What is the value of \(\sin\left(\frac{37\pi}{4}\right)\)?

1. TRANSLATE the problem information

• We need to find: \(\sin(37\pi/4)\)
• Given answer choices suggest this should equal a familiar sine value

2. INFER the approach

• Since \(37\pi/4\) is much larger than \(2\pi\), we need to use the periodicity property
• Sine repeats every \(2\pi\), so \(\sin(\mathrm{x}) = \sin(\mathrm{x} - 2\pi\mathrm{k})\) for any integer k
• Strategy: Reduce \(37\pi/4\) to an equivalent angle between 0 and \(2\pi\)

3. SIMPLIFY to find the equivalent angle

• Divide the angle by the period: \(37\pi/4 \div 2\pi = 37/8 = 4.625\)
• This means \(37\pi/4\) contains 4 full periods plus a remainder
• Calculate: \(4 \times 2\pi = 8\pi\) (the full periods)
• Find remainder: \(37\pi/4 - 8\pi = 37\pi/4 - 32\pi/4 = 5\pi/4\)
• Therefore: \(\sin(37\pi/4) = \sin(5\pi/4)\)

4. INFER the quadrant and sign

• \(5\pi/4 = 225°\), which places it in the third quadrant
• In the third quadrant, sine values are negative
• \(5\pi/4 = \pi + \pi/4\), so we can use the reference angle \(\pi/4\)

5. APPLY CONSTRAINTS to find the exact value

• Using the reference angle relationship: \(\sin(\pi + \theta) = -\sin(\theta)\)
• \(\sin(5\pi/4)\)
  \(= \sin(\pi + \pi/4)\)
  \(= -\sin(\pi/4)\)
  \(= -\sqrt{2}/2\)

Answer: A) \(-\sqrt{2}/2\)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students don't recognize they need to use periodicity and attempt to work with \(37\pi/4\) directly, leading to confusion about how to evaluate such a large angle. This causes them to get stuck and guess randomly among the answer choices.

Second Most Common Error:

Conceptual confusion about quadrant signs: Students successfully reduce to \(\sin(5\pi/4)\) but incorrectly determine the sign. They might think \(5\pi/4\) is in the fourth quadrant (where sine is negative) but miscalculate, or forget that sine is negative in the third quadrant. This may lead them to select Choice D (\(\sqrt{2}/2\)) with the wrong sign.

The Bottom Line:

This problem tests whether students can systematically handle large angles by combining periodicity with unit circle knowledge. The key insight is recognizing that reduction via periodicity transforms an intimidating large angle into a manageable standard position angle.