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Question #40416

Difficulty: 8/10 | 🟢 Free

The Question

In a triangle ABCABC, it is given that cos⁡(3A)+cos⁡(3B)+cos⁡(3C)=1\cos(3A) + \cos(3B) + \cos(3C) = 1 and ∠B>∠C\angle B > \angle C. If sin⁡(3B)+cos⁡(3B)=0\sin(3B) + \cos(3B) = 0, then the value of 2π∑k∈{A,B,C}(arcsin⁡(sin⁡3k)+arccos⁡(cos⁡3k))\frac{2}{\pi} \sum_{k \in \{A,B,C\}} \left( \arcsin(\sin 3k) + \arccos(\cos 3k) \right) is equal to:

Options

  • A.

    3

  • B.

    4

  • C.

    7

  • D.

    12

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

Double and Triple Angle Ratios|Conditional Identity Constraints|Composition Continuity

Why This Question Is Tricky

Common Mistakes

This question tests the intersection of conditional trigonometric identities and the domain restrictions of inverse trigonometric functions.

Students often blindly apply the heuristic arcsin⁡(sin⁡x)=x\arcsin(\sin x) = x and arccos⁡(cos⁡x)=x\arccos(\cos x) = x, treating them as purely algebraic cancellations.

By forcing the angles into specific quadrants via the identity cos⁡(3A)+cos⁡(3B)+cos⁡(3C)=1\cos(3A)+\cos(3B)+\cos(3C)=1 (which requires one angle to be 2π/32\pi/3), the triple angles 3A3A and 3B3B fall outside the principal domains [−π/2,π/2][-\pi/2, \pi/2] and [0,π][0, \pi].

This creates a 'continuity rupture' where the composite functions must be evaluated using their periodic definitions, trapping students who ignore the piecewise nature of inverse compositions.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterTrigonometric Functions - Class 11
ConceptDouble and Triple Angle Ratios
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree