30
$\begingroup$

How can I find the value of $$\sin^{24}\frac{\pi}{24} + \cos^{24}\frac{\pi}{24}$$ Specifically, is there some easy method that I am overlooking?

$\endgroup$
  • 1
    $\begingroup$ These answer's are really gonna make me cry :'D Well explained :) $\endgroup$ – Shashank Jun 8 '15 at 19:49
  • 6
    $\begingroup$ This is the best title I have ever seen on any MSE question. $\endgroup$ – theage Jun 8 '15 at 20:03
18
$\begingroup$

First note that this is the simplest solution that I know. It may or may not be the simplest one. Edit: You can take Micah's path for a simpler solution.

We know that: $$\cos{\frac{\pi}{6}} = \frac {\sqrt{3}}{2}\quad \Rightarrow 2\cos^2{\frac{\pi}{12}} - 1 = \frac {\sqrt{3}}{2}$$ $$\Rightarrow 2(2\cos^2{\frac{\pi}{24}} -1)^2 - 1= \frac {\sqrt{3}}{2}$$ $$\Rightarrow 8\cos^4{\frac{\pi}{24}}-8\cos^2{\frac{\pi}{24}} + 1 = \frac {\sqrt{3}}{2} $$ $$\Rightarrow 16\cos^4{\frac{\pi}{24}}-16\cos^2{\frac{\pi}{24}} + 2 - \sqrt{3} = 0 $$ $$\Rightarrow \cos^2{\frac{\pi}{24}} = \frac {2+\sqrt{2+\sqrt{3}}}{4}\quad \Rightarrow \sin^2{\frac{\pi}{24}} = \frac {2-\sqrt{2+\sqrt{3}}}{4} $$ Now, $$\sin^{24}{\frac{\pi}{24}} + \cos^{24}{\frac{\pi}{24}} = (\sin^2{\frac{\pi}{24}})^{12} + (\cos^2{\frac{\pi}{24}})^{12}$$ $$= \left( \frac {2-\sqrt{2+\sqrt{3}}}{4}\right)^{12} + \left( \frac {2+\sqrt{2+\sqrt{3}}}{4}\right)^{12} $$ $$= \dfrac{1}{4^{12}} \left[ \left( 2-\sqrt{2+\sqrt{3}} \right)^{12} + \left( 2+\sqrt{2+\sqrt{3}} \right)^{12} \right] $$ $$= \dfrac{2}{4^{12}} \left[ 2^{12} + \begin{pmatrix} 12 \\ 2 \end{pmatrix} (2^{10}) (\sqrt{2+\sqrt{3}})^{2} + \begin{pmatrix} 12 \\ 4 \end{pmatrix} (2^{8}) (\sqrt{2+\sqrt{3}})^{4} \\ \quad +\begin{pmatrix} 12 \\ 6 \end{pmatrix} (2^{6}) (\sqrt{2+\sqrt{3}})^{6} + \begin{pmatrix} 12 \\ 8 \end{pmatrix} (2^{4}) (\sqrt{2+\sqrt{3}})^{8} \\ \quad + \begin{pmatrix} 12 \\ 10 \end{pmatrix} (2^{2}) (\sqrt{2+\sqrt{3}})^{10} + \begin{pmatrix} 12 \\ 12 \end{pmatrix} (\sqrt{2+\sqrt{3}})^{12} \right] $$ $$= \dfrac{2}{4^{12}} \left[ 2^{12} + \begin{pmatrix} 12 \\ 2 \end{pmatrix} (2^{10}) (2+\sqrt{3}) + \begin{pmatrix} 12 \\ 4 \end{pmatrix} (2^{8}) (2+\sqrt{3})^{2} \\ \quad +\begin{pmatrix} 12 \\ 6 \end{pmatrix} (2^{6}) (2+\sqrt{3})^{3} + \begin{pmatrix} 12 \\ 8 \end{pmatrix} (2^{4}) (2+\sqrt{3})^{4} \\ \quad + \begin{pmatrix} 12 \\ 10 \end{pmatrix} (2^{2}) (2+\sqrt{3})^{5} + \begin{pmatrix} 12 \\ 12 \end{pmatrix} (2+\sqrt{3})^{6} \right] $$ $$= \dfrac{1}{8388608} \left[ 4096 + (66)(1024)(2+\sqrt{3}) +(495)(256)(7+4\sqrt{3}) \\ \quad +(924)(64)(26+15\sqrt{3}) +(495)(16)(97+56\sqrt{3}) \\ \quad +(66)(4)(362+209\sqrt{3}) + (1351+780\sqrt{3}) \right] $$ $$= \dfrac {3428999+1960980\sqrt{3}}{8388608}$$

$\endgroup$
  • $\begingroup$ This is the path I was going to take. Thank you! $\endgroup$ – Sonya Jun 8 '15 at 20:32
  • $\begingroup$ No problem, but if you're looking for a shorter solution, Micah has the shorter/reasonable one. $\endgroup$ – Yagna Patel Jun 8 '15 at 20:33
39
$\begingroup$

I would use complex exponentials. We have: \begin{align} \sin^{24}\frac{\pi}{24}+\cos^{24}\frac{\pi}{24}&= \left(\frac{e^{i\frac{\pi}{24}}-e^{-i\frac{\pi}{24}}}{2i}\right)^{24} +\left(\frac{e^{i\frac{\pi}{24}}+e^{-i\frac{\pi}{24}}}{2}\right)^{24}\\ &=\frac{1}{2^{24}}\left(\sum_{k=0}^{24}\binom{24}{k} (-1)^k e^{i \frac{24-2k}{24}\pi}\right) +\frac{1}{2^{24}}\left(\sum_{k=0}^{24}\binom{24}{k} e^{i \frac{24-2k}{24}\pi}\right)\\ &= \frac{1}{2^{23}}\sum_{l=0}^{12} \binom{24}{2l} e^{i \frac{12 - 2l}{12}\pi} \end{align} after canceling every other term in the two sums. We can rewrite this last expression in terms of trig functions again, as $$ \frac{1}{2^{22}} \left(\sum_{m=0}^{5} \binom{24}{2m} \cos \left(\pi-\frac{m\pi}{6}\right) + \frac{1}{2} \binom{24}{12}\right) $$ (here we're folding the sum in half and taking advantage of the fact that $\binom{n}{k}=\binom{n}{n-k}$, which is why the middle term is anomalous).

Now, as $\cos x=-\cos(\pi - x)$, this collapses to a reasonable number of terms: $$ \frac{1}{2^{22}}\left[\frac{1}{2}\binom{24}{12} - 1 + \left(\binom{24}{10} - \binom{24}{2}\right) \cos \frac{\pi}{6} + \left(\binom{24}{8} - \binom{24}{4}\right) \cos \frac{\pi}{3} \right] $$ which simplifies into $$ \frac{1}{2^{22}} \left(\frac{3428999}{2} + 1960980\frac{\sqrt{3}}{2}\right) =\frac{3428999 + 1960980 \sqrt{3}}{2^{23}} $$

$\endgroup$
  • 1
    $\begingroup$ This is a neat solution. $\endgroup$ – Nit Jun 8 '15 at 20:28
  • 1
    $\begingroup$ Great Solution. +1 $\endgroup$ – Yagna Patel Jun 8 '15 at 20:42
31
$\begingroup$

Let: $$ a_n = \sin^{2n}\frac{\pi}{24}+\cos^{2n}\frac{\pi}{24}.\tag{1}$$ Then trivially $a_0=2,a_1=1$ and: $$ a_n - a_1 a_{n-1} = -\left(\sin^2\frac{\pi}{24}\cos^2\frac{\pi}{24}\right) a_{n-2}\tag{2} $$ so that $a_{12}$ can be computed in a few steps through the recurrence: $$ a_n = a_{n-1}-\frac{2-\sqrt{3}}{16} a_{n-2}.\tag{3}$$

$\endgroup$
8
$\begingroup$

A note: A general formula that is applicable is \begin{align} \sin^{4n}\left(\frac{\pi}{4n}\right) + \cos^{4n}\left( \frac{\pi}{4n}\right) = \frac{1}{4^{2n-1}} \left[ \sum_{r=0}^{n-1} \binom{4n}{2r} \cos\left(1 - \frac{r}{n} \right) \pi \, + \frac{1}{2} \binom{4n}{2n} \right]. \end{align} When $n=6$ this reduces to \begin{align} \sin^{24}\left(\frac{\pi}{24}\right) + \cos^{24}\left( \frac{\pi}{24}\right) = \frac{1}{4^{11}} \left[ \sum_{r=0}^{5} \binom{24}{2r} \cos\left(1 - \frac{r}{6} \right) \pi \, + \frac{1}{2} \binom{24}{12} \right]. \end{align} The remaining details have been given by Micah's solution.

$\endgroup$
  • 3
    $\begingroup$ Reference please. $\endgroup$ – marty cohen Jun 8 '15 at 22:51
  • $\begingroup$ @Leucippus Sir please could you evaluate the formula for $n=11?$ I've was unable to evaluate tt myself. $\endgroup$ – Ishan Jul 15 '15 at 13:40
  • $\begingroup$ @Leucippus Or Sir could you please show me how to put the formula into Wolfram Alpha so that I could use it in future for different values of $n?$ $\endgroup$ – Ishan Jul 15 '15 at 13:44
  • $\begingroup$ @BetterWorld As it would seem Wolfram Alpha does not like to compute this series. I tested it a few times. In general just type in the latex version of formulas in WA and it will compute what it can. $\endgroup$ – Leucippus Jul 15 '15 at 14:32
  • $\begingroup$ @Leucippus Thank you Sir. Sir could you please suggest a value other than 11 for which this series could computed? Many thanks Sir. $\endgroup$ – Ishan Jul 15 '15 at 16:08
6
$\begingroup$

Considering that the answer is apparently $$ \frac{7 \left(489857+280140 \sqrt{3}\right)}{8388608}, $$ I doubt there's an easy way.

You know $\sin{\frac{\pi}{6}}=\frac{1}{2}$, $\cos{\frac{\pi}{6}}=\frac{\sqrt{3}}{2}$, so using the half-angle formulae $$ 2\sin^2{\frac{x}{2}} = 1-\cos{x},\\ 2\cos^2{\frac{x}{2}} = 1+\cos{x} $$ will get us to $\sin{\frac{\pi}{24}}$ and $\cos{\frac{\pi}{24}}$, when applied twice.

In particular, $$ 4\sin^4{\frac{x}{4}} = \left(1-\cos{\frac{x}{2}}\right)^2 = 1+\cos^2{\frac{x}{2}}-2\cos{\frac{x}{2}} \\ 4\cos^4{\frac{x}{4}} = \left(1+\cos{\frac{x}{2}}\right)^2 = 1+\cos^2{\frac{x}{2}}+2\cos{\frac{x}{2}} $$ (At this point we see we have the nice identity $$4(\sin^4{\theta}+\cos^4{\theta}) = 2(1+\cos^2{2\theta}) = 3+\cos{4\theta},$$ although sadly this is not of much help to us here.)

Therefore we need to raise everything to the power $6$, then divide by $4^6=2^{12}$. Set $a=\cos{(x/2)}$, then we have $$ 4^6 ( \sin^{24}{\frac{x}{4}} + \cos^{24}{\frac{x}{4}} ) = (1-a)^{12}+(1+a)^{12} $$ At this point we break out the binomial theorem, and notice that quite a lot cancels: all the odd terms, in fact. Then $$ \sin^{24}{\frac{x}{4}} + \cos^{24}{\frac{x}{4}} = \frac{1}{2^5} \left( 1 + a^6 + \binom{12}{2}(a^2 + a^{10}) + \binom{12}{4}(a^4+a^8) + \binom{12}{6}a^6 \right) = \frac{1}{2^{11}}(1+a^{12}+66(a^2+a^{10})+495(a^4+a^8)+924a^6) $$ Right, now the good news is that we don't need to know what $a$ is, just $a^2$. Indeed, the formula above gives $$ \cos^2{\frac{\pi}{12}} = \frac{1}{2}\left(1+\cos{\frac{\pi}{6}}\right) = \frac{2+\sqrt{3}}{4}. $$ Therefore you just have to find $a^n$ for even $n$ up to $12$.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

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