1

Find all the solution to the equation

$\cos 4\theta=\cos 3\theta$. Hence or otherwise, show that the roots of the equation
$$8x^3+4x^2-4x-1=0$$ are $$\cos \frac{2\pi}{7},\cos \frac{4\pi}{7},\cos \frac{6\pi}{7}$$

Therefore prove that $$\sec \frac{2\pi}{7}+\sec \frac{4\pi}{7}+\sec \frac{6\pi}{7}=-4$$ I am slightly stuck! I have solved the first equation using a graphical calculator and found that it is $(2n\pi)/7$ where n is any positive integer. I have then algebraically rewritten $cos(4\theta) = cos(3\theta)$ using double angle identities.

Now I am stuck... Is there are mistake in the question? What would be your model answer?

Thank you so very much.

Dietrich Burde
  • 130,978
vgupt
  • 310
  • 3
    Please type out the problem instead of using a picture. – Théophile Mar 31 '21 at 17:36
  • For example $\cos(8\pi/7) = \cos(6\pi/7)$. There are only three distinct roots of the equation. – GEdgar Mar 31 '21 at 17:41
  • 1
    include more of your attempt please – Siong Thye Goh Mar 31 '21 at 17:47
  • 1
    https://math.stackexchange.com/questions/638874/factor-z7-1-into-linear-and-quadratic-factors-and-prove-that-cos-pi-7-c https://math.stackexchange.com/questions/2139075/prove-that-cos-pi-7-is-root-of-equation-8x3-4x2-4x1-0 – lab bhattacharjee Mar 31 '21 at 17:48
  • With closed eyes, rewriting the trigonometric equation in terms of $\cos\theta$ (or $\sin\theta$ ?) will lead you to the given polynomial. –  Apr 01 '21 at 07:53

1 Answers1

5

$\cos(x) = \cos(y) \Leftrightarrow x = 2n\pi \pm y, n\in \Bbb Z$

So,

$4\theta = 2n\pi \pm3\theta \Rightarrow \begin{cases}\theta = 2n\pi\\\text{ or} \\\theta = \dfrac{2n\pi}{7}\end{cases}$

Now using compound angle formula,

$\begin{align}&\cos4\theta = \cos3\theta \\\Rightarrow &2\cos^2(2\theta )-1 = 4\cos^3\theta-3\cos\theta \\\Rightarrow &2(2\cos^2\theta-1)^2 -1 = 4\cos^3\theta-3\cos\theta \\ \Rightarrow &8\cos^4\theta-8\cos^2\theta+1 = 4\cos^3\theta-3\cos\theta \\ \Rightarrow&8\cos^4\theta-4\cos^3\theta- 8\cos^2\theta+3\cos\theta+1 = 0\end{align}$

We don't need the case where $\theta = 2n\pi$, that'll lead to $\cos\theta =1$.

So, we can factor out $\cos\theta -1 $ from the 4th order equation in $\cos$ (Perform long division or any other suitable method)

$\Rightarrow (\cos\theta-1)(8\cos^3\theta+4\cos^2\theta-4\cos\theta-1) = 0$

$\Rightarrow 8\cos^3\theta+4\cos^2\theta-4\cos\theta-1 =0$ or $\boxed{8x^3+4x^2-4x-1 = 0}$.

As $\cos$ has a periodicity of $2\pi$, we only consider for $n = \pm1,\pm2,\pm3$ or $\cos\frac{\pm2\pi}{7} = \cos\frac{2\pi}{7},\cos\frac{\pm4\pi}{7}=\cos\frac{4\pi}{7},\cos\frac{\pm6\pi}{7}=\cos\frac{6\pi}{7} $.

So the roots are $\cos\frac{2\pi}{7},\cos\frac{4\pi}{7},\cos\frac{6\pi}{7}$

If $a,b,c$ are roots of $Ax^3 +Bx^2+Cx+D = 0$, then

$ab +bc + ca = \frac{C}{A}$ and $abc = -\frac{D}{A} \Rightarrow \frac{ab+bc+ca}{abc}= \frac1a+\frac1b+\frac 1c = -\frac CD$

As $\frac1{\cos x} = \sec x$, here,

$$\frac1a+\frac 1b +\frac 1c =\sec\frac{2\pi}7 +\sec\frac{4\pi}7+\sec\frac{6\pi}7- \frac{-4}{-1} = 4$$

19aksh
  • 12,768
  • 1
    Why $\cos\theta =1$ have to be avoided(purporse of factoring out $ \cos\theta -1$)?. What is the consequence of not doing so? – Ian Mar 31 '21 at 20:38
  • 3
    @Ian We have a 4th order equation $8x^4-4x^3- 8x^2+3x+1 = 0, (x=\cos\theta)$with roots $\cos\theta$ where $\theta= 2n\pi \text{ or } \frac{2n\pi}{7}$. I've factorized $\cos\theta-1$, because we already know that it's a root of the biquadratic equation(from the solution of the first part). So that'll lead to either $\cos\theta-1=0$ or $8\cos^3\theta+4\cos^2\theta-4\cos\theta-1=0$. Now if $\theta \ne 2n\pi$, then only $8\cos^3\theta+4\cos^2\theta-4\cos\theta-1$ will be $0$ and that'll provide the desired result. – 19aksh Apr 01 '21 at 01:24
  • 1
    Now I understand. I appreciate your feedback – Ian Apr 01 '21 at 02:18
  • 1
    Thank you =) @ian – 19aksh Apr 01 '21 at 02:22
  • 2
    @Ak. out of curiosity - if we were given just the polynomial how would we know to use the trigonometric equation to solve the problem? Also, I don't quite understand why we only consider the cases n=1,2, and 3 only. Thanks. [A brilliant answer by the way :)] – vgupt Apr 01 '21 at 07:14
  • @vgupt Thank you! For the 1st qn... If they've given only the algebraic equation we could not have found the solution that easily. But here, it is given, *Hence, show that...* so we need to use the given trigonometric equality to solve the algebraic eqn. – 19aksh Apr 01 '21 at 07:36
  • For the 2nd qn,.. we have $\theta = \frac{2n\pi}{7}$, Let's take $n=4$, this will lead to $\theta = \frac{8\pi}{7} = 2\pi -\frac{6\pi}{7}$ and $\cos(2\pi- x) =\cos(x)$.. So, $\cos(\frac{8\pi}{7})= \cos(\frac{6\pi}{7})$..Similarly for other values of $n$, you can find angles such that it is equivalent to $\cos$ of $\frac{2\pi}{7},\frac{4\pi}{7},\frac{6\pi}{7}$.. It is because, the cosine has a periodicity of $2\pi$ – 19aksh Apr 01 '21 at 07:36