Question

Let \(\vec{a}_k = (\tan \theta_k) \hat{i} + \hat{j}\) and \(\vec{b}_k = \hat{i} - (\cot \theta_k) \hat{j}\), where \(\theta_k = \frac{2^{k-1}\pi}{2^n+1}\), for some \(n \in \mathbb{N}\), \(n>5\). Then the value of \(\frac{\sum_{k=1}^{n} |\vec{a}_k|^2}{\sum_{k=1}^{n} |\vec{b}_k|^2}\) is ____.

Answer 1

Correct Answer: 1

Step 1: Understanding the Concept:
We are given two sequences of vectors \(\vec{a}_k\) and \(\vec{b}_k\).
We first need to compute their squared magnitudes, \(|\vec{a}_k|^2\) and \(|\vec{b}_k|^2\).
Then, we will find the sum of these squared magnitudes over \(k = 1\) to \(n\) and compute their ratio.
Step 2: Key Formula or Approach:
Calculate the squared magnitudes:
\(|\vec{a}_k|^2 = \tan^2 \theta_k + 1^2 = \sec^2 \theta_k\).
\(|\vec{b}_k|^2 = 1^2 + (-\cot \theta_k)^2 = \csc^2 \theta_k\).
Let \(S_a = \sum_{k=1}^n \sec^2 \theta_k\) and \(S_b = \sum_{k=1}^n \csc^2 \theta_k\).
We use the trigonometric identity connecting secant and cosecant:
\(\sec^2 \theta + \csc^2 \theta = \frac{1}{\cos^2 \theta} + \frac{1}{\sin^2 \theta} = \frac{\sin^2 \theta + \cos^2 \theta}{\sin^2 \theta \cos^2 \theta} = \frac{4}{(2 \sin \theta \cos \theta)^2} = 4 \csc^2 (2\theta)\).
Rearranging this identity gives a recursive relationship: \(\sec^2 \theta = 4 \csc^2 (2\theta) - \csc^2 \theta\).
Step 3: Detailed Explanation:
Notice that the angles are defined as \(\theta_k = \frac{2^{k-1}\pi}{2^n+1}\).
This means that \(2\theta_k = 2 \left( \frac{2^{k-1}\pi}{2^n+1} \right) = \frac{2^k\pi}{2^n+1} = \theta_{k+1}\).
Substitute this into our derived identity:
\(\sec^2 \theta_k = 4 \csc^2 \theta_{k+1} - \csc^2 \theta_k\).
Now, substitute this into the sum \(S_a\):
\(S_a = \sum_{k=1}^n \sec^2 \theta_k = \sum_{k=1}^n (4 \csc^2 \theta_{k+1} - \csc^2 \theta_k)\).
Split the sum into two parts:
\(S_a = 4 \sum_{k=1}^n \csc^2 \theta_{k+1} - \sum_{k=1}^n \csc^2 \theta_k\).
Let's analyze the first sum: \(\sum_{k=1}^n \csc^2 \theta_{k+1} = \csc^2 \theta_2 + \csc^2 \theta_3 + \dots + \csc^2 \theta_{n+1}\).
We know that \(\theta_{n+1} = \frac{2^n\pi}{2^n+1}\).
We can rewrite \(\theta_{n+1}\) as: \(\theta_{n+1} = \frac{(2^n+1-1)\pi}{2^n+1} = \pi - \frac{\pi}{2^n+1} = \pi - \theta_1\).
Since \(\csc^2(\pi - \theta) = \csc^2 \theta\), we have \(\csc^2 \theta_{n+1} = \csc^2 \theta_1\).
Therefore, \(\sum_{k=1}^n \csc^2 \theta_{k+1} = \sum_{k=2}^{n+1} \csc^2 \theta_k = (\sum_{k=2}^n \csc^2 \theta_k) + \csc^2 \theta_1 = \sum_{k=1}^n \csc^2 \theta_k\).
This entire sum is exactly equal to \(S_b\).
Step 4: Final Answer:
Substituting this back into the equation for \(S_a\):
\(S_a = 4 S_b - S_b = 3 S_b\).
We are looking for the ratio \(\frac{S_a}{S_b}\):
\(\frac{\sum_{k=1}^n |\vec{a}_k|^2}{\sum_{k=1}^n |\vec{b}_k|^2} = \frac{3 S_b}{S_b} = 3\).