\pi}{3}\), use the identity \(\tan(\pi - \alpha) = -\tan(\alpha)\)
• \(\tan(\frac{2\pi}{3}) = \tan(\pi - \frac{\pi}{3}) = -\tan(\frac{\pi}{3}) = -\sqrt{3}\)

Answer: A (\(-\sqrt{3}\))

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak SIMPLIFY skill: Students attempt to work directly with \(-\frac{70\pi}{3}\) without using the periodic property of tangent, or they make arithmetic errors when adding multiples of \(\pi\).

Many students see the large angle and either give up immediately or try to convert to degrees and work from there, getting lost in the conversion process. Others might correctly identify that they need to add multiples of \(\pi\) but make calculation errors like adding \(23\pi\) instead of \(24\pi\), leading to an incorrect reduced angle.

This may lead them to select Choice C (\(\frac{\sqrt{3}}{3}\)) or causes confusion and guessing.

Second Most Common Error:

Missing conceptual knowledge: Students don't remember that \(\tan(\pi - \alpha) = -\tan(\alpha)\) or confuse it with other trigonometric identities.

After correctly reducing to \(\frac{2\pi}{3}\), they might evaluate it as \(\tan(\frac{\pi}{3}) = \sqrt{3}\) instead of \(-\tan(\frac{\pi}{3}) = -\sqrt{3}\), forgetting about the supplementary angle relationship.

This may lead them to select Choice D (\(\sqrt{3}\)).

The Bottom Line:

The key challenge is recognizing that working with large angles requires using periodic properties, and then executing the angle reduction accurately while remembering the correct trigonometric identities for supplementary angles.