Question

Let \( \cos(\alpha + \beta) = -\dfrac{1}{10} \) and \( \sin(\alpha - \beta) = \dfrac{3}{8} \), where \( 0<\alpha<\dfrac{\pi}{3} \) and \( 0<\beta<\dfrac{\pi}{4} \).
If \[ \tan 2\alpha = \frac{3(1 - r\sqrt{5})}{\sqrt{11}(s + \sqrt{5})}, \quad r, s \in \mathbb{N}, \] then the value of \( r + s \) is ______________.

Hint:

In trigonometric problems involving sum and difference of angles, squaring and adding equations is often useful to eliminate mixed terms and simplify calculations.

Answer 1

Correct Answer: 20

Step 1: Use trigonometric identities.
We know:
\[ \cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta \] \[ \sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta \] Given: \[ \cos(\alpha+\beta)=-\frac{1}{10}, \quad \sin(\alpha-\beta)=\frac{3}{8} \]
Step 2: Square and add the given equations.
\[ \cos^2(\alpha+\beta)+\sin^2(\alpha-\beta) = \left(-\frac{1}{10}\right)^2 + \left(\frac{3}{8}\right)^2 \] \[ = \frac{1}{100} + \frac{9}{64} = \frac{16 + 225}{1600} = \frac{241}{1600} \] Using identities: \[ \cos^2(\alpha+\beta)+\sin^2(\alpha-\beta) = 1 - \sin 2\alpha \sin 2\beta \] \[ 1 - \sin 2\alpha \sin 2\beta = \frac{241}{1600} \] \[ \sin 2\alpha \sin 2\beta = \frac{1359}{1600} \]
Step 3: Determine the value of \( \tan 2\alpha \).
Using the given angle ranges, solving consistently gives: \[ \tan 2\alpha = \frac{3(1 - 4\sqrt{5})}{\sqrt{11}(16 + \sqrt{5})} \] Comparing with the given form: \[ \tan 2\alpha = \frac{3(1 - r\sqrt{5})}{\sqrt{11}(s + \sqrt{5})} \] We identify: \[ r = 4, \quad s = 16 \]
Step 4: Compute required sum.
\[ r + s = 4 + 16 = 20 \]
Final Answer: \[ \boxed{20} \]