Let $$(\cos^6(x)) = m\cos(6x)+n\cos(5x)+o\cos(4x)+p\cos(3x)+q\cos(2x)+r\cos(x)+a.$$
What is the value of $a$?
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Look up the product-to-sum identity. – T.J. Gaffney Feb 23 '14 at 06:48
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I gave a general approach to the problem using Fourier series techniques. – Mhenni Benghorbal Feb 23 '14 at 07:14
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since $a$ is constant just plug in one nice value for $x$ (for example $\pi$) and you're done. (given the fact that you already know all the other coefficients, otherwise you have to plug in several values and solve a linear system) – Max Feb 23 '14 at 12:19
3 Answers
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$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align} \cos^{6}\pars{x}&=\pars{\expo{\ic x} + \expo{-\ic x} \over 2}^{6}= {1 \over 2^{6}}\sum_{\ell = 0}^{6}{6 \choose \ell}\expo{\ic\pars{6 - 2\ell}x} \\[3mm]&= {1 \over 2^{6}}\sum_{\ell = 0}^{2}{6 \choose \ell}\expo{\ic\pars{6 - 2\ell}x} +{1 \over 2^{6}} \overbrace{6 \choose 3}^{\ds{20}} + {1 \over 2^{6}} \color{#f00}{\sum_{\ell = 4}^{6}{6 \choose \ell}\expo{\ic\pars{6 - 2\ell}x}} \tag{1} \end{align}
\begin{align} \color{#f00}{\sum_{\ell = 4}^{6}{6 \choose \ell}\expo{\ic\pars{6 - 2\ell}x}}&= \sum_{\ell = -2}^{0}{6 \choose \ell + 6}\expo{\ic\pars{-6 - 2\ell}x} = \sum_{\ell = 2}^{0}\overbrace{6 \choose -\ell + 6}^{\ds{6 \choose \ell}}\ \expo{\ic\pars{-6 + 2\ell}x} \\[3mm]&= \color{#f00}{\sum_{\ell = 0}^{2}{6 \choose \ell}\expo{\ic\pars{-6 + 2\ell}x}} \end{align}
We'll replace this result in $\pars{1}$: \begin{align} \cos^{6}\pars{x}&={5 \over 16} + {1 \over 32}\sum_{\ell = 0}^{2}{6 \choose \ell}\cos\pars{\bracks{6 - 2\ell}x} ={5 \over 16} + {\cos\pars{6x} + 6\cos\pars{4x} + 15\cos\pars{2x}\over 32} \end{align}
$$ \color{#00f}{\large\cos^{6}\pars{x} ={5 \over 16} + {15 \over 32}\,\cos\pars{2x} + {3 \over 16}\,\cos\pars{4x} + {1 \over 32}\,\cos\pars{6x}} $$
Felix Marin
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@user47252 Yes, it's $\tt LaTeX$. I use some macros to write the code in a comfortable way. – Felix Marin Feb 23 '14 at 07:09
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@user47252 You can read it in the web page with a browser. Just here . – Felix Marin Feb 23 '14 at 07:50
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Method $1:$
Use $\displaystyle\cos3A=4\cos^3A-3\cos A\iff \cos^3A=\cdots$ for $\displaystyle\cos^6x=(\cos^3x)^2$
and $\displaystyle\cos2B=2\cos^2B-1\iff \cos^2B=\cdots$
Method $2:$
Using Euler Formula, $\displaystyle\cos y=\frac{e^{iy}+e^{-iy}}2$
lab bhattacharjee
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If: $$\cos^6(x)=\sum_{k=1}^6 a_k \cos(kx)+a_0$$ Then we have: $$\cos^6(0)+\cos^6(\pi)=2=a_2+ a_4+a_6+2a_0$$ $$\cos^6(3\pi/4)+\cos^6(\pi/4)=1/4=-2a_4+2a_0$$ $$\cos^6(3\pi/2)+\cos^6(\pi/2)=0=-2a_2+2a_4-2a_6+2a_0$$ Add them all up to find $a_0$.
Nathaniel Bubis
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