# Trigonometry to the 24th power

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?

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

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

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

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 $$\frac{1}{2^{22}} \left(\frac{3428999}{2} + 1960980\frac{\sqrt{3}}{2}\right) =\frac{3428999 + 1960980 \sqrt{3}}{2^{23}}$$

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

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: $$a_n = a_{n-1}-\frac{2-\sqrt{3}}{16} a_{n-2}.\tag{3}$$

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.

• Reference please. – marty cohen Jun 8 '15 at 22:51
• @Leucippus Sir please could you evaluate the formula for $n=11?$ I've was unable to evaluate tt myself. – Ishan Jul 15 '15 at 13:40
• @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 $n?$ – Ishan Jul 15 '15 at 13:44
• @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. – Leucippus Jul 15 '15 at 14:32
• @Leucippus Thank you Sir. Sir could you please suggest a value other than 11 for which this series could computed? Many thanks Sir. – Ishan Jul 15 '15 at 16:08

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 $12$.