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What is the value of $\cos\left(\frac{2\pi}{7}\right)$ ? I don't know how to calculate it.

idm
  • 12,049

3 Answers3

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I suggest you have a look at http://mathworld.wolfram.com/TrigonometryAnglesPi7.html which clearly explains the problem.

As you will see, $\cos\left(\frac{2\pi}{7}\right)$ is the solution of $$8 x^3+4 x^2-4 x-1=0$$ Using Cardano, you will get $$\cos\left(\frac{2\pi}{7}\right)=\frac{1}{6} \left(-1+\frac{7^{2/3}}{\sqrt[3]{\frac{1}{2} \left(1+3 i \sqrt{3}\right)}}+\sqrt[3]{\frac{7}{2} \left(1+3 i \sqrt{3}\right)}\right)$$

If you want to approximate it, even very accurately, you could expand $\cos(x)$ as a Taylor series at $x=\frac{\pi}{3}$ which gives $$\cos(x)=\frac{1}{2}-\frac{1}{2} \sqrt{3} \left(x-\frac{\pi }{3}\right)-\frac{1}{4} \left(x-\frac{\pi }{3}\right)^2+\frac{\left(x-\frac{\pi }{3}\right)^3}{4 \sqrt{3}}+\frac{1}{48} \left(x-\frac{\pi }{3}\right)^4+\cdots$$

Using the first term, you will get $0.6295570974$, and adding terms $0.6239620836$, $0.6234788344$, $0.6234892692$, $0.6234898099$, $0.6234898021$ for an exact value equal to $0.6234898019$.

I cannot resist to provide the approximation $\cos(x)=\frac{\pi^2-4x^2}{\pi^2+x^2}$ which would give an estimate of $\frac{33}{53}$.

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The value of $\cos\left(\frac{2\pi}{7}\right)$ is $\sin\left(\frac{3\pi}{14}\right)$. Although this might not help much, it does demonstrate that one method of computing a trigonometric function is to use a trigonometric Identity to simplify the expression to some known values of $\cos$ and $\sin$ (and, maybe other trig functions).

Another method is to use a Taylor Expansion (Maclaurin Series) and compute enough terms to satisfy your desired accuracy and precision. The Taylor Series for $\cos\left(x\right)$ is: $$ \cos\left(x\right) = \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}x^{2n} $$ But, if you are needing a quick answer, use your calculator and compute by pushing buttons until you get the answer: $$ \cos\left(\frac{2\pi}{7}\right) = 0.62349+ $$

K7PEH
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It is a solution to a cubic equation.
Two reasons: Firstly, $$\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}+...+\cos\frac{14\pi}{7}=0$$ and that simplifies to $\cos2\pi/7+\cos4\pi/7+\cos6\pi/7=-1/2$.
Secondly, $\cos2\theta=2\cos^2\theta-1$ and $\cos3\theta=4\cos^3\theta-3\cos\theta$.
There is a formula to solve cubic equations.

Empy2
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