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For a non negative integer $n,$ let $$f(n)=\frac{\sum^{n}_{k=0}\sin\left(\frac{k+1}{n+2}\pi\right)\sin\left(\frac{k+2}{n+2}\pi\right)}{\sum^{n}_{k=0}\sin^2\left(\frac{k+1}{n+2}\pi\right)}$$

Assuming $\cos^{-1}(x)$ takes values in $ [0,\pi]$. Which of the following options is/are correct.

$(a)\ f(4)=\frac{\sqrt{3}}{2}$

$(b)\ $If $\alpha =\tan(\cos^{-1}(f(6))$, Then $\alpha^2+2\alpha-1=0$

$(c)\ \sin(7\cos^{-1}(f(5))=0$

$(d)\ \displaystyle \lim_{n\rightarrow\infty}f(n)=\frac{1}{2}$

If $\displaystyle \frac{\pi}{n+2}=z$, then \begin{align*}f(n)&=\frac{\sum^{n}_{k=0}2\sin((k+2)z)\sin((k+1)z)}{\sum^{n}_{k=0}2\sin^2(k+1)}\\ &=\frac{\sum^{n}_{k=0}\cos(z)-\cos((2k+3)z)}{\sum^{n}_{k=0}(1-\cos(2k+2)z)}\\ &=\frac{(n+1)\cos(z)-\sum^{n}_{k=0}\cos((2k+3)z)}{\sum^{n}_{k=0}(1-\cos(2k+2)z)}.\end{align*}

How do I simplify it after that?

Please have a look

qianqian07
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Priti Bisht
  • 421

4 Answers4

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HINT: $$\sum\limits^{n}_{k=0}2\sin\frac{\pi}{n+2}\cos\frac{(2k+3)\pi}{n+2}=\sum\limits^{n}_{k=0}(\sin\frac{(2k+4)\pi}{n+2}-\sin\frac{(2k+2)\pi}{n+2})$$ $$\sum\limits^{n}_{k=0}2\sin\frac{\pi}{n+2}\cos\frac{(2k+2)\pi}{n+2}=\sum\limits^{n}_{k=0}(\sin\frac{(2k+3)\pi}{n+2}-\sin\frac{(2k+1)\pi}{n+2})$$

Krystal Justin
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Let $z:=\displaystyle \frac{\pi}{n+2}$ as you did.

The numerator of $f(n)$ can be written as $$\begin{align}&\sum^{n}_{k=0}\sin\left((k+1)z\right)\sin\left((k+2)z\right) \\\\&=\frac 12\sum^{n}_{k=0}2\sin\left((k+1)z\right)\sin\left((k+2)z\right) \\\\&=\frac 12\sum_{k=0}^{n}\bigg(\cos(z)-\cos((2k+3)z)\bigg) \\\\&=\frac 12(n+1)\cos(z)-\frac 12\sum_{k=0}^{n}\cos((2k+3)z)\tag1\end{align}$$

Here, Krystal Justin's hint is helpful.

$$\begin{align}(1)&=\frac 12(n+1)\cos(z)-\frac 12\cdot\frac{1}{2\sin(z)}\sum_{k=0}^{n}2\sin(z)\cos((2k+3)z) \\\\&=\frac 12(n+1)\cos(z)-\frac 12\cdot\frac{1}{2\sin(z)}\color{red}{\sum_{k=0}^{n}\bigg(\sin((2k+4)z)-\sin((2k+2)z)\bigg)}\tag2\end{align}$$

The red part can be simplified by the idea of telescoping sums.

$$\begin{align}(2)&=\frac 12(n+1)\cos(z)-\frac 12\cdot\frac{1}{2\sin(z)}\color{red}{\bigg(\sin((2n+4)z)-\sin(2z)\bigg)} \\\\&=\frac 12(n+1)\cos(z)-\frac{1}{4\sin(z)}(-\sin(2z)) \\\\&=\frac 12(n+1)\cos(z)+\frac{\cos(z)}{2} \\\\&=\frac{(n+2)}{2}\cos(z)\end{align}$$

The denominator of $f(n)$ can be written as $$\begin{align}&\sum^{n}_{k=0}\sin^2\left((k+1)z\right) \\\\&=\frac 12\sum^{n}_{k=0}2\sin^2\left((k+1)z\right) \\\\&=\frac 12\sum^{n}_{k=0}\left(1-\cos((2k+2)z)\right) \\\\&=\frac 12(n+1)-\frac 12\sum^{n}_{k=0}\cos((2k+2)z) \\\\&=\frac 12(n+1)-\frac 12\cdot\frac{1}{2\sin (z)}\sum^{n}_{k=0}2\sin(z)\cos((2k+2)z) \\\\&=\frac 12(n+1)-\frac{1}{4\sin (z)}\sum^{n}_{k=0}\left(\sin((2k+3)z)-\sin((2k+1)z)\right) \\\\&=\frac 12(n+1)-\frac{1}{4\sin (z)}\left(\sin((2n+3)z)-\sin(z)\right) \\\\&=\frac 12(n+1)-\frac{1}{4\sin (z)}\left(-2\sin z\right) \\\\&=\frac 12(n+2)\end{align}$$

Therefore, we have $$f(n)=\cos\left(\frac{\pi}{n+2}\right)$$

I think you can continue from here.

mathlove
  • 139,939
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I will try geometric summation method and I believe that this method is faster. Let $E=\exp(\frac{\pi i}{n+2})$ so that $E^{2n+4}=1$. Then, $$\begin{align} f(n)&=\frac{\sum_{k=0}^{n}(E^{k+1}-E^{-k-1})(E^{k+2}-E^{-k-2})}{\sum_{k=0}^{n}(E^{k}-E^{-k})^2}\\ &=\frac{E^3\sum_{k=0}^{n}E^{2n}+E^{-3}\sum_{k=0}^{n}E^{-2n}-\sum_{k=0}^{n}(E+E^{-1})}{E^2\sum_{k=0}^{n}E^{2n}+E^{-2}\sum_{k=0}^{n}E^{-2n}-\sum_{k=0}^{n}2}\\ &=\frac{-E-E^{-1}-(n+1)(E+E^{-1})}{-1-1-2(n+1)}\\ &=\frac{-2(n+2)\cos(\pi/(n+2))}{-2(n+2)}\\ &=\cos\left(\frac{\pi}{n+2}\right) \end{align}$$

Bob Dobbs
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Firstly, $\sin\left(\dfrac{n+2}{n+2}\pi\right)=0.\;$

Then, applying sums rearrangement, we can get $$P(n)=\sum^{\mathbf{n-1}}_{k=0}\sin\left(\frac{k+1}{n+2}\pi\right)\sin\left(\frac{k+2}{n+2}\pi\right) =\dfrac12\sum^{n-1}_{k=0}\left(\cos\left(\frac{\pi}{n+2}\right)+\cos\left(\frac{2k+3}{n+2}\pi\right)\right)$$ $$=\dfrac{n+2}2\cos\left(\frac{\pi}{n+2}\right) + \frac14\left(\sum^{n-1}_{k=0}\cos\left(\frac{2k+3}{n+2}\pi\right) +\sum^{n-1}_{k=0}\cos\left(\frac{2(n-1-k)+3}{n+2}\pi\right)\right)$$ $$=\dfrac{n+2}2\cos\left(\frac{\pi}{n+2}\right) + \frac14\sum^{n-1}_{k=0}\left(\cos\left(\frac{2k+3}{n+2}\pi\right)+\cos\left(\pi-\frac{2k+3}{n+2}\pi\right)\right) \mathbf{=\frac{n+2}2\cos\left(\frac{\pi}{n+2}\right)},$$

$$Q(n)=\sum^{\mathbf{n+1}}_{k=0}\sin^2\left(\frac{k+1}{n+2}\pi\right) =\frac12\sum^{n+1}_{k=0}\left(1-\cos\left(\frac{k+1}{n+2}2\pi\right)\right)$$ $$=\frac{n+2}2-\frac14\left(\sum^{n+1}_{k=0}\cos\left(\frac{k+1}{n+2}2\pi\right)+\sum^{n+1}_{k=0}\cos\left(\frac{{(n+1-k)}+1}{n+2}2\pi\right)\right)$$ $$=\frac{n+2}2-\frac14\sum^{n+1}_{k=0}\left(\cos\left(\frac{k+1}{n+2}2\pi\right)+\cos\left(\pi-\frac{k+1}{n+2}2\pi\right)\right)\mathbf{=\frac{n+2}2}.$$

Therefore, $$\mathbf{f(n)=\frac{P(n)}{Q(n)}=\cos\left(\frac\pi{n+2}\right)}.$$

Yuri Negometyanov
  • 28,026