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while trying to solve a problem, I stumbled upon a way of finding $\cos\left(\frac{\pi}{5}\right)$ using identities and the cubic formula. Is it possible to find other values of sine or cosine in a similar way ?

Consider $$\cos\left(\frac{\pi}{5}\right) - \cos\left(\frac{2\pi}{5}\right).$$ Using the difference of cosines identity, we have $$\cos\left(\frac{\pi}{5}\right) - \cos\left(\frac{2\pi}{5}\right) = -2\sin\left(\frac{3\pi}{10}\right)\sin\left(-\frac{\pi}{10}\right).$$

Now we change the RHS using the identity $\sin(x) = \cos\left(\frac{\pi}{2}-x\right)$ and the fact that $\sin(x)$ is odd.

$$-2\sin\left(\frac{3\pi}{10}\right)\sin\left(-\frac{\pi}{10}\right) = 2\cos\left(\frac{\pi}{5}\right) \cos\left(\frac{2\pi}{5}\right)$$ So, $$\cos\left(\frac{\pi}{5}\right) - \cos\left(\frac{2\pi}{5}\right) = 2\cos\left(\frac{\pi}{5}\right) \cos\left(\frac{2\pi}{5}\right)$$

Now we make use of the identity $\cos(2x)=2\cos^2(x)-1$. $$\cos\left(\frac{\pi}{5}\right) - 2\cos^2\left(\frac{\pi}{5}\right)+1 = 2\cos\left(\frac{\pi}{5}\right) \left(2\cos^2\left(\frac{\pi}{5}\right)-1\right)$$ Let $y=\cos\left(\frac{\pi}{5}\right)$ and we have $$y-y^2+1=2y(2y^2-1)$$ $$4y^3+2y^2-3y-1=0$$ which has the correct solution $$ y=\frac{\sqrt{5}+1}{4} =\cos\left(\dfrac{\pi}{5}\right) $$ One of the roots is also $\sin\left(\dfrac{\pi}{10}\right)$ which I'm guessing is because you end up with the same cubic if you apply the above to sin too.

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    $\begingroup$ There's no question $\endgroup$ – Jakobian Jun 7 '18 at 15:12
  • $\begingroup$ I added the question at the end. $\endgroup$ – Adam Jun 7 '18 at 15:14
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    $\begingroup$ I'm 100% sure there is a similar way to calculate arbitrary values of $\sin(\cdot)$ with your way. Another way I know would be to use the triple angle identity$$\sin 3\theta=3\sin\theta-4\sin^3\theta$$but you'll need to calculate what $\sin^3 3\theta$ is, which can be tedious. $\endgroup$ – Frank W. Jun 7 '18 at 15:17
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    $\begingroup$ $\cos\frac{2\pi}{n}$ is an algebraic number over $\mathbb{Q}$ width degree $\frac{1}{2}\varphi(n)$. In particular finding an explicit form for $\cos\frac{\pi}{5}$ only requires the quadratic formula. $\endgroup$ – Jack D'Aurizio Jun 7 '18 at 15:19
  • $\begingroup$ Is that the method with the 36-72-72 triangle ? $\endgroup$ – Adam Jun 7 '18 at 15:22
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Using complex numbers we can derive $\sin(\frac{\pi}{5})$ and $\cos(\frac{\pi}{5})$ $$ z^5=-1 \implies z^4-z^3+z^2-z+1=0 \implies z^2+\frac{1}{z^2}-z-\frac{1}{z}+1=0 $$ By substitution $t=z+\frac{1}{z}$ we get: $$ t^2-t-1=0 \implies t=\frac{1\pm\sqrt{5}}{2} $$ And because $\cos(\frac{\pi}{5})$ is positive, we may consider just $t=\frac{1+\sqrt{5}}{2}$

$$ z^2-z\frac{1+\sqrt{5}}{2}+1=0 \implies z=\frac{1+\sqrt{5}}{4}\pm i\frac{\sqrt{10-2\sqrt{5}}}{4} $$ And we get that $\cos(\frac{\pi}{5}) = \frac{1+\sqrt{5}}{4} $ and $\sin(\frac{\pi}{5})= \frac{\sqrt{10-2\sqrt{5}}}{4}$

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  • $\begingroup$ that's very cool! $\endgroup$ – Adam Jun 7 '18 at 18:02
  • $\begingroup$ I was quite confused how this arrived at $z^2-z\frac{1+\sqrt{5}}{2}+1=0$. But, for anyone confused, you're substituting the value of $t$ you've found into the equation for $z$, to solve for $z$. $\endgroup$ – Jam Jun 14 '18 at 12:10
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There is a slightly shorter way:

$$\sin\frac{\pi}{5}=\sin\frac{4\pi}{5}=2\sin\frac{2\pi}{5}\cos\frac{2\pi}{5}=4\sin\frac{\pi}{5}\cos\frac{\pi}{5}\cos\frac{2\pi}{5}$$

Hence

$$\cos\frac{\pi}{5}\cos\frac{2\pi}{5}=\frac14$$

You can also write this

$$\cos\frac{\pi}{5}\cos\frac{3\pi}{5}=-\frac14$$

Then, using the formula $2\cos a\cos b=\cos(a+b)+\cos(a-b)$, you have:

$$\frac12=2\cos\frac{\pi}{5}\cos\frac{2\pi}{5}=\cos\frac{\pi}{5}+\cos\frac{3\pi}{5}$$

Hence $\cos\frac{\pi}{5}$ and $\cos\frac{3\pi}{5}$ are the roots of $t^2-\frac12t-\frac14$. The rest is easy, and the positive root is $\cos\frac{\pi}5$.


To answer your question: in general, it's not possible to find values of $\cos\frac{\pi}{n}$ only by radicals, even though they are algebraic numbers. Even in the case you obtain an irreducible cubic equation (which can be solved by radicals), it's the "trigonometric case" here, which has no expression with real radicals (and a complex cubic root needs the trigonometric functions anyway). For instance, $\cos1^\circ$ can't be computed with real radicals. But $\cos3^\circ$ can.

Even if in general it's not possible, there are a few tricks you can apply, for instance

$$\cos\frac{\pi}{12}=\cos\left(\frac{\pi}{3}-\frac{\pi}{4}\right)$$

Also, if you can compute $\cos x$ by radicals, then you can compute $\cos\dfrac{x}{2^n}$ too.


So, which angles of the form $\frac{\pi}{n}$ leads to trigonometric functions computable by radicals? The answer is given by the Gauss-Wantzel theorem. For instance, one can compute $\cos\dfrac{\pi}{17}$. However, the computation is not obvious, see http://mathworld.wolfram.com/TrigonometryAnglesPi17.html

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Everything begins with finding $\sin\left(\frac{\pi}{10}\right)$

Okay... Lets say $x=\frac{\pi}{10}$

$$5x=\frac{\pi}{2}$$ $$2x=\frac{\pi}{2}-3x$$ $$\sin(2x)=\sin\left(\frac{\pi}{2}-3x\right)$$ $$\sin(2x)=\cos(3x)$$ $$2\sin(x)\cos(x)=4\cos^3(x)-3\cos(x)$$ $$2\sin(x)\cos(x)-4\cos^3(x)+3\cos(x)=0$$ $$\cos(x)\left(2\sin(x)-4\cos^2(x)+3\right)=0$$ $$2\sin(x)-4\cos^2(x)+3=0$$ $$2\sin(x)-4(1-sin^2(x))+3=0$$ $$2\sin(x)-4+4\sin^2(x)+3=0$$ $$4\sin^2(x)+2\sin(x)-1=0$$ Now use the formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ $$a=4,b=2,c=-1$$ $$=\frac{-2\pm \sqrt{2^2-4(4)(-1)}}{2(4)}$$ $$=\frac{-2\pm \sqrt{20}}{8}$$ $$\sin\left(\frac{\pi}{10}\right)=\frac{-1\pm \sqrt{5}}{4}$$

One another way of finding the value of $\cos\left(\frac{\pi}{5}\right)$

$$\cos\left(\frac{\pi}{5}\right)=\cos\left(2\left(\frac{\pi}{10}\right)\right)$$ $$=1-2\left(\sin\frac{\pi}{10}\right)^2$$ Since $\sin\left(\frac{\pi}{10}\right)=\frac{\sqrt{5}-1}{4}$ $$=1-2\left(\frac{\sqrt{5}-1}{4}\right)^2$$ $$=1-\frac{(\sqrt{5}-1)^2}{8}$$ $$=\frac{8-5-1+2\sqrt{5}}{8}$$ $$=\frac{2+2\sqrt{5}}{8}$$ $$\cos\left(\frac{\pi}{5}\right)=\frac{1+\sqrt{5}}{4}$$ Now lets find $\sin\left(\frac{\pi}{5}\right)$ $$\sin\left(\frac{\pi}{5}\right)=\sqrt{1-\cos^2\left(\frac{\pi}{5}\right)}$$ $$=\sqrt{1-\left(\frac{1+\sqrt{5}}{4}\right)^2}$$ $$=\sqrt{1-\frac{1+5+2\sqrt{5}}{16}}$$ $$=\sqrt{\frac{10-2\sqrt{5}}{16}}$$ $$\sin\left(\frac{\pi}{5}\right)=\frac{\sqrt{10-2\sqrt{5}}}{4}$$

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  • $\begingroup$ But how would you find $\sin(\pi /5)$ ? $\endgroup$ – Adam Jun 7 '18 at 18:03
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    $\begingroup$ @Adam I added it in my answer $\endgroup$ – tien lee Jun 7 '18 at 18:21
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    $\begingroup$ How do you find $\sin(\pi/10)$? $\endgroup$ – Blue Jun 7 '18 at 18:24
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    $\begingroup$ @Blue I added it in my answer. Finding $\sin(\pi/10)$ was really challenging and it took a bit to find out the way $\endgroup$ – tien lee Jun 7 '18 at 18:43
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    $\begingroup$ Yes, I should have done it in that way too... $\endgroup$ – tien lee Jun 7 '18 at 19:26
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Not so cool. The following may be cooler.

Let $\Delta ABC$ be a triangle with $\angle BAC=36^o$, $\angle ABC=\angle ACB=72^o.$ $BD$ bisects $\angle ABC$. $BE\perp AC$. You would find that $AD=BD=BC$ and $\Delta ABC \sim \Delta BDC.$ Thus, as the figure shows, we may obtain $$\frac{x}{2y}=\frac{x+2y}{x}.$$ Thus $$\left(\frac{x}{y}\right)^2-2\left(\frac{x}{y}\right)-4=0.$$

Solve it. We have $$\frac{x}{y}=1+\sqrt{5}.$$

Another negative root is not what we want.Thus, $$\cos 36^o=\frac{AE}{AB}=\frac{x+y}{x+2y}=\dfrac{\dfrac{x}{y}+1}{\dfrac{x}{y}+2}=\frac{1+\sqrt{5}}{4}.$$

enter image description here

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