What is the value of \(\mathrm{cos}\left(\frac{-37\pi}{2}\right)\)?

1. INFER the approach for negative angles

• Since we have \(\cos(-37\pi/2)\), recognize that cosine is an even function
• This means \(\cos(-x) = \cos(x)\), so \(\cos(-37\pi/2) = \cos(37\pi/2)\)
• This eliminates the negative sign and simplifies our work

2. INFER the need to use periodicity

• The angle \(37\pi/2 = 18.5\pi\) is much larger than \(2\pi\)
• Since cosine repeats every \(2\pi\), we need to find an equivalent angle between 0 and \(2\pi\)
• Strategy: Find how many complete periods of \(2\pi\) fit into \(37\pi/2\)

3. SIMPLIFY using the period of cosine

• Calculate: \(37\pi/2 \div 2\pi = 37\pi/2 \times 1/2\pi = 37/4 = 9.25\)
• This means \(37\pi/2 = 9.25 \times 2\pi = 9\) complete periods \(+ 0.25\) periods
• Convert the remainder: \(0.25 \times 2\pi = \pi/2\)
• Therefore: \(37\pi/2 = 18\pi + \pi/2\)

4. INFER the final equivalent angle

• Since \(18\pi\) represents exactly 9 complete periods of \(2\pi\), it doesn't change the cosine value
• So \(\cos(37\pi/2) = \cos(18\pi + \pi/2) = \cos(\pi/2)\)

5. APPLY unit circle knowledge

• From the unit circle: \(\cos(\pi/2) = 0\)
• Therefore: \(\cos(-37\pi/2) = 0\)

Answer: C (0)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students may not recognize that they need to use the even function property first, or they might try to work directly with the negative angle, making the arithmetic more complicated.

Without using \(\cos(-x) = \cos(x)\), they might attempt complex calculations with the negative angle or get confused about the sign of their final answer. This often leads to confusion and guessing among the answer choices.

Second Most Common Error:

Poor SIMPLIFY execution: Students make arithmetic errors when finding how many complete periods fit into \(37\pi/2\), such as incorrectly calculating \(37/4\) or not properly converting the decimal remainder back to radians.

For example, if they miscalculate and think the equivalent angle is \(3\pi/2\) instead of \(\pi/2\), they would get \(\cos(3\pi/2) = 0\) (which happens to still be correct) or make other angle errors that lead to Choice A (1) or Choice D (-1).

The Bottom Line:

This problem tests whether students can systematically reduce complex angles using fundamental trigonometric properties rather than being intimidated by the large numbers. The key insight is recognizing that no matter how large or negative the angle, the same basic properties (even functions and periodicity) always apply.