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Trigonometry Product of a lot of cosines

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
Hello MHB. I need help with the follwoing:

Evaluate $\displaystyle P=\prod_{j=1}^{1004}\cos\left(\frac{2\pi j}{2009}\right)$.

I think it would be easier to first evaluate $\displaystyle Q=\prod_{j=1}^{2008}\cos\left(\frac{2\pi j}{2009}\right)$. Knowing the value of $Q$ we can easily find the value of $P$.

Can anybody help?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,774
Hello MHB. I need help with the follwoing:

Evaluate $\displaystyle P=\prod_{j=1}^{1004}\cos\left(\frac{2\pi j}{2009}\right)$.

I think it would be easier to first evaluate $\displaystyle Q=\prod_{j=1}^{2008}\cos\left(\frac{2\pi j}{2009}\right)$. Knowing the value of $Q$ we can easily find the value of $P$.

Can anybody help?
Hi caffeinemachine! :)

When you write \(\displaystyle \cos\left(\frac{2\pi j}{2009}\right) = \frac 1 2 \left(e^{2\pi i j/2009} + e^{-2\pi i j/2009}\right)\), you can simplify P into a sum that turns out to be a geometric series with almost everything canceling...
 

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
Hi caffeinemachine! :)

When you write \(\displaystyle \cos\left(\frac{2\pi j}{2009}\right) = \frac 1 2 \left(e^{2\pi i j/2009} + e^{-2\pi i j/2009}\right)\), you can simplify P into a sum that turns out to be a geometric series with almost everything canceling...
Thank you I Like Serena. This solved the problem. :)
Wonder how can we do it without invoking Euler's Formula.
 

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,676
Hello MHB. I need help with the follwoing:

Evaluate $\displaystyle P=\prod_{j=1}^{1004}\cos\left(\frac{2\pi j}{2009}\right)$.

I think it would be easier to first evaluate $\displaystyle Q=\prod_{j=1}^{2008}\cos\left(\frac{2\pi j}{2009}\right)$. Knowing the value of $Q$ we can easily find the value of $P$.

Can anybody help?
Hi caffeinemachine,

I like Serena
's proposed method is great, of course, but I've another method in mind that I want to share with you.

\(\displaystyle P=\prod_{j=1}^{1004}\cos\left(\frac{2\pi j}{2009}\right)\)

\(\displaystyle P=\cos\left(\frac{2\pi }{2009}\right)\cos\left(\frac{4\pi }{2009}\right)\cos\left(\frac{6\pi }{2009}\right)\cdots\cos\left(\frac{2008\pi }{2009}\right)\)

Let

\(\displaystyle Q=\prod_{j=1}^{1004}\sin\left(\frac{2\pi j}{2009}\right)\)

\(\displaystyle Q=\sin\left(\frac{2\pi }{2009}\right)\sin\left(\frac{4\pi }{2009}\right)\sin\left(\frac{6\pi }{2009}\right)\cdots\sin\left(\frac{2008\pi }{2009}\right)\)

Hence

\(\displaystyle PQ=\frac{1}{2^{1004}}\sin\left(\frac{4\pi }{2009}\right)\sin\left(\frac{8\pi }{2009}\right)\sin\left(\frac{12\pi }{2009}\right)\cdots\sin\left(\frac{4016\pi }{2009}\right)\)

\(\displaystyle PQ=\frac{1}{2^{1004}}\small\left(\sin\left(\frac{4\pi }{2009}\right)\sin\left(\frac{8\pi }{2009}\right)\sin\left(\frac{12\pi }{2009}\right)\cdots\sin\left(\frac{2008\pi }{2009}\right)\right)\left(\sin\left(\frac{2012\pi }{2009}\right)\sin\left(\frac{2016\pi }{2009}\right)\sin\left(\frac{2020\pi }{2009}\right)\cdots\sin\left(\frac{4016\pi }{2009}\right)\right)\)

Note that

\(\displaystyle \sin\left(\frac{2012\pi }{2009}\right)=\sin\left(2\pi-\frac{2006\pi }{2009}\right)=-\sin\left(\frac{2006\pi }{2009}\right)\)

and

\(\displaystyle \sin\left(\frac{2016\pi }{2009}\right)=\sin\left(2\pi-\frac{2002\pi }{2009}\right)=-\sin\left(\frac{2002\pi }{2009}\right)\) and so on and so forth,

it's obvious that

\(\displaystyle PQ=\frac{1}{2^{1004}}(-1)^{502}\sin\left(\frac{2\pi }{2009}\right)\sin\left(\frac{4\pi }{2009}\right)\sin\left(\frac{6\pi }{2009}\right)\cdots\sin\left(\frac{2008\pi }{2009}\right)\)

\(\displaystyle PQ=\frac{1}{2^{1004}}Q\)

\(\displaystyle \therefore P=\frac{1}{2^{1004}}\)

P.S. You might want to read this thread as well!http://mathhelpboards.com/trigonome...s-2a-cos-3a-cos-999a-if-=-2pi-1999-a-253.html
 

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
Hi caffeinemachine,

I like Serena
's proposed method is great, of course, but I've another method in mind that I want to share with you.

\(\displaystyle P=\prod_{j=1}^{1004}\cos\left(\frac{2\pi j}{2009}\right)\)

\(\displaystyle P=\cos\left(\frac{2\pi }{2009}\right)\cos\left(\frac{4\pi }{2009}\right)\cos\left(\frac{6\pi }{2009}\right)\cdots\cos\left(\frac{2008\pi }{2009}\right)\)

Let

\(\displaystyle Q=\prod_{j=1}^{1004}\sin\left(\frac{2\pi j}{2009}\right)\)

\(\displaystyle Q=\sin\left(\frac{2\pi }{2009}\right)\sin\left(\frac{4\pi }{2009}\right)\sin\left(\frac{6\pi }{2009}\right)\cdots\sin\left(\frac{2008\pi }{2009}\right)\)

Hence

\(\displaystyle PQ=\frac{1}{2^{1004}}\sin\left(\frac{4\pi }{2009}\right)\sin\left(\frac{8\pi }{2009}\right)\sin\left(\frac{12\pi }{2009}\right)\cdots\sin\left(\frac{4016\pi }{2009}\right)\)

\(\displaystyle PQ=\frac{1}{2^{1004}}\small\left(\sin\left(\frac{4\pi }{2009}\right)\sin\left(\frac{8\pi }{2009}\right)\sin\left(\frac{12\pi }{2009}\right)\cdots\sin\left(\frac{2008\pi }{2009}\right)\right)\left(\sin\left(\frac{2012\pi }{2009}\right)\sin\left(\frac{2016\pi }{2009}\right)\sin\left(\frac{2020\pi }{2009}\right)\cdots\sin\left(\frac{4016\pi }{2009}\right)\right)\)

Note that

\(\displaystyle \sin\left(\frac{2012\pi }{2009}\right)=\sin\left(2\pi-\frac{2006\pi }{2009}\right)=-\sin\left(\frac{2006\pi }{2009}\right)\)

and

\(\displaystyle \sin\left(\frac{2016\pi }{2009}\right)=\sin\left(2\pi-\frac{2002\pi }{2009}\right)=-\sin\left(\frac{2002\pi }{2009}\right)\) and so on and so forth,

it's obvious that

\(\displaystyle PQ=\frac{1}{2^{1004}}(-1)^{502}\sin\left(\frac{2\pi }{2009}\right)\sin\left(\frac{4\pi }{2009}\right)\sin\left(\frac{6\pi }{2009}\right)\cdots\sin\left(\frac{2008\pi }{2009}\right)\)

\(\displaystyle PQ=\frac{1}{2^{1004}}Q\)

\(\displaystyle \therefore P=\frac{1}{2^{1004}}\)

P.S. You might want to read this thread as well!http://mathhelpboards.com/trigonome...s-2a-cos-3a-cos-999a-if-=-2pi-1999-a-253.html
That's masterful. And very nicely typed in LaTeX. I was wondering why anemone has not yet answered this question. :)
 

kaliprasad

Well-known member
Mar 31, 2013
1,308
Hi caffeinemachine,

I like Serena
's proposed method is great, of course, but I've another method in mind that I want to share with you.

\(\displaystyle P=\prod_{j=1}^{1004}\cos\left(\frac{2\pi j}{2009}\right)\)

\(\displaystyle P=\cos\left(\frac{2\pi }{2009}\right)\cos\left(\frac{4\pi }{2009}\right)\cos\left(\frac{6\pi }{2009}\right)\cdots\cos\left(\frac{2008\pi }{2009}\right)\)

Let

\(\displaystyle Q=\prod_{j=1}^{1004}\sin\left(\frac{2\pi j}{2009}\right)\)

\(\displaystyle Q=\sin\left(\frac{2\pi }{2009}\right)\sin\left(\frac{4\pi }{2009}\right)\sin\left(\frac{6\pi }{2009}\right)\cdots\sin\left(\frac{2008\pi }{2009}\right)\)

Hence

\(\displaystyle PQ=\frac{1}{2^{1004}}\sin\left(\frac{4\pi }{2009}\right)\sin\left(\frac{8\pi }{2009}\right)\sin\left(\frac{12\pi }{2009}\right)\cdots\sin\left(\frac{4016\pi }{2009}\right)\)

\(\displaystyle PQ=\frac{1}{2^{1004}}\small\left(\sin\left(\frac{4\pi }{2009}\right)\sin\left(\frac{8\pi }{2009}\right)\sin\left(\frac{12\pi }{2009}\right)\cdots\sin\left(\frac{2008\pi }{2009}\right)\right)\left(\sin\left(\frac{2012\pi }{2009}\right)\sin\left(\frac{2016\pi }{2009}\right)\sin\left(\frac{2020\pi }{2009}\right)\cdots\sin\left(\frac{4016\pi }{2009}\right)\right)\)

Note that

\(\displaystyle \sin\left(\frac{2012\pi }{2009}\right)=\sin\left(2\pi-\frac{2006\pi }{2009}\right)=-\sin\left(\frac{2006\pi }{2009}\right)\)

and

\(\displaystyle \sin\left(\frac{2016\pi }{2009}\right)=\sin\left(2\pi-\frac{2002\pi }{2009}\right)=-\sin\left(\frac{2002\pi }{2009}\right)\) and so on and so forth,

it's obvious that

\(\displaystyle PQ=\frac{1}{2^{1004}}(-1)^{502}\sin\left(\frac{2\pi }{2009}\right)\sin\left(\frac{4\pi }{2009}\right)\sin\left(\frac{6\pi }{2009}\right)\cdots\sin\left(\frac{2008\pi }{2009}\right)\)

\(\displaystyle PQ=\frac{1}{2^{1004}}Q\)

\(\displaystyle \therefore P=\frac{1}{2^{1004}}\)

P.S. You might want to read this thread as well!http://mathhelpboards.com/trigonome...s-2a-cos-3a-cos-999a-if-=-2pi-1999-a-253.html
I am unable to access the link above.