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Question #41231

Difficulty: 8/10 | 🟢 Free

The Question

Let $z_1, z_2, \dots, z_{10}$ be the non-real roots of the equation $z^{11} = 1$ . Let $\theta_k = \text{Arg}(z_k)$ , where $\text{Arg}(z)$ denotes the principal argument of $z$ such that $-\pi < \text{Arg}(z) \le \pi$ . Evaluate the sum: $S = \sum_{k=1}^{10} \left( \cos^{-1}(\cos \theta_k) + \sin^{-1}(\sin \theta_k) \right)^2$

Options

A.
$\frac{50\pi^2}{11}$
B.
$\frac{160\pi^2}{11}$
C.
$\frac{80\pi^2}{11}$
D.
$\frac{25\pi^2}{11}$

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

Inverse Function Periodicity|Roots of Unity Properties|Domain and Range Nuance

Why This Question Is Tricky

Common Mistakes

This question tests the nuanced interaction between the principal arguments of complex roots of unity and the restricted ranges of inverse trigonometric functions.

A common trap is assuming $\cos^{-1}(\cos \theta) = \theta$ and $\sin^{-1}(\sin \theta) = \theta$ for all $\theta$ , which ignores the branch cuts of these functions.

Students often blindly apply this identity to the arguments of the roots of unity, failing to recognize that for arguments in the second and third quadrants (e.g., $\pm 6\pi/11$ ), the inverse sine function maps to $\pi - \theta$ or $-\pi - \theta$ .

By squaring the sum of these functions, the question prevents trivial cancellations that might otherwise mask this conceptual error, forcing a rigorous evaluation of the piecewise definitions of inverse trigonometric mappings.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Inverse Trigonometric Functions - Class 12
Concept	Inverse Function Periodicity
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free