The problem asks for the number of solutions, denoted by n(S), for the equation $\cos\theta \cos\frac{5\theta}{2} = \cos 7\theta \cos\frac{7\theta}{2}$ within the interval $\theta \in [-\pi, \pi]$.
Solving the Trigonometric Equation
We use the product-to-sum trigonometric identity: $2 \cos A \cos B = \cos(A-B) + \cos(A+B)$ Applying this to both sides of the given equation:
- Left Side: $\cos\theta \cos\frac{5\theta}{2}$ $ \implies \frac{1}{2} \left( \cos\left(\theta - \frac{5\theta}{2}\right) + \cos\left(\theta + \frac{5\theta}{2}\right) \right) $ $ = \frac{1}{2} \left( \cos\left(-\frac{3\theta}{2}\right) + \cos\left(\frac{7\theta}{2}\right) \right) $ $ = \frac{1}{2} \left( \cos\left(\frac{3\theta}{2}\right) + \cos\left(\frac{7\theta}{2}\right) \right) $ (since $\cos(-x) = \cos x$)
- Right Side: $\cos 7\theta \cos\frac{7\theta}{2}$ $ \implies \frac{1}{2} \left( \cos\left(7\theta - \frac{7\theta}{2}\right) + \cos\left(7\theta + \frac{7\theta}{2}\right) \right) $ $ = \frac{1}{2} \left( \cos\left(\frac{7\theta}{2}\right) + \cos\left(\frac{21\theta}{2}\right) \right) $
Equating the two sides:
$ \frac{1}{2} \left( \cos\left(\frac{3\theta}{2}\right) + \cos\left(\frac{7\theta}{2}\right) \right) = \frac{1}{2} \left( \cos\left(\frac{7\theta}{2}\right) + \cos\left(\frac{21\theta}{2}\right) \right) $Simplifying gives:
$ \cos\left(\frac{3\theta}{2}\right) = \cos\left(\frac{21\theta}{2}\right) $Finding General Solutions
The general solution for $\cos A = \cos B$ is $A = 2n\pi \pm B$, where n is an integer.
Applying this:
$ \frac{3\theta}{2} = 2n\pi \pm \frac{21\theta}{2} $Case 1: Using the plus sign
$ \frac{3\theta}{2} = 2n\pi + \frac{21\theta}{2} $ $ \frac{3\theta}{2} - \frac{21\theta}{2} = 2n\pi $ $ -\frac{18\theta}{2} = 2n\pi $ $ -9\theta = 2n\pi $ $ \theta = -\frac{2n\pi}{9} $Case 2: Using the minus sign
$ \frac{3\theta}{2} = 2n\pi - \frac{21\theta}{2} $ $ \frac{3\theta}{2} + \frac{21\theta}{2} = 2n\pi $ $ \frac{24\theta}{2} = 2n\pi $ $ 12\theta = 2n\pi $ $ \theta = \frac{2n\pi}{12} = \frac{n\pi}{6} $Counting Solutions in the Interval $[-\pi, \pi]$
Solutions from Case 1: $ \theta = -\frac{2n\pi}{9} $
We require $-\pi \le -\frac{2n\pi}{9} \le \pi$. Dividing by $\pi$ and multiplying by $-9/2$ gives $4.5 \ge n \ge -4.5$, or $-4.5 \le n \le 4.5$. The integer values for n are: -4, -3, -2, -1, 0, 1, 2, 3, 4. This gives 9 distinct solutions.
Solutions from Case 2: $ \theta = \frac{n\pi}{6} $
We require $-\pi \le \frac{n\pi}{6} \le \pi$. Dividing by $\pi$ and multiplying by 6 gives $-6 \le n \le 6$. The integer values for n are: -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6. This gives 13 distinct solutions.
Finding Unique Solutions
The total number of potential solutions is $9 + 13 = 22$. We must subtract any common solutions found in both cases.
Common solutions occur when $ -\frac{2n\pi}{9} = \frac{m\pi}{6} $ for integers n and m.
$ -\frac{2n}{9} = \frac{m}{6} \implies -12n = 9m \implies 4n = -3m $.This implies n must be a multiple of 3 and m must be a multiple of 4.
From Case 1, possible values for n that are multiples of 3 are: -3, 0, 3.
- If $n = -3$, $\theta = -\frac{2(-3)\pi}{9} = \frac{6\pi}{9} = \frac{2\pi}{3}$.
- If $n = 0$, $\theta = 0$.
- If $n = 3$, $\theta = -\frac{2(3)\pi}{9} = -\frac{6\pi}{9} = -\frac{2\pi}{3}$.
These three values ($-\frac{2\pi}{3}, 0, \frac{2\pi}{3}$) are also present in the solutions from Case 2.
Therefore, there are 3 common solutions.
Calculating the Cardinality n(S)
The total number of unique solutions, n(S), is:
$ \text{n(S)} = (\text{Solutions from Case 1}) + (\text{Solutions from Case 2}) - (\text{Common Solutions}) $ $ \text{n(S)} = 9 + 13 - 3 $ $ \text{n(S)} = 19 $Thus, there are 19 solutions in the set S.
Similar Questions
Let $\cos(\alpha + \beta) = -\frac{1}{10}$ and $\sin(\alpha - \beta) = \frac{3}{8}$, where $0 < \alpha < \frac{\pi}{3}$ and $0 < \beta < \frac{\pi}{4}$. If $\tan 2\alpha = \frac{3(1 - r\sqrt{5})}{\sqrt{11}(s + \sqrt{5})}, r, s \in \mathbb{N}$, then $r+s$ is equal to ________.
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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 _____.
- If $k = \tan\left(\frac{\pi}{4} + \frac{1}{2}\cos^{-1}\left(\frac{2}{3}\right)\right) + \tan\left(\frac{1}{2}\sin^{-1}\left(\frac{2}{3}\right)\right)$, then the number of solutions of the equation $\sin^{-1}(kx-1) = \sin^{-1}x - \cos^{-1}x$ is ________
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Important Questions from Trigonometry
Let $\cos(\alpha + \beta) = -\frac{1}{10}$ and $\sin(\alpha - \beta) = \frac{3}{8}$, where $0 < \alpha < \frac{\pi}{3}$ and $0 < \beta < \frac{\pi}{4}$. If $\tan 2\alpha = \frac{3(1 - r\sqrt{5})}{\sqrt{11}(s + \sqrt{5})}, r, s \in \mathbb{N}$, then $r+s$ is equal to ________.
- The number of elements in the set $\{x \in [0, 180^\circ] : \tan(x + 100^\circ) = \tan(x + 50^\circ) \tan x \tan(x - 50^\circ)\}$ is _______.
- If $\frac{\pi}{4} + \sum_{p=1}^{11} \tan^{-1} \left( \frac{2^{p-1}}{1 + 2^{2p-1}} \right) = \alpha$, then $\tan \alpha$ is equal to _________.
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 _____.
- If $k = \tan\left(\frac{\pi}{4} + \frac{1}{2}\cos^{-1}\left(\frac{2}{3}\right)\right) + \tan\left(\frac{1}{2}\sin^{-1}\left(\frac{2}{3}\right)\right)$, then the number of solutions of the equation $\sin^{-1}(kx-1) = \sin^{-1}x - \cos^{-1}x$ is ________
