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 ________.
Problem Analysis
We are given trigonometric values for $\cos(\alpha + \beta)$ and $\sin(\alpha - \beta)$, along with ranges for $\alpha$ and $\beta$. We need to find the value of $r+s$ based on the expression for $\tan(2\alpha)$, where $r, s \in \mathbb{N}$.
- Given: $\cos(\alpha + \beta) = -\frac{1}{10}$
- Given: $\sin(\alpha - \beta) = \frac{3}{8}$
- Ranges: $0 < \alpha < \frac{\pi}{3}$ and $0 < \beta < \frac{\pi}{4}$
- Target expression: $\tan(2\alpha) = \frac{3(1 - r\sqrt{5})}{\sqrt{11}(s + \sqrt{5})}$
- Goal: Find $r+s$.
Calculate Intermediate Trigonometric Values
First, find the necessary sine, cosine, and tangent values for the sum and difference of the angles.
For $\alpha + \beta$:
Since $\cos(\alpha + \beta) = -\frac{1}{10}$ and $0 < \alpha + \beta < \frac{7\pi}{12}$ (Quadrant I or II), $\sin(\alpha + \beta)$ must be positive.
$\sin(\alpha + \beta) = \sqrt{1 - \cos^2(\alpha + \beta)} = \sqrt{1 - \left(-\frac{1}{10}\right)^2} = \sqrt{1 - \frac{1}{100}} = \sqrt{\frac{99}{100}} = \frac{\sqrt{99}}{10} = \frac{3\sqrt{11}}{10}$
$\tan(\alpha + \beta) = \frac{\sin(\alpha + \beta)}{\cos(\alpha + \beta)} = \frac{3\sqrt{11}/10}{-1/10} = -3\sqrt{11}$
For $\alpha - \beta$:
Since $\sin(\alpha - \beta) = \frac{3}{8}$ and $-\frac{\pi}{4} < \alpha - \beta < \frac{\pi}{3}$, and the sine value is positive, $\alpha - \beta$ must be in Quadrant I ($0 < \alpha - \beta < \frac{\pi}{3}$). Therefore, $\cos(\alpha - \beta)$ must be positive.
$\cos(\alpha - \beta) = \sqrt{1 - \sin^2(\alpha - \beta)} = \sqrt{1 - \left(\frac{3}{8}\right)^2} = \sqrt{1 - \frac{9}{64}} = \sqrt{\frac{55}{64}} = \frac{\sqrt{55}}{8}$
$\tan(\alpha - \beta) = \frac{\sin(\alpha - \beta)}{\cos(\alpha - \beta)} = \frac{3/8}{\sqrt{55}/8} = \frac{3}{\sqrt{55}}}$
Determine $\tan(2\alpha)$
Use the tangent addition formula $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$, where $A = \alpha+\beta$ and $B = \alpha-\beta$. Thus, $2\alpha = A+B$.
$\tan(2\alpha) = \tan((\alpha+\beta) + (\alpha-\beta)) = \frac{\tan(\alpha+\beta) + \tan(\alpha-\beta)}{1 - \tan(\alpha+\beta)\tan(\alpha-\beta)}$
Substitute the calculated values:
Numerator: $\tan(\alpha+\beta) + \tan(\alpha-\beta) = -3\sqrt{11} + \frac{3}{\sqrt{55}} = \frac{-3\sqrt{11}\sqrt{55} + 3}{\sqrt{55}} = \frac{-3\sqrt{11}\sqrt{5}\sqrt{11} + 3}{\sqrt{55}} = \frac{-3(11)\sqrt{5} + 3}{\sqrt{55}} = \frac{3(1 - 11\sqrt{5})}{\sqrt{55}}$
Denominator: $1 - \tan(\alpha+\beta)\tan(\alpha-\beta) = 1 - (-3\sqrt{11})\left(\frac{3}{\sqrt{55}}\right) = 1 + \frac{9\sqrt{11}}{\sqrt{55}} = 1 + \frac{9\sqrt{11}}{\sqrt{5}\sqrt{11}} = 1 + \frac{9}{\sqrt{5}} = \frac{\sqrt{5} + 9}{\sqrt{5}}$
Combine numerator and denominator:
$\tan(2\alpha) = \frac{\frac{3(1 - 11\sqrt{5})}{\sqrt{55}}}{\frac{9 + \sqrt{5}}{\sqrt{5}}} = \frac{3(1 - 11\sqrt{5})}{\sqrt{55}} \times \frac{\sqrt{5}}{9 + \sqrt{5}}$
Simplify $\frac{\sqrt{5}}{\sqrt{55}} = \frac{\sqrt{5}}{\sqrt{5}\sqrt{11}} = \frac{1}{\sqrt{11}}$:
$\tan(2\alpha) = \frac{3(1 - 11\sqrt{5})}{\sqrt{11}(9 + \sqrt{5})}$
Solve for $r$ and $s$
Compare the calculated $\tan(2\alpha)$ with the given expression:
Calculated: $\tan(2\alpha) = \frac{3(1 - 11\sqrt{5})}{\sqrt{11}(9 + \sqrt{5})}$
Given: $\tan(2\alpha) = \frac{3(1 - r\sqrt{5})}{\sqrt{11}(s + \sqrt{5})}$
By direct comparison, we identify:
- $r = 11$
- $s = 9$
Both $r=11$ and $s=9$ are natural numbers ($\mathbb{N}$), satisfying the condition.
Final Calculation
Calculate the required sum $r+s$:
$r+s = 11 + 9 = 20$
Similar Questions
- 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 _______.
- Number of solutions of $\sqrt{3}\cos 2\theta + 8\cos \theta + 3\sqrt{3} = 0, \theta \in [-3\pi, 2\pi]$ is :
- Let $\alpha$ and $\beta$ respectively be the maximum and the minimum values of the function $f(\theta) = 4\left(\sin^4\left(\frac{7\pi}{2} - \theta\right) + \sin^4(11\pi + \theta)\right) - 2\left(\sin^6\left(\frac{3\pi}{2} - \theta\right) + \sin^6(9\pi - \theta)\right), \theta \in \mathbf{R}$. Then $\alpha + 2\beta$ is equal to :
- The vertices B and C of a triangle ABC lie on the line $\frac{x}{1} = \frac{1 - y}{-2} = \frac{z - 2}{3}$. The coordinates of A and B are $(1, 6, 3)$ and $(4, 9, \alpha)$ respectively and C is at a distance of 10 units from B. The area (in sq. units) of $\Delta ABC$ 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 _________.
- If $\text{S} = \left\{\theta \in [-\pi, \pi] : \cos\theta \cos\frac{5\theta}{2} = \cos 7\theta \cos\frac{7\theta}{2}\right\}$, then $\text{n(S)}$ is equal to ___________.
- If $\text{A} = \frac{\sin 3^\circ}{\cos 9^\circ} + \frac{\sin 9^\circ}{\cos 27^\circ} + \frac{\sin 27^\circ}{\cos 81^\circ}$ and $\text{B} = \tan 81^\circ - \tan 3^\circ$, then $\frac{\text{B}}{\text{A}}$ 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 ________
- Let the maximum value of $(\sin^{-1} x)^2 + (\cos^{-1} x)^2$ for $x \in \left[-\frac{\sqrt{3}}{2}, \frac{1}{\sqrt{2}}\right]$ be $\frac{m}{n} \pi^2$, where $\gcd(m, n) = 1$. Then $m + n$ is equal to __________.
Important Questions from Trigonometry
- 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 _______.
- Number of solutions of $\sqrt{3}\cos 2\theta + 8\cos \theta + 3\sqrt{3} = 0, \theta \in [-3\pi, 2\pi]$ is :
- Let $\alpha$ and $\beta$ respectively be the maximum and the minimum values of the function $f(\theta) = 4\left(\sin^4\left(\frac{7\pi}{2} - \theta\right) + \sin^4(11\pi + \theta)\right) - 2\left(\sin^6\left(\frac{3\pi}{2} - \theta\right) + \sin^6(9\pi - \theta)\right), \theta \in \mathbf{R}$. Then $\alpha + 2\beta$ is equal to :
- The vertices B and C of a triangle ABC lie on the line $\frac{x}{1} = \frac{1 - y}{-2} = \frac{z - 2}{3}$. The coordinates of A and B are $(1, 6, 3)$ and $(4, 9, \alpha)$ respectively and C is at a distance of 10 units from B. The area (in sq. units) of $\Delta ABC$ 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 _________.
