Difficulty: Easy
Correct Answer: 0
Explanation:
Introduction / Context:
This question is designed to test your knowledge of complementary angles in trigonometry. In many aptitude exams, angles like 37° and 53° appear together because they add up to 90°, making them complements. The most important identity here is the co-function relationship:
sin(90° − θ) = cos θ and cos(90° − θ) = sin θ.
Once you recognize 53° = 90° − 37°, you can rewrite cos 53° in terms of sin 37°. That immediately turns the expression into a subtraction of two identical quantities, giving a clean result. These problems are meant to reward pattern recognition rather than heavy calculation. A common mistake is to treat cos 53° and sin 37° as unrelated and attempt approximate decimal values, which is unnecessary and can introduce rounding errors. The correct approach is purely identity-based and produces an exact value instantly.
Given Data / Assumptions:
Concept / Approach:
Rewrite cos 53° using the complementary identity:
cos 53° = cos(90° − 37°) = sin 37°.
Then subtract sin 37° from itself. This gives an exact value without any computation of sin or cos numerically.
Step-by-Step Solution:
1) Note that 53° = 90° − 37°.
2) Apply the co-function identity:
cos(90° − θ) = sin θ
3) Therefore:
cos 53° = cos(90° − 37°) = sin 37°
4) Substitute into the expression:
cos 53° − sin 37° = sin 37° − sin 37° = 0
Verification / Alternative check:
If you approximate, sin 37° ≈ 0.6018 and cos 53° ≈ 0.6018 as well, so the difference is approximately 0. This supports the exact identity-based result, but the identity method is the correct exact approach.
Why Other Options Are Wrong:
• 1: would only happen if cos 53° were 1 and sin 37° were 0, which is impossible.
• 2 cos 53° or 2 sin 37°: these correspond to addition, not subtraction of equal terms.
• sin 53°: is a different co-function (sin 53° = cos 37°), not the given expression.
Common Pitfalls:
• Not noticing that 37° and 53° are complementary.
• Using incorrect identity like cos(90° − θ) = cos θ (false).
• Attempting decimal approximations and losing exactness.
Final Answer:
0
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