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Question #40455

Difficulty: 8/10 | 🟢 Free

The Question

Let $S_1 = \sum_{k=1}^{10} \sin^2\left(\frac{k\pi}{12}\right)$ , $S_2 = \sum_{k=1}^{10} \cos^2\left(\frac{k\pi}{12}\right)$ , and $S_3 = \sum_{k=1}^{10} \sin\left(\frac{k\pi}{12}\right)\cos\left(\frac{k\pi}{12}\right)$ . Find the value of $\lfloor S_1 S_2 - S_3^2 \rfloor$ , where $\lfloor x \rfloor$ denotes the greatest integer less than or equal to $x$ .

Options

A.
23
B.
24
C.
25
D.
26

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Bridge

Practice problems that bridge the gap between foundational theory and advanced practical application.

Concepts Behind This Question

Lagrange Identity Equivalents|Lagrange Identity|Greatest Integer Function Traps

Why This Question Is Tricky

Common Mistakes

This problem tests the application of Lagrange's Identity to condense a complex product of trigonometric series into a single pairwise sum.

The core trap lies in the final evaluation: the exact value is $24.5 - \frac{\sqrt{3}}{4}$ .

Students who loosely approximate $\sqrt{3}$ or fail to rigorously bound the fractional part might incorrectly assume the subtraction drops the value below 24 (yielding 23) or overestimates it (yielding 25).

Rigorous inequality bounding (recognizing $\frac{\sqrt{3}}{4} < 0.5$ ) is required to confidently conclude the floor is exactly 24, penalizing those who rely on rough guesswork near integer boundaries.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Trigonometric Functions - Class 11
Concept	Lagrange Identity Equivalents
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free