Subject: Mathematics — Limits
Question: For non-negative integers $n$, let
$f(n) = \frac{\sum_{k = 0}^{n}\mspace{2mu}\mspace{2mu} sin\left( \frac{k + 1}{n + 2}\pi \right)sin\left( \frac{k + 2}{n + 2}\pi \right)}{\sum_{k = 0}^{n}\mspace{2mu}\mspace{2mu}\sin^{2}\left( \frac{k + 1}{n + 2}\pi \right)}$\
Assuming $\cos^{- 1}x$ takes values in $\lbrack 0,\pi\rbrack$, and
$\alpha = tan\left( \cos^{- 1}f(6) \right)$, then find the value of
$\alpha^{2} + 2\alpha - 1$
Answer: 0 (zero)
Solution: $f(n) = \frac{\sum_{k = 0}^{n}\mspace{2mu} 2sin\left( \frac{k + 1}{n + 2}\pi \right)sin\left( \frac{k + 2}{n + 2}\pi \right)}{\sum_{k = 0}^{n}\mspace{2mu} 2\sin^{2}\left( \frac{k + 1}{n + 2} \right)\pi}$,\
where $n$ is non negative integer\
$${= \frac{\sum_{k = 0}^{n}\mspace{2mu}\left\lbrack cos\left( \frac{\pi}{n + 2} \right) - cos\frac{(2k + 3)\pi}{n + 2} \right\rbrack}{\sum_{k = 0}^{n}\mspace{2mu}\left\lbrack 1 - cos\frac{2(k + 1)\pi}{n + 2} \right\rbrack}
}{= \frac{(n + 1)cos\left( \frac{\pi}{n + 2} \right) - \left\lbrack cos\frac{3\pi}{n + 2} + cos\frac{5\pi}{n + 2} + \ldots\ldots\ldots + cos\frac{(2n + 3)\pi}{n + 2} \right\rbrack}{n + 1 - \left\lbrack cos\frac{2\pi}{n + 2} + cos\frac{4\pi}{n + 2} + \ldots\ldots + cos\frac{2(n + 1)\pi}{n + 2} \right\rbrack}
}{\frac{(n + 1)cos\left( \frac{\pi}{n + 2} \right) - \frac{\frac{sin(n + 1)\pi}{n + 2}}{sin\left( \frac{\pi}{n + 2} \right)} \cdot cos\frac{(2n + 6)\pi}{2(n + 2)}}{sin(n + 1)\pi}
}{n + 1 - \frac{\frac{sin(n + 1)\pi}{n + 2}}{sin\left( \frac{\pi}{n + 2} \right)} \cdot cos\frac{(2n + 4)\pi}{2(n + 2)}
}{= \frac{(n + 1)cos\frac{\pi}{n + 2} + cos\frac{\pi}{n + 2}}{n + 1 + 1} = \frac{(n + 2)cos\left( \frac{\pi}{n + 2} \right)}{n + 2}
}{\therefore f(n) = cos\left( \frac{\pi}{n + 2} \right)
}{\lim_{n \rightarrow \infty}\mspace{2mu} f(n) = \lim_{n \rightarrow \infty}\mspace{2mu} cos\left( \frac{\pi}{n + 2} \right) = 1
}{f(4) = cos\left( \frac{\pi}{4 + 2} \right) = cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}
}$$If $\alpha = tan\left( \cos^{- 1}f(6) \right)$\
$$= tan\left( \cos^{- 1}\left( cos\frac{\pi}{8} \right) \right) = tan\frac{\pi}{8}
$$Now,
$tan\frac{\pi}{4} = 1 \Rightarrow \frac{2tan\frac{\pi}{8}}{1 - \tan^{2}\pi/8} = 1$\
$${\Rightarrow \frac{2\alpha}{1 - \alpha^{2}} = 1 \Rightarrow \alpha^{2} + 2\alpha - 1 = 0
}{\sin\left( 7\cos^{- 1}{f(5)} \right) = \sin\left( 7\cos^{- 1}\left( \cos\frac{\pi}{7} \right) \right) = \sin\left( 7 \times \frac{\pi}{7} \right)}$$
$sin\ \pi = 0$