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yellowhighlighterr · Jul 16, 2023

I've gotten the first part of the question, but its the last bit that I stumble with. I followed the same steps to solve it but to no avail...

Any help??

tywebb · Jul 16, 2023

scaryshark09 · Jul 16, 2023

how do you do the first part?

tywebb · Jul 16, 2023

scaryshark09 said:
how do you do the first part?

Here is one way:

$\begin{aligned}&1+\color{red}{2\cos\left(\frac{2\pi}7\right)}+\color{green}{2\cos\left(\frac{4\pi}7\right)}+\color{blue}{2\cos\left(\frac{6\pi}7\right)}\\&=1+\color{red}{\cos\left(\frac{2\pi}7\right)}+\color{green}{\cos\left(\frac{4\pi}7\right)}+\color{blue}{\cos\left(\frac{6\pi}7\right)}+\color{blue}{\cos\left(\frac{8\pi}7\right)}+\color{green}{\cos\left(\frac{10\pi}7\right)}+\color{red}{\cos\left(\frac{12\pi}7\right)}\\&=\text{Re}\left(1+e^{2\pi i/7}+e^{4\pi i/7}+e^{6\pi i/7}+e^{8\pi i/7}+e^{10\pi i/7}+e^{12\pi i/7}\right)\\&=\text{Re}\left(\frac{e^{14\pi i/7}-1}{e^{2\pi i/7}-1}\right)\\&=0\end{aligned}$

$\therefore\ \cos\left(\frac{2\pi}7\right)+\cos\left(\frac{4\pi}7\right)+\cos\left(\frac{6\pi}7\right)=-\frac12$

tywebb · Jul 16, 2023

Note this method can be generalised to establish

$\sum\limits_{k=1}^n\cos\left(\frac{2k\pi}{2n+1}\right)=-\frac12$

tywebb · Jul 16, 2023

There are of course other methods, some listed here: https://math.stackexchange.com/ques...-frac2-pi7-cos-frac4-pi7-cos-frac6-pi7-frac12

I suspect there is more to the question yellowhighlighterr didn't put there.

It is unlikely such question would be asked unscaffolded.

tywebb · Jul 16, 2023

tywebb said:
I suspect there is more to the question yellowhighlighterr didn't put there.

I found the rest. It is Cambridge Ext. 1 Year 11 Ex. 17G Q7

and here is the solution:

yellowhighlighterr · Jul 16, 2023

tywebb said:
$\text{Well you know }\cos(x)=-\cos(\pi-x)$

$\begin{aligned}\text{So }\textstyle\cos(\frac{\pi}{7})+\cos(\frac{3\pi}{7})+\cos(\frac{5\pi}{7})&=\textstyle-\cos(\pi-\frac{\pi}{7})-\cos(\pi-\frac{3\pi}{7})-\cos(\pi-\frac{5\pi}{7})\\&=\textstyle-\cos(\frac{6\pi}{7})-\cos(\frac{4\pi}{7})-\cos(\frac{2\pi}{7})\\&=\textstyle-(\cos(\frac{2\pi}{7})+\cos(\frac{4\pi}{7})+\cos(\frac{6\pi}{7}))\\&=\textstyle-(-\frac{1}{2})\\&=\textstyle\frac{1}{2}\end{aligned}$

aaaah ok. thanks m8

An evaluation (1 Viewer)

yellowhighlighterr

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tywebb

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scaryshark09

∞∆ who let 'em cook dis long ∆∞

tywebb

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yellowhighlighterr

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