I tried to solve a question which led to

$$4 \sin\frac{\pi}{5} \sin\frac{3\pi}{5} = \sqrt{5}$$

But, I got stuck to prove that.

Is there any easy solution?

  • 1
    $\begingroup$ Faravish: If you are familiar with complex roots of unity, then this answer gives you an answer without needing to solve any quadratics. There $x=e^{2\pi i/5}$, so $(x^4-x)=-2i\sin(2\pi/5)=-2i\sin(3\pi/5)$, and $(x^2-x^3)=2i\sin(4\pi/5)=2i\sin(\pi/5)$. Therefore $$\sqrt5=(x^2-x^3)(x^4-x)=4\sin\frac\pi5\sin\frac{3\pi}5.$$ $\endgroup$ – Jyrki Lahtonen Sep 20 '13 at 19:35

Note that

$$\sin{\frac{3 \pi}{5}} = \sin{\frac{2 \pi}{5}}$$

then use a double-angle and triple-angle forumla:

$$\sin{2 x} = 2 \sin{x} \cos{x}$$ $$\sin{3 x} = 3 \sin{x} - 4 \sin^3{x}$$

In this case, $x=\pi/5$. Setting the above two equations equal to each other results in the quadratic equation in question:

$$2 \cos{x} = 3 - 4 (1-\cos^2{x}) = 4 \cos^2{x}-1$$

which means that

$$\cos{x} = \frac{1 + \sqrt{5}}{4}$$ $$\sin{x} = \frac{\sqrt{10-2\sqrt{5}}}{4}$$

Then, using $\sin{2 x} = \sin{3 x}$ as noted above, we have

$$4 \sin{x} \sin{2 x} = 8 \sin^2{x} \cos{x} = 8 \frac{10-2 \sqrt{5}}{16} \frac{1+\sqrt{5}}{4} = \frac{10-10-2 \sqrt{5}+10\sqrt{5}}{8} = \sqrt{5}$$

  • $\begingroup$ true, and surely accepted. I think there is no way to solve without engaging polynomial equations. $\endgroup$ – faravish Sep 20 '13 at 12:54

Using this,

$\sin5x=16\sin^5x-20\sin^3x+5\sin x$

Let $\sin5x=0\implies 5x=n\pi$ where $n$ is any integer $\displaystyle\implies x=\frac{n\pi}5$

Clearly, $\displaystyle\sin\frac{\pi}5, \sin\frac{3\pi}5$ satisfy $16\sin^5x-20\sin^3x+5\sin x=0$

Again, if $\sin x=0,x=m\pi$ where $m$ is any integer

$\implies \displaystyle\sin\frac{\pi}5, \sin\frac{3\pi}5$ satisfy $16\sin^4x-20\sin^2x+5=0$

$\displaystyle\implies \sin^2x=\frac{20\pm\sqrt{20^2-4\cdot16\cdot5}}{2\cdot16}=\frac{5\pm\sqrt5}8$

Now, as $\displaystyle0<\frac\pi5,\frac{3\pi}5<\pi;\sin\frac{\pi}5, \sin\frac{3\pi}5>0,$ the values of $\displaystyle \sin\frac{\pi}5, \sin\frac{3\pi}5 $ will be $\displaystyle\sqrt{\frac{5\pm\sqrt5}8}$

$\displaystyle\implies \sin\frac{\pi}5\cdot\sin\frac{3\pi}5=\sqrt{\frac{5+\sqrt5}8}\cdot\sqrt{\frac{5-\sqrt5}8}=\frac{\sqrt{(5+\sqrt5)(5-\sqrt5)}}8=\frac{\sqrt5}4$

  • $\begingroup$ @faravish, how about this one? $\endgroup$ – lab bhattacharjee Sep 20 '13 at 14:15
  • $\begingroup$ so advanced. I'm looking for a way working even for fools! That's fine except for I cannot memorize it nor easier way. $\endgroup$ – faravish Sep 20 '13 at 15:00
  • $\begingroup$ @faravish, sorry for making things complex. If you know Quadratic Equation & All-Sin-Tan-Cos Rule (mathsrevision.net/sin-cos-and-tan), you can have a look into the edited version $\endgroup$ – lab bhattacharjee Sep 20 '13 at 18:14

I will find the $\cos\frac{2\pi}{5}$ with basic complex number theory.

Let $z = \cos(\frac{2\pi}{5}) + i \sin(\frac{2\pi}{5})$, then $z^5=1$. (de Moivre's Theorem) Let's find the solution of this quintic equation.

This equation is factored into $(z-1)(z^4 + z^3 + z^2 + z +1 ) = 0$. $z=1$ is trivial solution. Now let $u = z + z^4$, and $v= z^2 + z^3$. then $u + v = -1$, and $uv = z^3 + z^4 + z^6 + z^7 = z + z^2 + z^3 + z^4 = -1$. So $t^2 + t - 1 = 0$ has two solutions $u$ and $v$. Solving the quadratic equation, we get $t = \frac{-1 \pm \sqrt{5}}{2}$.

Geometrically $u$ has positive real part, and $v$ has negative real part. so $z + z^4 = \frac{-1 + \sqrt{5}}{2}$. Since $z$ and $z^4$ are conjugate, the real part of $z$ is the half of it, i.e. $\cos\frac{2\pi}{5}=\frac{-1 + \sqrt{5}}{4}$.

Now, $\cos(\frac{2\pi}{5}) = 1-2\sin^2(\frac{\pi}{5})$, so you can find $\sin\frac{\pi}{5}$.

In this way you can find $\cos\frac{2\pi}{17}$, by decreasing the order of equation 16th to 8th, 4th, .... You need to take $u$ and $v$ in special way...

  • $\begingroup$ very creative way. did you find by yourself or found from somewhere $\endgroup$ – faravish Sep 22 '13 at 7:23
  • $\begingroup$ Thanks. Not by myself, but original article was written in korean so i didnt linked here. $\endgroup$ – dust05 Sep 22 '13 at 7:46
  • $\begingroup$ Here is an original link. I read this article several years ago, and now math equations are not available.... bomber0.byus.net/index.php/2009/10/28/1599 $\endgroup$ – dust05 Sep 22 '13 at 7:50

Using $2\sin A\sin B=\cos(A-B)-\cos(A+B),$


As $\frac{144-72}2=36,$


Using $2\sin B\cos A=\sin(A+B)-\sin(A-B),$

$\displaystyle \frac{2\sin36^\circ\cos72^\circ+2\sin36^\circ\cos144^\circ}{2\sin36^\circ}$ $\displaystyle=\frac{\sin108^\circ-\sin36^\circ+2\sin180^\circ-\sin72^\circ}{2\sin36^\circ}=-\frac12$ as $\sin108^\circ=\sin(180^\circ-72^\circ)=\sin72^\circ$

$\displaystyle\cos72^\circ+\cos144^\circ=-\frac12\ \ \ \ (1)$

Again using $\sin2A=2\sin A\cos A$, $\displaystyle\cos72^\circ\cos144^\circ=\frac{(2\sin72^\circ\cos72^\circ)\cos144^\circ}{2\sin72^\circ} =\frac{\sin144^\circ\cos144^\circ}{2\sin72^\circ}$ $\displaystyle=\frac{2\sin144^\circ\cos144^\circ}{2\cdot2\sin72^\circ}=\frac{\sin288^\circ}{4\sin72^\circ}=\frac{\sin(360^\circ-72^\circ)}{4\sin72^\circ}=\frac{-\sin72^\circ}{4\sin72^\circ}=-\frac14\ \ \ \ (2)$

$\displaystyle\implies \cos72^\circ-\cos144^\circ=+\sqrt{(\cos72^\circ+\cos144^\circ)^2-4\cos72^\circ\cos144^\circ}$

$\displaystyle =+\sqrt{\left(-\frac12\right)^2-4\left(-\frac14\right)}$ using $(1),(2)$

$\displaystyle =\frac{\sqrt5}2$

$\displaystyle\implies 2\sin36^\circ\sin108^\circ=\frac{\sqrt5}2$

Alternatively, using this or this, $\displaystyle\cos36^\circ=\frac{\sqrt5+1}4$

Use $\cos2A=2\cos^2A-1$ to find $\cos72^\circ=\frac{\sqrt5-1}4$

and $\cos144^\circ=\cos(180^\circ-36^\circ)=-\cos36^\circ$

  • $\begingroup$ @faravish, please have a look into the first method $\endgroup$ – lab bhattacharjee Oct 5 '13 at 13:35

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