# TrigonometryProving a trigonometric equation

#### Lisa91

##### New member
I want to prove this equation:
$$\cos\left(\frac{2\pi}{5}\right)+\cos\left( \frac{4\pi}{5}\right)+\cos\left( \frac{6\pi}{5}\right)+\cos\left(\frac{8\pi}{5} \right) = -1$$.

I've no idea how to begin. I think it may be related to complex numerd but I don't know how to combine them.

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#### MarkFL

Staff member
Re: trigonometry equation

You may want to consider the following sum-to-product identity:

$\displaystyle \cos(\alpha)+\cos(\beta)=2\cos\left(\frac{\alpha+ \beta}{2} \right)\cos\left(\frac{\alpha-\beta}{2} \right)$

#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar
Re: trigonometry equation

Also note that $\cos(2\pi/5)=\cos(8\pi/5)$ and $\cos(4\pi/5)=\cos(6\pi/5)$.

#### MarkFL

Staff member
Re: trigonometry equation

Also note that $\cos(2\pi/5)=\cos(8\pi/5)$ and $\cos(4\pi/5)=\cos(6\pi/5)$.
Yes, this comes from the identity:

$\displaystyle \cos(2\pi-\theta)=\cos(\theta)$

#### Lisa91

##### New member
Re: trigonometry equation

I got $$4\cos(\frac{2\pi}{5})\cos(\frac{2\pi}{5})\cos(\frac{\pi}{5})$$. I wanted to make it easier by using the following equation: $$2\cos^{2}(x)-1 = \cos(2x)$$.
Now I have $$16t^{3}-10t^{2}+4t+1=0$$, t means $$\cos(\frac{\pi}{5})$$. I don't know whether it's ok and how to solve it.

EDIT: OK, I've just spotted your answers, I'll try to make use of them.

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#### Lisa91

##### New member
Re: trigonometry equation

I got $$4\cos(\frac{4\pi}{5})\cos(\frac{2\pi}{5}) = -1$$. What shall I do with it?

#### MarkFL

Staff member
Re: trigonometry equation

You should find:

$\displaystyle \cos\left(\frac{2\pi}{5} \right)\cos\left(\frac{\pi}{5} \right)=\frac{1}{4}$

Now, to find the value of $\displaystyle \cos\left(\frac{\pi}{5} \right)$, we may employ a Chebyshev polynomial where if $\displaystyle x=\cos\left(\frac{\pi}{5} \right)$ we then have:

$\displaystyle 16x^5-20x^3+5x+1=0$

Using the rational roots theorem and division to factor, we may write:

$\displaystyle (x+1)(4x^2-2x-1)^2=0$

Given we must have $\displaystyle 0<x<1$ (we have a first quadrant angle) use the quadratic formula to find the appropriate root.

Next, use a double-angle identity for cosine to find $\displaystyle \cos\left(\frac{2\pi}{5} \right)$. You should then be able to show:

$\displaystyle \cos\left(\frac{2\pi}{5} \right)\cos\left(\frac{\pi}{5} \right)=\frac{1}{4}$

#### Lisa91

##### New member
Re: trigonometry equation

Are you sure it's $$\frac{1}{4}$$? I think there should be $$- \frac{1}{4}$$. So $$-1< x<0$$. I got two more roots $$x = \frac{1-\sqrt{5}}{4}$$ and $$x = \frac{1+\sqrt{5}}{4}$$.

How did you get $$4\cos(\frac{2\pi}{5})\cos(\frac{\pi}{5})$$? I got $$4\cos(\frac{4\pi}{5})\cos(\frac{2\pi}{5})$$...

#### Opalg

##### MHB Oldtimer
Staff member
Re: trigonometry equation

I want to prove this equation:
$$\cos(\frac{2\pi}{5})+\cos( \frac{4\pi}{5})+\cos( \frac{6\pi}{5})+\cos(\frac{8\pi}{5}) = -1$$.

I've no idea how to begin. I think it may be related to complex numbers but I don't know how to combine them.
One way to do this using complex numbers is to notice that the numbers $e^{2k\pi i/5}\ (k=0,1,2,3,4)$ are the roots of the equation $z^5-1=0.$ The sum of the roots of that equation is $0$ (because the coefficient of $z^4$ in the equation is $0$). So $1+e^{2\pi i/5}+e^{4\pi i/5}+e^{6\pi i/5}+e^{8\pi i/5} = 0$. Take the real part of that, to get the result you want.

As a bonus, you could take the imaginary part of that last equation and get the fact that $\sin\bigl(\tfrac{2\pi}{5}\bigr)+\sin\bigl( \tfrac{4\pi}{5}\bigr)+\sin\bigl( \tfrac{6\pi}{5}\bigr)+\sin\bigl(\tfrac{8\pi}{5} \bigr) = 0.$

#### Lisa91

##### New member
Re: trigonometry equation

So $1+e^{2\pi i/5}+e^{4\pi i/5}+e^{6\pi i/5}+e^{8\pi i/5} = 0$. Take the real part of that, to get the result you want.
How to take the real part of it?

#### Fernando Revilla

##### Well-known member
MHB Math Helper
Re: trigonometry equation

Another way. Denoting:

$A=\cos \dfrac{2\pi}{5}+\cos \dfrac{4\pi}{5}+\cos \dfrac{6\pi}{5}+\cos \dfrac{8\pi}{5}$
$B=\sin \dfrac{2\pi}{5}+\sin \dfrac{4\pi}{5}+\sin \dfrac{6\pi}{5}+\sin \dfrac{8\pi}{5}$

we get a geometric progression:

$A+iB=e^{\frac{2\pi}{5}i}+e^{\frac{4\pi}{5}i}+ e^{\frac{6\pi}{5}i}+e^{\frac{8\pi}{5}i}$
$=\dfrac{e^{\frac{2\pi}{5}i}\left(e^{\frac{8\pi}{5}i}-1\right)}{e^{\frac{2\pi}{5}i}-1}$
$=\dfrac{e^{\frac{2\pi}{5}i}e^{\frac{4\pi}{5}i}}{e^{\frac{\pi}{5}i}}\;\dfrac{e^{\frac{4\pi}{5}i}-e^{\frac{-4\pi}{5}i}}{e^{\frac{\pi}{5}i}-e^{\frac{-\pi}{5}i}}$
$=e^{\pi i}\dfrac{\sin \frac{4\pi}{5}}{\sin \frac{\pi}{5}}$
$=-\dfrac{\sin\left(\pi-\frac{\pi}{5}\right)}{\sin \frac{\pi}{5}}$
$=-1$

So,

$\cos \dfrac{2\pi}{5}+\cos \dfrac{4\pi}{5}+\cos \dfrac{6\pi}{5}+\cos \dfrac{8\pi}{5}=-1$
$\sin \dfrac{2\pi}{5}+\sin \dfrac{4\pi}{5}+\sin \dfrac{6\pi}{5}+\sin \dfrac{8\pi}{5}=0$

#### Deveno

##### Well-known member
MHB Math Scholar
Re: trigonometry equation

let's take the scenic route (i love to look out the window and wave to the birds and trees and stuff....)!

what is the average of $\dfrac{2\pi}{5}$ and $\dfrac{3\pi}{5}$?

um...well...lemme see...it's:

$$\frac{1}{2}\left(\frac{2\pi}{5} + \frac{3\pi}{5}\right) = \frac{5\pi}{10} = \frac{\pi}{2}$$.

this tells us that:

$\displaystyle \sin\left(\frac{2\pi}{5}\right) = \sin\left(\frac{3\pi}{5}\right)$

(since $\frac{2\pi}{5}$ and $\frac{3\pi}{5}$ are the same distance away from $\frac{\pi}{2}$ and sine is symmetric about $\frac{\pi}{2}$, draw this on a circle, and you will see what i mean).

hopefully you know these two formulas (formulae? formuli? formu-la-ha-ha-ha-ha-ha-ha...oops, sorry):

$\sin(2x) = 2\sin(x)\cos(x)$
$\sin(3x) = 4\sin(x)\cos^2(x) - \sin(x)$

(i'll wait while you verify this. ok, time's up! stop!!!).

since these are equal when $x = \frac{\pi}{5}$, we have (using $y$ instead of $\frac{\pi}{5}$...why? why y? oh why o why?, well y naught? seriously, i kill myself):

$2\sin(y)\cos(y) = 4\sin(y)\cos^2(y) - \sin(y)$

and we have some serious cancellation going on, now (and not those frivolous cancellations you might find on "other" math sites). in particular $\sin(y) \neq 0$, so we can divide it out, leaving:

$2\cos(y) = 4\cos^2(y) - 1$

using a "$u$-substitution" (because i CAN. so there), of: $u = \cos(y)$, this becomes the friendly neighborhood quadratic:

$4u^2 - 2u - 1 = 0$. we can solve this (or at least i can, because i know the formu-....ok, we won't go there), to get:

$\displaystyle u = \frac{1 \pm \sqrt{5}}{4}$

we want the positive solution, since $y$ is in the first quadrant. well, it's all downhill from here (so i'll just shut off the engine, and coast. wheeeee!).

since $\cos(2x) = 2\cos^2(x) - 1$, we have:

$\displaystyle \cos\left(\frac{2\pi}{5}\right) = 2\left(\frac{1+\sqrt{5}}{4}\right)^2 - 1 = \frac{\sqrt{5} - 1}{4}$

similarly, since $4 = 2\cdot 2$ (i KNEW that would come in handy some day!):

$\displaystyle \cos\left(\frac{4\pi}{5}\right) = 2\left(\frac{\sqrt{5} - 1}{4}\right)^2 - 1 = \frac{-\sqrt{5} - 1}{4}$

now since "others" (who aren't nearly as awesome as ME, so they shall remain nameless, in order to further promote my shameless self-promotion. i take cash, checks and all major credit cards) have pointed out that:

$\displaystyle \cos\left(\frac{2\pi}{5}\right) = \cos\left(\frac{8\pi}{5}\right);\ \cos\left(\frac{4\pi}{5}\right) = \cos\left(\frac{6\pi}{5}\right)$

when we add these four bad boys up (bad boys, bad boys, whatcha gonna do...whatcha gonna do when dey come for you....am i easily distracted? yes? no? maybe??) we get:

$\displaystyle \frac{\sqrt{5} -1}{4} + \frac{-\sqrt{5} -1}{4} + \frac{\sqrt{5} - 1}{4} + \frac{-\sqrt{5} - 1}{4} = -1$

TA DA!!!!

(this message brought to you by starbucksTM coffee. remember to drink responsibly).

(posthumous credit given to evgeny.makarov and MarkFL on advice of counsel).

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#### MarkFL

Staff member
Re: trigonometry equation

Are you sure it's $$\frac{1}{4}$$? I think there should be $$- \frac{1}{4}$$. So $$-1< x<0$$. I got two more roots $$x = \frac{1-\sqrt{5}}{4}$$ and $$x = \frac{1+\sqrt{5}}{4}$$.

How did you get $$4\cos(\frac{2\pi}{5})\cos(\frac{\pi}{5})$$? I got $$4\cos(\frac{4\pi}{5})\cos(\frac{2\pi}{5})$$...
I grouped the first and last terms and the two middle terms on the left to get:

$\displaystyle \left(\cos\left(\frac{8\pi}{5} \right)+\cos\left(\frac{2\pi}{5} \right) \right)+\left(\cos\left(\frac{6\pi}{5} \right)+\cos\left(\frac{4\pi}{5} \right) \right)=-1$

Now, using the sum-to-product identity I cited, we have:

$\displaystyle 2\cos(\pi)\cos\left(\frac{3\pi}{5} \right)+2\cos(\pi)\cos\left(\frac{\pi}{5} \right)=-1$

$\displaystyle 2\cos(\pi)\left(\cos\left(\frac{3\pi}{5} \right)+\cos\left(\frac{\pi}{5} \right) \right)=-1$

Now, using $\displaystyle cos(\pi)=-1$ and the sum-to-product identity again, we have:

$\displaystyle -2\left(2\cos\left(\frac{2\pi}{5} \right)\cos\left(\frac{\pi}{5} \right) \right)=-1$

$\displaystyle -4\cos\left(\frac{2\pi}{5} \right)\cos\left(\frac{\pi}{5} \right)=-1$

$\displaystyle \cos\left(\frac{2\pi}{5} \right)\cos\left(\frac{\pi}{5} \right)=\frac{1}{4}$

#### Lisa91

##### New member
Re: trigonometry equation

Thank you so much. Seems like I made a mistake while calculating.