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
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Important Questions from Trigonometry
- The number of solutions of $\tan^{-1} 4x + \tan^{-1} 6x = \frac{\pi}{6}$, where $-\frac{1}{2\sqrt{6}} < x < \frac{1}{2\sqrt{6}}$, is equal to
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Considering the principal values of inverse trigonometric functions, the value of the expression $\tan\left(2\sin^{-1}\left(\frac{2}{\sqrt{13}}\right) - 2\cos^{-1}\left(\frac{3}{\sqrt{10}}\right)\right)$ is equal to :
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