Question

Trigonometry Question Inside.

Answer 1

Evaluate \[\sum_{r = 1}^{n-1} \cos^2 \left( \cfrac{r \pi }{n} \right) \]

Answer 2

@mukushla @ganeshie8 @sleepyhead314 @dan815 @mathmale @Luigi0210 @ikram002p

Answer 3

i would evaluate cos^2 into 1/2(1+cos 2pi r /n )
since its periodic
\(\sum_{r = 1}^{n-1} \cos^2 \left( \cfrac{r \pi }{n} \right)=\frac{1}{2}\sum_{r}^{n-1} 1+\frac{1}{2}\sum_{r = 1}^{n-1} \cos \left( \cfrac{2r \ }{n} \right) \)

Answer 4

How did you get to that? And is it \(\left( \cfrac{2\pi r}{n} \right) \)

Answer 5

isnt it Trigonometric Identity ?
and sry i made a typo it
\( \sum_{r = 1}^{n-1} \cos^2 \left( \cfrac{r \pi }{n} \right)=\frac{1}{2}\sum_{ r}^{n-1} 1+\frac{1}{2}\sum_{r = 1}^{n-1} \cos \left( \cfrac{2\pi r \ }{n} \right) \)

Answer 6

Okay, fine! I always search for proofs of these identities. I will google it :)

So, any plans for the next step?

Answer 7

wth is this?!?

Answer 8

\( \frac{1}{2}\sum_{r = 1}^{n-1} \cos \left( \cfrac{ r \ }{n} \right)\)=\(\cos(\frac{1}{n} )+\cos(\frac{2}{n} )+\cos(\frac{3}{n} )+....\cos(\frac{n-1}{n} ) \)
and its sound like fourier series for a function :D

Answer 9

Hmm, m confused :(

This seems like a tough question to me. I've to go for now, if you have any ideas (the answer acc. to book is (n/2) - 1 ) regarding this question.. please post here, I wil check soon.

Answer 10

\[\large \sum_{r = 1}^{n-1} \cos^2 \left( \cfrac{r \pi }{n} \right) = \dfrac{1}{2}\left((n-1) + \sum_{r = 1}^{n-1} \cos \left( \cfrac{2r \pi }{n} \right) \right)\]

\[\large = \dfrac{1}{2}\left((n-1) + \mathbb{Real }\sum_{r = 1}^{n-1} e^ \left(i \cfrac{2r \pi }{n} \right) \right)\]

Answer 11

then what @ganeshie8 ?

Answer 12

that sum is just a geometric series, use the partial sum formula and evaluate

Answer 13

\[\large \sum_{r = 1}^{n-1} e^ \left(i \cfrac{2r \pi }{n} \right) = \sum_{r = 1}^{n-1}\left(e^ \left(i \cfrac{2 \pi }{n} \right)\right)^r \]

\[ \large = -1 + \sum_{r = 0}^{n-1}\left(e^ \left(i \cfrac{2 \pi }{n} \right)\right)^r \]

\[\large = -1 + \dfrac{1-e^{i2\pi}}{1-e^{i2\pi / n}} \]

\[\large = -1 + \dfrac{1-1}{1-e^{i2\pi / n}} \]

\[\large = -1 \]

Answer 14

nice !
so its 1/2(n-2) ?

Answer 15

yes !

Answer 16

Oh, very nice. @ganeshie8 - Thanks a lot! That's perfect.

Answer 17

While what the book did is :

\[\text{Sum} = \cfrac{1}{2} \sum_{r = 1}^{n-1} \left( 1+ \cos \cfrac{2r \pi}{n} \right) \\
= \cfrac{1}{2} (n-1) + \cfrac{1}{2} \left\{ \cos \cfrac{2\pi}{n} + \cos \cfrac{4\pi}{n} + ...+ \cos \cfrac{(2n-2)\pi}{n} \right\} \\
= \cfrac{1}{2} (n-1) + \cfrac{1}{2} \left\{ \cfrac{\sin (n-1) \cfrac{2\pi}{2n}}{\sin \cfrac{2n}{n.2}} . \cos \left\{ \cfrac{ 2\left(\cfrac{2\pi}{n} \right)+ (n-2) \cfrac{2\pi}{n} }{2} \right\} \right\} \\
\text{Using,} \\
\cos \alpha + \cos \left( \alpha +\beta \right) + ... + \cos \left( \alpha + (n-1) \beta \right) = \cfrac{\sin \cfrac{n \beta}{2}}{\sin \cfrac{\beta}{2}} . \cos \left\{ \cfrac{2\alpha + (n-1) \beta}{2} \right\} \\
= \cfrac{1}{2} (n-1) + \cfrac{1}{2} \left\{ \cfrac{\sin \cfrac{(n-1).\pi}{n} . \cos \pi}{\sin \pi/n} \right\} \\
= \cfrac{1}{2} (n-1) + \cfrac{1}{2} \left\{ \cfrac{\left( \sin \cfrac{\pi}{n} \right) (-1) }{\sin \left(\pi/n\right) }\right\} \\
= \cfrac{1}{2} (n-1) - \cfrac{1}{2} = \cfrac{n}{2} - 1 \\
\boxed{ \sum_{r=1}^{n-1} \cos^2 \left( \cfrac{r \pi}{n} \right) = \cfrac{n}{2} -1 }\]

Answer 18

Though, of course, @ganeshie8 's method is much easier, and good but both methods work. I was just not interested to go with the book's method as it is a long method, and it may consume a lot of time.

Answer 19

nice :) that looks like a very useful identity !