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A student of mine has trouble with the following, and so do I. The solution should be easy since it has been ask to première S students (equivalent to American 11th grade I guess).

The question is following : Simplify $P = \prod_{k=0}^n \cos(2^{-k})$.

Using the identity $\cos x = 2\cos^2\left(\frac{x}{2}\right) - 1$, we can express $P$ as $\prod_{k=0}^n f^{(n)}(\cos 1)$, where $f^{(n)}$ is the $n$th iteration of $t\mapsto \sqrt{\frac{t+1}{2}}$, but that's not really a simplification.

If we define the polynomials $g_0 = x$ and $g_{n+1} = xg_n(2 x^2-1)$, then we have $P = g_n(\cos 2^{-n})$, but again it's not really a simplification for a 16 year old student.

Do you have an idea ?

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up vote 4 down vote accepted

Hint: Multiply

$$ \frac{2^n \sin (2^{-n})}{2^n \sin (2^{-n})}. $$

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Clever, thanks ! – Lierre Apr 28 '13 at 9:30
+1: Very nice! – Blue Apr 28 '13 at 9:41

HINT: Using Euler's formula, $2\cos y=e^{iy}+e^{-iy},2i\sin y=e^{iy}-e^{-iy}$

$2\cos (2^{-k})=e^{i2^{-k}}+e^{-i2^{-k}}$

Putting $k=0,1,2\cdots, n-1,n$

$2\cos (2^{-0})=e^{i}+e^{-i}$

$2\cos (2^{-1})=e^{\frac i2}+e^{-\frac i2}$

$2\cos (2^{-2})=e^{\frac i{2^2}}+e^{-\frac i{2^2}}$

$\cdots $

$2\cos (2^{-n})=e^{i2^{-n}}+e^{-i2^{-n}}$

Now, $(e^{i}+e^{-i})(e^{\frac i2}+e^{-\frac i2})(e^{\frac i{2^2}}+e^{-\frac i{2^2}})\cdots (e^{i2^{-n}}+e^{-i2^{-n}})=\dfrac{(e^{2i}-e^{-2i})}{(e^{i2^{-n}}-e^{-i2^{-n}})}=\dfrac{2i\sin 2}{2i\sin 2^{-n}}$

So, $\prod_{0\le k\le n}2\cos (2^{-k})=\dfrac{\sin 2}{\sin 2^{-n}}$

or $2^{n+1}\prod_{0\le k\le n}\cos (2^{-k})=\dfrac{\sin 2}{\sin 2^{-n}}$

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Thanks, unfortunately Euler's formula is taught in Terminale S, not Première ;) – Lierre Apr 28 '13 at 9:44
@lab : how did you get the first equality after "Now"? – Stefan Smith Apr 29 '13 at 16:43
@StefanSmith, $a^2-b^2=(a+b)(a-b), a^4-b^4=(a^2+b^2)(a^2-b^2)=(a^2+b^2)(a+b)(a-b)$ $\implies a^{2^n}-b^{2^n}=(a^{2^{n-1}}+b^{2^{n-1}})(a^{2^{n-2}}+b^{2^{n-2}})(a^{2^{n-3}}+b‌​^{2^{n-3}})\cdots (a^2+b^2)(a+b)(a-b)$ Put $a=e^{i2^{-n}},b=e^{-i2^{-n}}$ – lab bhattacharjee Apr 30 '13 at 4:13
@labbhattacharjee : Thanks. – Stefan Smith Apr 30 '13 at 19:27

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