What is the Product of a Lot of Cosines?

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In summary, the conversation is about evaluating the product P, with Q as a helpful intermediate step, and two proposed methods for simplifying P. The first method involves using Euler's formula, while the second method involves using trigonometric identities and simplifying the product into a geometric series. Both methods ultimately lead to the solution of P being equal to 1 over 2 to the power of 1004.
  • #1
caffeinemachine
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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?
 
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  • #2
caffeinemachine said:
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...
 
  • #3
I like Serena said:
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.
 
  • #4
caffeinemachine said:
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/trigonometry-12/simplify-cos-cos-2a-cos-3a-cos-999a-if-%3D-2pi-1999-a-253.html
 
  • #5
anemone said:
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/trigonometry-12/simplify-cos-cos-2a-cos-3a-cos-999a-if-%3D-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. :)
 
  • #6
anemone said:
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/trigonometry-12/simplify-cos-cos-2a-cos-3a-cos-999a-if-%3D-2pi-1999-a-253.html

I am unable to access the link above.
 
  • #7
kaliprasad said:
I am unable to access the link above.

Try this link:

http://mathhelpboards.com/trigonometry-12/simplify-cos-cos-2a-cos-3a-cos-999a-if-2pi-1999-a-253.html
 

FAQ: What is the Product of a Lot of Cosines?

What is the "Product of a lot of cosines"?

The "Product of a lot of cosines" refers to a mathematical concept where the product of multiple cosine functions is calculated. It is commonly used in trigonometry and can help solve problems involving multiple angles and periodic functions.

How is the "Product of a lot of cosines" calculated?

The "Product of a lot of cosines" is calculated by multiplying together all the cosine functions in the given equation. For example, if the equation is cos(x) * cos(2x) * cos(3x), the product would be cos(x) * cos(2x) * cos(3x) = cos(x + 2x + 3x) = cos(6x).

What is the significance of the "Product of a lot of cosines" in science?

The "Product of a lot of cosines" has various applications in science, particularly in fields such as physics and engineering. It can be used to model periodic phenomena, such as sound waves, and can also help in solving problems involving multiple angles and oscillating systems.

Are there any special properties of the "Product of a lot of cosines"?

Yes, the "Product of a lot of cosines" has several special properties that can be useful in solving equations. One property is that the product of two cosine functions with different frequencies is equivalent to the sum of two cosine functions with the individual frequencies. This is known as the cosine product formula.

Can the "Product of a lot of cosines" be simplified?

Yes, the "Product of a lot of cosines" can be simplified using various trigonometric identities and formulas. For example, the cosine double-angle formula can be used to simplify a cosine function with a double angle in the argument. Additionally, the cosine sum and difference formulas can also be helpful in simplifying the product of multiple cosine functions.

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