Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 ?

share|improve this question

2 Answers 2

up vote 4 down vote accepted

Hint: Multiply

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

share|improve this answer
    
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 $2^{n+1}\prod_{0\le k\le n}\cos (2^{-k})=\dfrac{\sin 2}{\sin 2^{-n}}$

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.