Reach for the Summit - M-S2-A2

From calculator: $\sec 50^\circ + \tan 10^\circ = \sqrt 3$

Proof:

$\begin{aligned} \sec 50^\circ + \tan 10^\circ & = \frac 1{\cos 50^\circ} + \cot 80^\circ \\ & = \frac 1{\sin 40^\circ} + \frac 1{\tan (60^\circ + 20^\circ)} & \small \blue{\text{Let }t = \tan 20^\circ} \\ & = \frac {1+t^2}{2t} + \frac {1-\sqrt 3 t}{\sqrt 3 + t} & \small \blue{\text{Then }\tan 60^\circ = \frac {3t-t^3}{1-3t^2} = \sqrt 3} \\ & = \frac {\sqrt 3 + 3t - \sqrt 3 t^2 + t^3}{2t(\sqrt 3 +t)} & \small \blue{\implies t^3 = 3\sqrt 3 t^2 + 3t - \sqrt 3} \\ & = \frac {6t + 2\sqrt 3 t^2}{2t(\sqrt 3 +t)} \\ & = \frac {2\sqrt 3 t(\sqrt 3 + t)}{2t(\sqrt 3 +t)} \\ & = \sqrt 3 \end{aligned}$

From calculator: $\cos \frac {2\pi}7 + \cos \frac {4\pi}7 + \cos \frac {6\pi}7 = - \frac 12$

Proof: In general $\displaystyle \sum_{k=0}^{n-1} \cos \left(\frac {2k+1}{2n+1} \pi \right) = \frac 12$ . Then

$\begin{aligned} \cos \frac {2\pi}7 + \cos \frac {4\pi}7 + \cos \frac {6\pi}7 & = - \cos \left(\pi - \frac {2\pi}7 \right) - \cos \left(\pi - \frac {4\pi}7 \right) - \cos \left(\pi - \frac {6\pi}7 \right) \\ & = - \cos \frac {5\pi}7 - \cos \frac {3\pi}7 - \cos \frac \pi 7 \\ & = - \frac 12 \end{aligned}$

From calculator: $\tan 6^\circ \tan 42^\circ \tan 66^\circ \tan 78^\circ = 1$

Proof: Equation (35): $\displaystyle \prod_{k=1}^{\lfloor (n-1)/2 \rfloor} \tan \left( \frac {k \pi}n \right) = \begin{cases} \sqrt n & \text{if }n \text{ is odd} \\ 1 & \text{if }n \text{ is even} \end{cases}$ , where $\lfloor \cdot \rfloor$ denotes the floor function.

$\begin{aligned} \tan \frac \pi {30} \tan \frac {7\pi}{30} \tan \frac {11\pi}{30} \tan \frac {13\pi}{30} & = \cot \frac {7\pi}{15} \cot \frac {4\pi}{15} \cot \frac {2\pi}{15} \cot \frac \pi{15} \\ & = \frac 1{\tan \frac \pi{15} \tan \frac {2\pi}{15} \tan \frac {4\pi}{15} \tan \frac {7\pi}{15}} \\ & = \frac {\tan \frac {3\pi}{15} \tan \frac {5\pi}{15} \tan \frac {6\pi}{15}}{\prod_{k=1}^7 \tan \frac {k \pi}{15}} \\ & = \frac {\tan \frac \pi3 \prod_{k=1}^2 \tan \frac {k\pi}5}{\prod_{k=1}^7 \tan \frac {k \pi}{15}} = \frac {\sqrt 3 \cdot \sqrt 5}{\sqrt{15}} = 1 \end{aligned}$

Therefore the answer is $A = \sqrt 3 \left(-\frac 12\right)(1) \approx - 0.866025404$ . $\implies \lfloor 10000 A \rfloor = \boxed {-8661}$ .

The first factor is $\sqrt 3$ , the second is $-\dfrac {1}{2}$ , and the third is $1$ .

Their product is $-\dfrac {\sqrt 3}{2}$ and the answer is $\boxed {-8661}$ .

Answer to this problem :

Same .