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Prove that $\sin\frac{\pi}{14}$ is a root of $8x^3 - 4x^2 - 4x + 1=0$.

I have no clue how to proceed and tried to prove that the whole equation becomes $0$ when $\sin\frac{\pi}{14}$ is placed in place of $x$ but couldn't do anything further. I think the symbols might be different but can be the same. If it is correct, please help me to solve this; if the equation is wrong, then please modify it and solve it.

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See Euler's and de Moivre's formula. – Lucian Apr 28 '14 at 18:21
How can these formula solve this problem? – MAFIA36790 Apr 28 '14 at 19:01
up vote 4 down vote accepted

First, you have, for any integer $n>0$ and any real $a, b$ with $b \not \equiv 0 \pmod{2\pi}$,

$$\sum_{k=0}^{n-1} \sin (a+kb) = \frac{\sin \frac{nb}{2}}{\sin \frac{b}{2}} \sin \left( a+ (n-1)\frac{b}{2}\right)$$

I'll prove this formula a bit later, but it's a rather classic one.

With $n=7, a=\frac{\pi}{14} \textrm{and}\; b=\frac{2\pi}{7}$, you get

$$\sin \frac{\pi}{14} + \sin \frac{5\pi}{14} + \sin \frac{9\pi}{14} + \sin \frac{13\pi}{14} + \sin \frac{17\pi}{14} + \sin \frac{21\pi}{14} + \sin \frac{25\pi}{14}=0$$

Now, using $\sin (\theta \pm \pi)= - \sin \theta$, notice that $$\sin \frac{9\pi}{14} = \sin \frac{5\pi}{14}$$ $$\sin \frac{13\pi}{14} = \sin \frac{\pi}{14}$$ $$\sin \frac{17\pi}{14} = - \sin \frac{3\pi}{14}$$ $$\sin \frac{21\pi}{14} = - 1$$ $$\sin \frac{25\pi}{14} = - \sin \frac{3\pi}{14}$$

So, the equality becomes

$$\sin \frac{\pi}{14} - \sin \frac{3\pi}{14} + \sin \frac{5\pi}{14} = \frac{1}{2}$$

Using $\sin p - \sin q = 2 \sin \frac{p-q}{2} \cos \frac{p+q}{2}$, you get

$$\sin \frac{\pi}{14} + 2 \sin \frac{\pi}{14} \cos \frac{4\pi}{14} = \frac{1}{2}$$


$$\sin \frac{\pi}{14} \left(1 + 2 \cos \frac{4\pi}{14}\right) = \frac{1}{2}$$


$$\cos \frac{4\pi}{14} = \sin \left(\frac{\pi}{2} - \frac{4\pi}{14} \right) = \sin \frac{3\pi}{14}$$

And we have also $\sin 3\theta = 3\sin \theta - 4 \sin^3 \theta$, so

$$\cos \frac{4\pi}{14} = 3\sin \frac{\pi}{14} - 4\sin^3 \frac{\pi}{14}$$

Let $x=\sin \frac{\pi}{14}$, we have then the equation

$$x (1+6x-8x^3) = \frac{1}{2}$$


$$16 x^4-12x^2-2x+1=0$$

But $-\frac{1}{2}$ is a trivial root of $16 x^4-12x^2-2x+1$, so this polynomial is divisible by $2x+1$. And since obviously $\sin \frac{\pi}{14} \neq -\frac{1}{2}$, we can do polynomial division, and the equation becomes


And we are done.

As a side comment, by changing slightly the proof, starting from $\sin \frac{\pi}{14} - \sin \frac{3\pi}{14} + \sin \frac{5\pi}{14} = \frac{1}{2}$, you would discover that the other two roots are $-\sin \frac{3\pi}{14}$ and $\sin \frac{5\pi}{14}$.

Now, the proof of the sum

$$\sum_{k=0}^{n-1} \sin (a+kb) = \frac{\sin \frac{nb}{2}}{\sin \frac{b}{2}} \sin \left( a+ (n-1)\frac{b}{2}\right)$$

There is an almost trivial proof using complex numbers, but here we are asked not to use them, so we will do this by induction.

First, the formula is true for $n=1$, for it amounts to $ \sin a = \sin a$. Let's suppose it's true for $n$, we will compute

$$A=\frac{\sin \frac{(n+1)b}{2}}{\sin \frac{b}{2}} \sin \left( a+ n\frac{b}{2}\right) - \frac{\sin \frac{nb}{2}}{\sin \frac{b}{2}} \sin \left( a+ (n-1)\frac{b}{2}\right)$$

Using $2\sin \theta \sin \phi = \cos (\theta - \phi) - \cos (\theta + \phi)$, we have

$$A=\frac{1}{2\sin \frac{b}{2}} \left[\cos \left(\frac{b}{2} - a \right) - \cos \left(a+ (2n+1)\frac{b}{2} \right) -\\ \cos \left(\frac{b}{2} - a \right) + \cos \left(a+ (2n-1)\frac{b}{2} \right) \right]$$

$$A=\frac{1}{2\sin \frac{b}{2}} \left[ \cos \left(a+ (2n-1)\frac{b}{2} \right) - \cos \left(a+ (2n+1)\frac{b}{2} \right) \right]$$

And, since $\cos q - \cos p = 2\sin \frac{p+q}{2} \sin \frac{p-q}{2}$,

$$A=\frac{1}{\sin \frac{b}{2}} \left[ \sin (a+nb) \sin \frac{b}{2} \right] = \sin (a+nb)$$


$$\sum_{k=0}^{n} \sin (a+kb) = \sum_{k=0}^{n-1} \sin (a+kb) + \sin(a+nb)$$ $$=\frac{\sin \frac{nb}{2}}{\sin \frac{b}{2}} \sin \left( a+ (n-1)\frac{b}{2}\right) + A$$ $$=\frac{\sin \frac{(n+1)b}{2}}{\sin \frac{b}{2}} \sin \left( a+ n\frac{b}{2}\right)$$

And the induction step is proved, so the formula is true for all $n>0$.

The "complex numbers" proof of the sum runs like this:

$$S = \sum_{k=0}^{n-1} e^{i(a+kb)} = e^{ia} \sum_{k=0}^{n-1} z^k = e^{ia} \frac{z^n - 1}{z - 1}$$

with $z=e^{ib}$ (and $z\neq1$ because $b \not \equiv 0 \pmod{2\pi}$)


$$S = e^{ia} \frac{e^{inb} - 1}{e^{ib} - 1}$$

There is a well-known trick to simplify, write:

$$S = e^{ia} \frac{e^{inb/2}}{e^{ib/2}} \frac{e^{inb/2} - e^{-inb/2}}{e^{ib/2} - e^{ib/2}} = e^{ia} \frac{e^{inb/2}}{e^{ib/2}} \frac{\sin \frac{nb}{2}}{\sin \frac{b}{2}}$$ $$S = e^{i(a + (n-1)b/2)} \frac{\sin \frac{nb}{2}}{\sin \frac{b}{2}}$$

Thus, taking real and imaginary parts, you get:

$$\sum_{k=0}^{n-1} \cos (a+kb) = \frac{\sin \frac{nb}{2}}{\sin \frac{b}{2}} \cos \left( a+ (n-1)\frac{b}{2}\right)$$

$$\sum_{k=0}^{n-1} \sin (a+kb) = \frac{\sin \frac{nb}{2}}{\sin \frac{b}{2}} \sin \left( a+ (n-1)\frac{b}{2}\right)$$

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Sir,where have u got this formula?Can u tell when to use this formula? – MAFIA36790 Apr 29 '14 at 9:39
The sum formula? I added the derivation from complex exponential in the answer, it's very straightforward. It's not easy to give an advice on when to use it: basically, when you have a sum of sines (or cosines) you can look whether it may help to use it. – Jean-Claude Arbaut Apr 29 '14 at 9:58
Here, from the initial equation $8x^3-4x^2-4x+1=0$, I got $\sin \pi/14 - \sin 3\pi/14 + \sin 5\pi/14=1/2$, then it took me some time to see I had to "expand" this to the full sum above, in order to prove this one. – Jean-Claude Arbaut Apr 29 '14 at 10:00
A very useful answer. – MAFIA36790 Apr 29 '14 at 12:31
Wow, terrific answer. – rubik Apr 29 '14 at 14:15

Use the sine's definition in terms of complex exponentials:

\begin{align} & 8x^3-4x^2-4x+1 \\[10pt] = {} & 8\left(\frac{e^{i\pi/14}-e^{-i\pi/14}}{2i}\right)^3 - 4\left(\frac{e^{i\pi/14}-e^{-i\pi/14}}{2i}\right)^2 -4 \frac{e^{i\pi/14}-e^{-i\pi/14}}{2i} +1 \end{align}

Then recall that $(a+b)^3=a^3+3a^2b+3ab^2+b^3$ and $(a+b)^2 = a^2+2ab+b^2$.

And $(2i)^3 = -8i$ and $(2i)^2=-4$, and $1/i=-i$.

Then turn the crank.

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Ok i understand u ve used the exponential definition,but cannot this problem be solved by using trigonometrical identities? – MAFIA36790 Apr 28 '14 at 19:22
It probably can be solved that way, but I'd have to play with it a bit to be sure of the specifics. – Michael Hardy Apr 28 '14 at 19:38


Let $\displaystyle\cos4x=-\cos3x=\cos(\pi+3x)\implies4x=2n\pi\pm(\pi+3x)$ where $n$ is any integer

'+' $\displaystyle\implies x=(2n+1)\pi\equiv\pi\pmod{2\pi}\implies\cos x=-1$

'-' $\displaystyle\implies x=\frac{(2n-1)\pi}7$

Now $\displaystyle\cos4x=-\cos3x$

Using Multiple angle formula,

$\iff8c^4-8c^2+1=-(4c^3-3c)\iff8c^4+4c^3-8c^2-3c+1=0\ \ \ \ (1)$ where $c=\cos x$

Clearly, the roots of $(1)$ are $\cos x,$ where $\displaystyle x=\pi, \frac{(2n-1)\pi}7 $ where $n\equiv0,1,2\pmod3$

So, the equation whose roots are $\displaystyle\cos\frac{(2n-1)\pi}7 $ where $n\equiv0,1,2\pmod3$ is $$\frac{8c^4+4c^3-8c^2-3c+1}{c+1}=0$$

Here $n=2$

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Let $\displaystyle z^7=-1=\cos\pi+i\sin\pi$

Using this, $\displaystyle z=\cos\frac{(2n+1)\pi}7+i\sin\frac{(2n+1)\pi}7$ where $n\equiv-3,-2,-1,0,1,2,3\pmod7$

So, the equation whose roots are $\displaystyle z=\cos\frac{(2n+1)\pi}7+i\sin\frac{(2n+1)\pi}7$ where $n\equiv-3,-2,-1,0,1,2\pmod7$ is $$\frac{z^7+1}{z+1}=0\iff z^6-z^5+z^4-z^3+z^2-z+1=0$$

Like this divide either sides by $z^3\ne0$ to get $$z^3+\frac1{z^3}-\left(z^2+\frac1{z^2}\right)+z+\frac1z-1=0$$


$$\iff \left(z+\frac1z\right)^3-\left(z+\frac1z\right)^2-2\left(z+\frac1z\right)+1=0 $$

Observe that $\displaystyle n\equiv-2,1\pmod7\implies z+\frac1z=2\cos\frac{3\pi}7$

Similarly for $\displaystyle n\equiv-1,0\pmod7; \equiv-3,2\pmod7$

So, $\displaystyle2\cos\frac{3\pi}7$ is a root of $$u^3-u^2-2u+1=0 $$

$\displaystyle\implies\cos\frac{3\pi}7$ is a root of $?$

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@user36790, How about this? – lab bhattacharjee Apr 29 '14 at 16:16

Like my other answers $$\sin\frac\pi{14}=\cos\left(\frac\pi2-\frac\pi{14}\right)=\cos\frac{3\pi}7$$

Using this, $$\cos\frac{\pi}7-\cos\frac{2\pi}7+\cos\frac{3\pi}7=0$$

Now $\displaystyle\cos\frac{\pi}7=\cos\left(\pi-\frac{6\pi}7\right)=-\cos\left(2\cdot\frac{3\pi}7\right)$ (use Double Angle formula $\displaystyle\cos2A=2\cos^2A-1$)

$\displaystyle\cos\frac{2\pi}7=-\cos\left(\pi+\frac{2\pi}7\right)=-\cos\left(3\cdot\frac{3\pi}7\right)$ (use $\displaystyle\cos3A=4\cos^3A-3\cos A$ formula)

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@user36790, How about this? – lab bhattacharjee Apr 29 '14 at 16:33

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