Let (+) = -110 and (-) = 38, where 0

Question

Let $\cos(\alpha+\beta) = -\dfrac1{10}$ and $\sin(\alpha-\beta) = \dfrac38$, where $0 < \alpha < \dfrac\pi3$ and $0 < \beta < \dfrac\pi4$. If $	an 2\alpha = \dfrac{3(1-r\sqrt5)}{\sqrt{11}(s+\sqrt5)}$, where $r,s \in \mathbb{N}$, then $r+s$ is equal to \_\_\_\_\_\_\_\_\_\_\_

Answer: 20

Accepted Answer

Since $\cos(\alpha+\beta) = -\tfrac1{10}$, $\sin(\alpha+\beta) = \sqrt{1-\tfrac1{100}} = \tfrac{3\sqrt{11}}{10}$ (positive, as $\alpha+\beta \in (0, \tfrac{7\pi}{12})$), so:

$$\tan(\alpha+\beta) = \frac{3\sqrt{11}/10}{-1/10} = -3\sqrt{11}$$

Since $\sin(\alpha-\beta) = \tfrac38$, $\cos(\alpha-\beta) = \sqrt{1-\tfrac9{64}} = \tfrac{\sqrt{55}}{8}$ (positive, as $\alpha-\beta \in (-\tfrac\pi4,\tfrac\pi3)$), so:

$$\tan(\alpha-\beta) = \frac{3}{\sqrt{55}}$$

Using $\tan 2\alpha = \tan[(\alpha+\beta)+(\alpha-\beta)] = \dfrac{\tan(\alpha+\beta)+\tan(\alpha-\beta)}{1-\tan(\alpha+\beta)\tan(\alpha-\beta)}$:

$$\tan 2\alpha = \frac{-3\sqrt{11} + \dfrac{3}{\sqrt{55}}}{1 + \dfrac{9\sqrt{11}}{\sqrt{55}}}$$

Multiplying numerator and denominator by $\sqrt{55}$:

$$\text{Numerator} = -3\sqrt{11}\cdot\sqrt{55} + 3 = -3\sqrt{605} + 3 = -33\sqrt5 + 3 = 3(1-11\sqrt5)$$
$$\text{Denominator} = \sqrt{55} + 9\sqrt{11} = \sqrt{11}(\sqrt5 + 9)$$

So $\tan 2\alpha = \dfrac{3(1-11\sqrt5)}{\sqrt{11}(9+\sqrt5)}$, giving $r=11,\ s=9$, so $r+s = 20$.

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