Product of cosines: $\prod_{r=1}^{7} \cos \left(\frac{r\pi}{15}\right)$

Evaluate

$$\prod_{r=1}^{7} \cos \left({\dfrac{r\pi}{15}}\right)$$

I tried trigonometric identities of product of cosines, i.e, $$\cos\text{A}\cdot\cos\text{B} = \dfrac{1}{2}[ \cos(A+B)+\cos(A-B)]$$

but I couldn't find the product.

Any help will be appreciated.
Thanks.

• Hint: Consider the square of the product. And consider the leading coefficient of the Chebyshev Polynomial $\cos(15x)$ when written in terms of $\cos(x)$. – GohP.iHan Jul 6 '15 at 14:59
• @GohP.iHan Can you please elaborate your method in a solution, it looks interesting. Thanks. – Henry Jul 6 '15 at 15:20
• If you square it, then it is of the form of $-\cos(\pi/15) \cos(2\pi/15)\ldots\cos(14\pi/15)$. Now consider $\cos(15x) = 0$ which has all the 14 roots. Writing it as the Chebyshev polynomial, $0=\cos(15x) = 2^{14}\cos^{15}(x) -\ldots -1$. By Vieta's formula, the product is simply $1/2^{14}$, take its square root, and voilà! – GohP.iHan Jul 6 '15 at 15:57
• @GohP.iHan Thank you so much! It is enlightening to know various approaches to the same problem :) – Henry Jul 6 '15 at 18:24

Let $\displaystyle\text{C}=\prod_{r=1}^{7}\cos{\left(\dfrac{r\pi}{15}\right)}$

and

$\displaystyle\text{S}=\prod_{r=1}^{7}\sin{\left(\dfrac{r\pi}{15}\right)}$

Now,

$\text{C}\cdot\text{S}=\left(\sin{\dfrac{\pi}{15}} \cdot \cos{\dfrac{\pi}{15}}\right) \cdot \left(\sin{\dfrac{2\pi}{15}}\cdot\cos{\dfrac{2\pi}{15}}\right)\cdot \ldots \cdot\left(\sin{\dfrac{7\pi}{15}} \cdot \cos{\dfrac{7\pi}{15}}\right)$

$\implies \text{C}\cdot\text{S}= \dfrac{1}{2^7} \left(2\sin{\dfrac{\pi}{15}} \cdot \cos{\dfrac{\pi}{15}}\right) \cdot \left(2\sin{\dfrac{2\pi}{15}}\cdot\cos{\dfrac{2\pi}{15}}\right)\cdot \ldots \cdot\left(2\sin{\dfrac{7\pi}{15}} \cdot \cos{\dfrac{7\pi}{15}}\right)$

$\implies \text{C}\cdot\text{S}= \dfrac{1}{2^7} \ \sin{\dfrac{2\pi}{15}}\cdot\sin{\dfrac{4\pi}{15}}\cdot \ldots \cdot\sin{\dfrac{14\pi}{15}}$

$\{\because \sin(2x) = 2\sin (x) \cos (x) \}$

$\implies \text{C}\cdot\text{S}= \dfrac{1}{2^7} \ \sin{\dfrac{\pi}{15}}\cdot\sin{\dfrac{2\pi}{15}} \cdot \ldots \cdot \sin{\dfrac{7\pi}{15}} \\\\ \{\because \sin(\pi-x)=\sin(x)\}$

$\implies \text{C}\cdot\text{S}= \dfrac{1}{2^7} \cdot \text{S}$

since $\text{S} \neq 0$,

$\therefore \boxed{\text{C}=\dfrac{1}{2^7}}$

• +1 Looks like this can be generalized to $$\prod_{r=1}^n \cos\frac{r\pi}{2n+1} = \frac{1}{2^n}$$. – Darth Geek Jul 6 '15 at 14:29
• (+1) very nice approach, it reminds the Gauss lemma for quadratic reciprocity - in particular, the proof that $$\left(\frac{2}{p}\right)=(-1)^{\frac{p^2-1}{8}}.$$ – Jack D'Aurizio Jul 6 '15 at 14:57
• @Ishu Using math.stackexchange.com/questions/159214/… then: $$\frac{2n}{2^{2n}} = \prod_{r=1}^{2n} \sin \frac{r\pi}{2n+1} = \left(\prod_{r=1}^{n} \sin \frac{r\pi}{2n+1}\right)\left(\prod_{r=n+1}^{2n} \sin \frac{r\pi}{2n+1}\right) = \left(\prod_{r=1}^{n} \sin \frac{r\pi}{2n+1}\right)\left(\prod_{r=1}^{n} \sin \frac{(n+r)\pi}{2n+1}\right)$$ And since $\sin \dfrac{(n+r)\pi}{2n+1} = \sin \dfrac{(n+1-r)\pi}{2n+1}$ then $$\dfrac{2n}{2^{2n}} = \left(\prod_{r=1}^{n} \sin \frac{r\pi}{2n+1}\right)^2$$ Therefore the product of sines is $\dfrac{\sqrt{2n}}{2^n}$. – Darth Geek Jul 6 '15 at 15:22
• EDIT: It should be $\dfrac{2n+1}{2^{2n}} = \ldots$ and the product of sines is therefore $\dfrac{\sqrt{2n+1}}{2^n}$. – Darth Geek Jul 6 '15 at 19:18

I think it's worth noting the product is also evaluable just remembering, besides the well known $\displaystyle \cos\frac{\pi}{3}=\frac{1}{2},$ the somewhat nice $$\displaystyle\cos\frac{\pi}{5}=\frac{1+\sqrt{5}}{4}=\frac{\phi}{2}$$ (where $\phi$ is the golden section) and iterating the sum/difference formula for the cosine and the product formula you mention. Your product is, once we rearrange factors and simplify fractions, equal to $${\color\red{\cos\frac{\pi}{3}\cdot\cos\frac{\pi}{5}}}\cdot\color\orange{\cos\frac{2\pi}{5}} \cdot\color\navy{\cos\frac{\pi}{15}\cdot\cos\frac{4\pi}{15}}\cdot\color\green{\cos\frac{2\pi}{15}\cdot\cos\frac{7\pi}{15}} \\ ={\color\red{\frac{\phi}{4}}}\color\orange{\left(2\cos^2\frac{\pi}{5}-1\right)}\color\navy{\frac{1}{2}\left(\cos\frac{\pi}{3}+\cos\left(-\frac{\pi}{5}\right)\right)}\color\green{\frac{1}{2}\left(\cos\left(-\frac{\pi}{3}\right)+\cos\frac{3\pi}{5}\right)} \\ = \frac{\phi}{16}\left(\frac{\phi^2}{2}-1\right)\frac{\phi+1}{2}\left(\frac{1}{2}+\cos\frac{\pi}{5}\cdot\left(2\cos^2\frac{\pi}{5}-1\right)-2\cos\frac{\pi}{5}\left(1-\cos^2\frac{\pi}{5}\right)\right)\\=\frac{\phi(\phi^2-1)}{64}\left(\frac{1}{2}+\frac{\phi(\phi^2-2)}{4}-\frac{\phi(4-\phi^2)}{4}\right)\\=\frac{\phi^2}{64}\left(\frac{1}{2}+\frac{\phi(\phi-1)-\phi(3-\phi)}{4}\right)\\=\frac{\phi+1}{64}\left(\frac{1}{2}+\frac{2-2\phi}{4}\right)=\frac{\phi+1}{128}-\frac{\phi^2-1}{128}=\frac{\phi+1}{128}-\frac{\phi}{128}=\frac{1}{128}.$$

Since an elegant solution has already been provided, I will go for an overkill.

From the Fourier cosine series of $\log\cos x$ we have: $$\log\cos x = -\log 2-\sum_{n\geq 1}^{+\infty}\frac{(-1)^n\cos(2n x)}{n}\tag{1}$$ but for any $n\geq 1$ we have: $$15\nmid n\rightarrow\sum_{k=1}^{7}\cos\left(\frac{2n k \pi}{15}\right) = -\frac{1}{2},\quad 15\mid n\rightarrow\sum_{k=1}^{7}\cos\left(\frac{2n k \pi}{15}\right) = 7\tag{2}$$ so: $$\sum_{k=1}^{7}\log\cos\frac{k\pi}{15} = -7\log 2+\frac{1}{2}\sum_{n\geq 1}\frac{(-1)^n}{n}-\frac{15}{2}\sum_{n\geq 1}\frac{(-1)^n}{15n}=-7\log 2\tag{3}$$ and by exponentiating the previous line:

$$\prod_{k=1}^{7}\cos\left(\frac{\pi k}{15}\right) = \color{red}{\frac{1}{2^7}}.\tag{4}$$

• Can you tell me how did you arrive at $(2)$ ? – Henry Jul 6 '15 at 15:23
• @Samurai: write $\cos z$ as $\text{Re}(e^{iz})$. Then the sum becomes a trivial sum if $15\mid n$, and half a sum of $15$-th roots of unity if $15\nmid n$. – Jack D'Aurizio Jul 6 '15 at 15:25

Note: Here's another variation inspired by an answer to this question.

We consider the roots of unity $e^{\frac{2\pi i k}{15}}, 0\leq k < 15$ of the polynomial

$$p(z)=z^{15}-1=\prod_{k=0}^{14}(z-e^{\frac{2\pi i k}{15}})$$

We obtain

\begin{align*} -p(-z)=z^{15}+1&=\prod_{k=0}^{14}(z+e^{\frac{2\pi i k}{15}})\\ &=(z+1)\prod_{k=1}^{7}\left[(z+e^{\frac{2\pi i k}{15}})(z+e^{-\frac{2\pi i k}{15}})\right]\\ \end{align*}

Evaluating the polynomial $-p(-z)$ at $z=1$ gives

\begin{align*} 1&=\prod_{k=1}^{7}\left[(1+e^{\frac{2\pi i k}{15}})(1+e^{-\frac{2\pi i k}{15}})\right]\\ &=\prod_{k=1}^{7}\left[(e^{-\frac{\pi i k}{15}}+e^{\frac{\pi i k}{15}})e^{\frac{\pi i k}{15}}(e^{\frac{\pi i k}{15}}+e^{-\frac{\pi i k}{15}})e^{-\frac{\pi i k}{15}}\right]\\ &=\prod_{k=1}^{7}(e^{\frac{\pi i k}{15}}+e^{-\frac{\pi i k}{15}})^2\tag{1}\\ &=\prod_{k=1}^{7}\left(2\cos\left(\frac{k \pi}{15}\right)\right)^2\tag{2}\\ \end{align*}

In (1) we use the formula $\cos(z)=\frac{1}{2}\left(e^{iz}+e^{-iz}\right)$.

We conclude from (2) \begin{align*} \prod_{k=1}^{7}\cos\left(\frac{k\pi}{15}\right)=\frac{1}{2^7} \end{align*}

Note: Writing $-p(-z)$ as

\begin{align*} -p(-z)=\prod_{k=1}^{7}\left[z^2+\left(e^{\frac{2\pi i k}{15}}+e^{-\frac{2\pi i k}{15}}\right)z+1\right]\\ \end{align*}

and evaluating the polynomial $-p(-z)$ at $z=i$ we obtain the related formula

\begin{align*} \prod_{k=1}^{7}\cos\left(\frac{2k\pi}{15}\right)=\frac{1}{2^7} \end{align*}

Doubling the argument does not change the value of the product.

I like the answer by @Steven Gregory because of the way the dominos fall and it seems the only one presented that a precalculus student could hope to find. Using the reflection and multiplication formulas for the gamma function a fairly compact proof is possible. \begin{align}\prod_{r=1}^7\cos\frac{r\pi}{15} & = \prod_{r=1}^7\sin\left(\frac{\pi}2-\frac{r\pi}{15}\right) = \prod_{r=0}^6\sin\left(\frac{\pi}2-\frac{7\pi}{15}+\frac{r\pi}{15}\right) \\ & = \prod_{r=0}^6\sin\left(\frac{\pi}{30}+\frac{r\pi}{15}\right) = \prod_{r=0}^6\frac{\pi}{\Gamma\left(\frac1{30}+\frac{r}{15}\right)\Gamma\left(\frac{29}{30}-\frac{r}{15}\right)} & \tag{1} \\ & = \frac{\pi^7}{\prod_{r=0}^6\Gamma\left(\frac1{30}+\frac{r}{15}\right)\prod_{s=8}^{14}\Gamma\left(\frac1{30}+\frac{s}{15}\right)} = \frac{\pi^7\,\Gamma\left(\frac1{30}+\frac7{15}\right)}{\prod_{r=0}^{14}\Gamma\left(\frac1{30}+\frac{r}{15}\right)} & \tag{2} \\ & = \frac{\pi^7\,\Gamma\left(\frac12\right)}{(2\pi)^{(15-1)/2}15^{\frac12-15\left(\frac1{30}\right)}\Gamma\left(15\left(\frac1{30}\right)\right)} & \tag{3} \\ & = \frac1{2^7} \end{align} $(1)$ Using the reflection formula for the gamma function: $\Gamma(x)\Gamma(1-x)=\frac{\pi}{\sin(\pi x)}$.
$(2)$ Reordering one product and multiplying and dividing by $\Gamma\left(\frac1{30}+\frac7{15}\right)$.
$(3)$ Using the multiplication formula for the gamma function: $$\prod_{k=0}^{n-1}\Gamma\left(x+\frac{k}n\right)=(2\pi)^{(n-1)/2}n^{\frac12-nx}\Gamma(nx)$$

• (+1) Nice. But can you explain how do you find the gamma products? IMO the accepted answer is the most elegant for anyone, precalc or calculus student. – Henry Mar 21 '16 at 4:04
• Hope my edits clarified the situation a bit. This is kind of a juicy product in that at first glance it just looks outrageous, but there are actually many disparate methods of attack, all of which seem to work. – user5713492 Mar 21 '16 at 6:29

Using de Moivre's formula,

$$\cos(2n+1)x+i\sin(2n+1)x=(\cos x+i\sin x)^{2n+1}=\sum_{r=0}\binom{2n+1}r(\cos x)^{2n+1-r}(i\sin x)^r$$

Equating the real parts, $$\cos(2n+1)x$$

$$=(\cos x)^{2n+1}-\binom{2n+1}2(\cos x)^{2n+1-2}(\sin x)^2+\binom{2n+1}r(\cos x)^{2n+1-4}(\sin x)^4+\cdots+\binom{2n+1}{2n-2}(-1)^{n-1}(\cos x)^3(\sin x)^{2n-2}+(-1)^n\cos x\sin^{2n}x$$

$$=(\cos x)^{2n+1}-\binom{2n+1}2(\cos x)^{2n+1-2}(1-\cos^2x)+\binom{2n+1}r(\cos x)^{2n+1-4}(1-\cos^2x)^2+\cdots+\binom{2n+1}{2n-2}(-1)^{n-1}(\cos x)^3(1-\cos^2x)^{n-1}+(-1)^n\cos x(1-\cos^2x)^nx$$

$$\implies\cos(2n+1)x=(\cos x)^{2n+1}\left[1+\binom{2n+1}2+\binom{2n+1}4+\cdots+\binom{2n+1}{2n}\right]+\cdots+\binom{2n+1}{2n}(-1)^n\cos x$$

Now $\displaystyle1+\binom{2n+1}2+\binom{2n+1}4+\cdots+\binom{2n+1}{2n}=\sum_{r=0}^n\binom{2n+1}{2r}=\sum_{r=0}^n\binom{2n+1}{2n+1-2r}=\dfrac{(1+1)^{2n+1}}2=2^{2n}$

$$\implies\cos(2n+1)x=2^{2n}(\cos x)^{2n+1}+\cdots+(-1)^n(2n+1)\cos x$$

Now if $\cos(2n+1)x=\cos A,(2n+1)x=2m\pi\pm A$ where $m$ is any integer

and $x=\dfrac{2m\pi\pm A}{2n+1}$ where $m\equiv0,\pm1,\pm2\cdots,\pm n\pmod{2n+1}$

So, the roots of $$2^{2n}(\cos x)^{2n+1}+\cdots+(-1)^n(2n+1)\cos x-\cos A=0\ \ \ \ (1)$$ are $\cos\dfrac{2m\pi\pm A}{2n+1}$ where $m\equiv0,\pm1,\pm2\cdots,\pm n\pmod{2n+1}$

Using Vieta's formula, $$\cos\dfrac A{2n+1}\prod_{m=1}^n\cos\dfrac{2m\pi\pm A}{2n+1}=\dfrac{\cos A}{2^{2n}}$$

If $A=0,$ $$\prod_{m=1}^n\cos^2\dfrac{2m\pi}{2n+1}=\dfrac1{2^{2n}}$$

Now $\cos\dfrac{2m\pi}{2n+1}=-\cos\left(\pi-\dfrac{2m\pi}{2n+1}\right)=-\cos\dfrac{(2n+1-2m)\pi}{2n+1}$

$$\implies(-1)^n\prod_{m=0}^{2n}\cos\dfrac{m\pi}{2n+1}=\dfrac1{2^{2n}}$$

As $\displaystyle\cos\dfrac{(2n+1-m)\pi}{2n+1}=\cdots=-\cos\dfrac{m\pi}{2m+1}$

$\displaystyle\implies(-1)^n\prod_{m=1}^n\cos^2\dfrac{m\pi}{2n+1}(-1)^n=\dfrac1{2^{2n}}\implies\prod_{m=1}^n\cos^2\dfrac{m\pi}{2n+1}=\dfrac1{2^{2n}}$

Again $0\le\dfrac{m\pi}{2n+1}\le\dfrac\pi2\iff0\le2m\le2n+1\iff0\le m\le n$

$\displaystyle\implies\prod_{m=1}^n\cos\dfrac{m\pi}{2n+1}=\dfrac1{2^n}$ as $\cos\dfrac{m\pi}{2n+1}>0$ as $0<m\le n$

This is definitely not the way to do it. But it sure was a lot of fun to do.

$$\cos {\dfrac{7\pi}{15}} = \cos{\left(\pi - \dfrac{7\pi}{15}\right)} = -\cos{\dfrac{8\pi}{15}}$$

$$\sin{\dfrac{16\pi}{15}} = \sin{\left(\pi + \dfrac{\pi}{15}\right)} = -\sin{\dfrac{\pi}{15}}$$

$$\sin{\dfrac{12\pi}{15}} = \sin{\left(\pi - \dfrac{3\pi}{15}\right)} = \sin{\dfrac{3\pi}{15}}$$

\begin{align} y & = \prod_{r=1}^{7} \cos {\frac{r\pi}{15}}\\ y & = \cos{\frac{\pi}{15}} \cos{\frac{2\pi}{15}} \cos{\frac{3\pi}{15}} \cos{\frac{4\pi}{15}} \cos{\frac{5\pi}{15}}\cos{\frac{6\pi}{15}} \cos{\frac{7\pi}{15}}\\ 2^7y\,\sin{\frac{\pi}{15}} & = 2^6 \left(2\sin{\frac{\pi}{15}} \cos{\frac{\pi}{15}}\right) \cos{\frac{2\pi}{15}} \cos{\frac{3\pi}{15}} \cos{\frac{4\pi}{15}} \cos{\frac{5\pi}{15}}\cos{\frac{6\pi}{15}} \cos{\frac{7\pi}{15}}\\ 2^7y\,\sin{\frac{\pi}{15}} & = 2^5 \left(2\sin{\frac{2\pi}{15}} \cos{\frac{2\pi}{15}}\right) \cos{\frac{3\pi}{15}} \cos{\frac{4\pi}{15}} \cos{\frac{5\pi}{15}}\cos{\frac{6\pi}{15}} \cos{\frac{7\pi}{15}}\\ 2^7y\,\sin{\frac{\pi}{15}} & = 2^4 \left(2\sin{\frac{4\pi}{15}}\right) \cos{\frac{3\pi}{15}} \left( \cos{\frac{4\pi}{15}}\right) \cos{\frac{5\pi}{15}}\cos{\frac{6\pi}{15}} \cos{\frac{7\pi}{15}}\\ 2^7y\,\sin{\frac{\pi}{15}} & = 2^3 \left(2\sin{\frac{8\pi}{15}} \right) \cos{\frac{3\pi}{15}} \cos{\frac{5\pi}{15}}\cos{\frac{6\pi}{15}} \left(-\cos{\frac{8\pi}{15}}\right)\\ 2^7y\,\sin{\frac{\pi}{15}} & = 2^3 \left(-\sin{\frac{16\pi}{15}}\right) \cos{\frac{3\pi}{15}} \cos{\frac{5\pi}{15}}\cos{\frac{6\pi}{15}}\\ 2^7y & = 2^3\cos{\frac{3\pi}{15}} \cos{\frac{5\pi}{15}}\cos{\frac{6\pi}{15}}\\ 2^7\sin{\frac{3\pi}{15}}y & = 2^2\left(2 \sin{\frac{3\pi}{15}}\cos{\frac{3\pi}{15}}\right) \cos{\frac{5\pi}{15}}\cos{\frac{6\pi}{15}}\\ 2^7\sin{\frac{3\pi}{15}}y & = 2\left(2 \sin{\frac{6\pi}{15}}\right) \cos{\frac{5\pi}{15}} \left( \cos{\frac{6\pi}{15}} \right)\\ 2^7\sin{\frac{3\pi}{15}}y & = 2\sin{\frac{12\pi}{15}} \cos{\frac{5\pi}{15}}\\ 2^7\sin{\frac{3\pi}{15}}y & = 2\sin{\frac{3\pi}{15}} \cos{\frac{\pi}{3}}\\ 2^7y & = 1\\ y & = \dfrac{1}{2^7}\\ \end{align}