# TrigonometryProduct of a lot of cosines

#### caffeinemachine

##### Well-known member
MHB Math Scholar
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
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
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
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

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

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

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

#### caffeinemachine

##### Well-known member
MHB Math Scholar
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

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

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

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

That's masterful. And very nicely typed in LaTeX. I was wondering why anemone has not yet answered this question.

##### Well-known member
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

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

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

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