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I just thought a way to find $\sin\frac{2π}{7}$.

Considering the equation $x^7=1$

$⇒(x-1)(x^6+x^5+x^4+x^3+x^2+x+1)=0$

$⇒(x-1)[(x+\frac1 x)^3+(x+\frac1 x)^2-2(x+\frac1 x)-1]=0$

We can then get the 7 solutions of x, but the steps will be very complicated, especially when solving cubic equation, and expressing x as a+bi. The imaginary part of the second root of x will be $\sin\frac{2π}{7}$.

Besides this troublesome way, are any other approach? Thank you.

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It is an approach that works very nicely with $5$. With $7$, not so good, though it is a good way to prove that the regular $7$-gon is not Euclidean constructible. I think that if you use the Cardano formula, you end up needing a cube root of a complex number, and that cube root cannot be found without knowing the required sine or a close relative. –  André Nicolas Feb 23 '13 at 6:21
    
needs to be -1 at the end in the square brackets. Edited. –  Will Jagy Feb 23 '13 at 6:26
    
The other bad news is that your cubic $u^3 + u^2 - 2 u - 1$ has three irrational real roots, which means there is no pretty way to separate the real and imaginary parts in Cardano's formula, en.wikipedia.org/wiki/Casus_irreducibilis –  Will Jagy Feb 23 '13 at 6:32
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@ Will Jagy, You're right... –  ᴊ ᴀ s ᴏ ɴ Feb 23 '13 at 6:39
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+1: Might I recommend that you study algebraic number theory some time in the future. This type of calculations show up inside cyclotomic fields. –  Jyrki Lahtonen Feb 23 '13 at 6:55
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2 Answers

up vote 3 down vote accepted

Just for laughs, we can at least in principle compute $\cos{(\pi/7)}$ by observing that

$$\sin{\frac{3 \pi}{7}} = \sin{\frac{4 \pi}{7}}$$

Using a combination of double-angle forumlae, we end up with a cubic equation for $\cos{(\pi/7)}$:

$$8 \cos^3{\frac{\pi}{7}} - 4 \cos^2{\frac{\pi}{7}} - 4 \cos{\frac{\pi}{7}}+1=0$$

This equation has one real solution which is $\cos{(\pi/7)}$. The bad news is that the expression is unwieldy at best:

$$\cos{\frac{\pi}{7}}=\frac{1}{6} \left(1+\frac{7^{2/3}}{\sqrt[3]{\frac{1}{2} \left(-1+3 i\sqrt{3}\right)}}+\sqrt[3]{\frac{7}{2} \left(-1+3 i \sqrt{3}\right)}\right)$$

The imaginary part of this expression is of course zero. The real part, however, ends up being expressed in terms of a sine and cosine of another angle, and I think the point of an exercise like this is to not do that. Anyway, I hope this adds to the discussion above.

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I considered figuring out how to derive this equation with double/triple angle sine/cosine formulas. +1 for saving me the trouble. –  Jyrki Lahtonen Feb 23 '13 at 7:02
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Using this or Point#$24$ of this ,

$\sin 7x=7s-56s^3+112s^5-64s^7$ where $s=\sin x$

If $\sin 7x=0,7x=n\pi$ where $n$ is any integer.

So, $x=\frac{n\pi}7$ where $n=0,1,2,3,4,5,6$

So, the roots of $7s-56s^3+112s^5-64s^7=0--->(1)$ are $\sin\frac{n\pi}7$ where $n=0,1,2,\cdots 5,6$

So, the roots of $64s^6-112s^4+56s^2-7=0--->(2)$ are $\sin\frac{n\pi}7$ where $n=1,2,\cdots 5,6$

So, the roots of $64t^3-112t^2+56t-7=0 --->(3)$ are $\sin^2\frac{n\pi}7$ where $n=1,2,4$ or $3,5,6$ as $\sin \frac{(7-r)\pi}7=\sin(\pi-\frac{r\pi}7)=\sin\frac{r\pi}7$

If we choose $n=1,2,4$ observe that $\sin^2\frac{4\pi}7-\sin^2\frac{2\pi}7=2\sin\frac{\pi}7\cos\frac{3\pi}7>0$ (Using $\sin^2A-\sin^2B=\sin(A+B)\sin(A-B)$)

Similarly, $\sin^2\frac{2\pi}7-\sin^2\frac{\pi}7>0$

So, $\sin^2\frac{4\pi}7>\sin^2\frac{2\pi}7>\sin^2\frac{\pi}7$

Using Cardano's method, we can solve the Cubic equation $(3)$

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