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I know that $\cos(6\phi)\equiv32c^6-48c^4+18c^2-1$ where $c=\cos\phi$.

I also know that when $\cos(6\phi)=0$, then $\phi=\frac{k\pi}{12}$ ($k = 1,3,5,7,9,11$).

How do I find the exact value of: $$\cos\left(\frac{\pi}{12}\right) \cos\left(\frac{5\pi}{12}\right) \cos\left(\frac{7\pi}{12}\right) \cos\left(\frac{11\pi}{12}\right)$$

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4 Answers 4

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Mosquito-nuking linear algebraic solution: it can be shown that the tridiagonal matrix

$$\mathbf T=\frac12\begin{pmatrix}-1&1&&&&\\1&0&1&&&\\&1&0&1&&\\&&1&0&1&\\&&&1&0&1\\&&&&1&1\end{pmatrix}$$

has the eigenvalues $\cos\dfrac{(2k-1)\pi}{12},\quad k=1,\dots,6$. Since the product of the eigenvalues is equal to the determinant of the matrix,

$$\cos\frac{\pi}{12}\cos\frac{5\pi}{12}\cos\frac{7\pi}{12}\cos\frac{11\pi}{12}=\frac{\det\mathbf T}{\cos\tfrac{\pi}{4}\cos\tfrac{3\pi}{4}}=\frac{-\tfrac1{32}}{-\tfrac12}=\frac1{16}$$

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  • $\begingroup$ I s there a document, I can learn this method? Thanks. $\endgroup$
    – Mikasa
    Jun 2, 2013 at 16:28
  • $\begingroup$ Yes, look for the hyperlink in my post. $\endgroup$
    – J. M. ain't a mathematician
    Jun 2, 2013 at 16:29
  • $\begingroup$ +1 I love (when people admit that they are) mosquito nuking. $\endgroup$
    – Calvin Lin
    Jun 2, 2013 at 22:16
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From the formulas: $\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)\\\cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b)$

we have: $\cos(a)\cos(b)=\frac{1}{2}\big(\cos(a+b)+\cos(a-b)\big)$

Apply this formula:

$\cos\left(\frac{\pi}{12}\right)\cos\left(\frac{11\pi}{12}\right) \cos\left(\frac{5\pi}{12}\right) \cos\left(\frac{7\pi}{12}\right) \\=\frac{1}{2}\big(\cos(\pi)+\cos(\frac{10\pi}{12})\big)\frac{1}{2}\big(\cos(\pi)+\cos(\frac{2\pi}{12})\big)=\frac{1}{4}\big(-1-\cos(\frac{2\pi}{12})\big)\big(-1+\cos(\frac{2\pi}{12})\big)=\frac{1}{4}\big(1-\cos(\frac{2\pi}{12})^2\big)$

You certainly know: $\cos(\frac{2\pi}{12})=\cos(\frac{\pi}{6})=\frac{\sqrt{3}}{2}$

Result is: $\frac{1}{16}$

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  • $\begingroup$ use \dfrac instead of \frac .it will give better view $\endgroup$
    – iostream007
    Jun 2, 2013 at 16:26
  • $\begingroup$ I don't quite understand why you have found $\cos(a)\cos(b)$ ? $\endgroup$
    – maxmitch
    Jun 2, 2013 at 16:39
  • $\begingroup$ @max, add the two equations together... $\endgroup$
    – J. M. ain't a mathematician
    Jun 2, 2013 at 16:45
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As $\frac{11\pi}{12}=\pi-\frac\pi{12}$ and $\frac{7\pi}{12}=\pi-\frac{5\pi}{12}$

and $\cos(\pi-x)=-\cos x,$

$$\cos\left(\frac{\pi}{12}\right) \cos\left(\frac{5\pi}{12}\right) \cos\left(\frac{7\pi}{12}\right) \cos\left(\frac{11\pi}{12}\right)$$

$$=\cos\left(\frac{\pi}{12}\right) \cos\left(\frac{5\pi}{12}\right)\left( - \cos\left(\frac{5\pi}{12}\right)\right)\left( -\cos\left(\frac{\pi}{12}\right)\right)$$

$$=\left(\cos\left(\frac{\pi}{12}\right) \cos\left(\frac{5\pi}{12}\right)\right)^2$$

Now as $\frac{5\pi}{12}=\frac\pi2-\frac\pi{12}$ and $\sin2x=2\sin x\cos x,$

$$\cos\left(\frac{\pi}{12}\right) \cos\left(\frac{5\pi}{12}\right)=\cos\left(\frac{\pi}{12}\right) \sin\left(\frac{\pi}{12}\right)=\frac{\sin \frac\pi6}2=\frac{\frac12}2=\frac14$$

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HINT:

SO, the equation whose roots are $\cos\frac{r\pi}{12}$ where $r=3,9$ is

$$\left(c-\cos \frac{3\pi}{12}\right)\left(c-\cos \frac{9\pi}{12}\right)=0$$

$$\implies \left(c-\frac1{\sqrt2}\right)\left(c+\frac1{\sqrt2}\right)=0\text{ as }\cos \frac{9\pi}{12}=\cos\left(\pi- \frac{3\pi}{12}\right)=-\cos\frac{3\pi}{12}$$

$$\implies 2c^2-1=0$$

SO, the equation whose roots are $\cos\frac{r\pi}{12}$ where $r=1,5,7,11$ will be

$$\frac{32c^6-48c^4+18c^2-1}{2c^2-1}=0$$

Now, apply Vieta's Formulas 1,2,3

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