Prove that: $\sec^2 20^\circ +\sec^2 40^\circ +\sec^2 80^\circ = \textrm 36$

Question

My Attempt:

$$L.H.S=\sec^2 20^\circ + \sec^2 40^\circ +\sec^2 80^\circ$$ $$=\dfrac {1}{\cos^2 20°} +\dfrac {1}{\cos^2 40°} +\dfrac {1}{\cos^2 80°}$$ $$=\dfrac {\cos^2 40°.\cos^2 80°+\cos^2 20°.\cos^2 80°+\cos^2 20°.\cos^2 40°}{\cos^2 20°.\cos^2 40°.\cos^2 80°}$$.

I got paused here. Please help to prove this..

Write $3+\tan ^{2}20^{\circ}+\tan ^{2}40^{\circ}+\tan ^{2}80^{\circ}$. — 2017-02-10
With http://math.stackexchange.com/questions/175736/evaluate-tan-220-circ-tan-240-circ-tan-280-circ — 2017-02-10

user id:10400 106k · Accepted Answer

A clean start, not remotely obvious ( but easy enough to prove ), the roots of $$ x^3 - 3 x + 1 $$ are $$ 2 \cos \frac{2 \pi}{9}, \; \; 2 \cos \frac{4 \pi}{9}, \; \; 2 \cos \frac{8 \pi}{9}. $$ From page 174 in Reuschle (1875). The method used is due to Gauss.

I learned today that, in 1933, D. H. Lehmer published a two-page proof that $2 \cos (2k\pi/n)$ is always an algebraic integer. Mentioned in an article in the January (M. A. A.) Monthly by Y. Z. Gurtas.

We immediately get that the roots of $$ 8 x^3 - 6 x + 1, $$ therefore $$ x^3 - \frac{3}{4} x + \frac{1}{8} $$ are $$ \cos \frac{2 \pi}{9}, \; \; \cos \frac{4 \pi}{9}, \; \; \cos \frac{8 \pi}{9}. $$ Compare with the three requested proofs in the answer by Michael R.

Proof comes by taking $\omega$ as a primitive ninth root of unity, then taking $$ x = \omega + \frac{1}{\omega}. $$ Primitive means $$ \omega \neq 1, \; \; \omega^3 \neq 1, \omega^9 = 1 $$ Then $$ x^3 = \omega^3 + 3 \omega + \frac{3}{\omega} + \frac{1}{\omega^3}, $$ $$ x^3 - 3 x + 1 = \omega^3 + 1 + \frac{1}{\omega^3}, $$ $$ \omega^3 (x^3 - 3 x + 1) = \omega^6 + \omega^3 + 1, $$ $$ (\omega^3 - 1)\omega^3 (x^3 - 3 x + 1) = (\omega^3 - 1)(\omega^6 + \omega^3 + 1) = \omega^9 - 1 = 0. $$

These can be combined with double angle formulas/half angle formulas.

user id:190319 114k · Accepted Answer

Prove that $$\cos40^{\circ}+\cos80^{\circ}+\cos160^{\circ}=0,$$ $$\cos40^{\circ}\cos80^{\circ}+\cos40^{\circ}\cos160^{\circ}+\cos80^{\circ}\cos160^{\circ}=-\frac{3}{4}$$ and $$\cos40^{\circ}\cos80^{\circ}\cos160^{\circ}=-\frac{1}{8}.$$ Hence, $$\sec^2 20^\circ +\sec^2 40^\circ +\sec^2 80^\circ=\frac{\left(-\frac{3}{4}\right)^2}{\left(-\frac{1}{8}\right)^2}=36$$ Because $$\cos40^{\circ}+\cos80^{\circ}+\cos160^{\circ}=2\cos60^{\circ}\cos20^{\circ}+\cos160^{\circ}=\cos20^{\circ}+\cos160^{\circ}=0,$$ $$\cos40^{\circ}\cos80^{\circ}+\cos40^{\circ}\cos160^{\circ}+\cos80^{\circ}\cos160^{\circ}=$$ $$=\frac{1}{2}\left(\cos120^{\circ}+\cos40^{\circ}+\cos120^{\circ}+\cos160^{\circ}+\cos240^{\circ}+\cos80^{\circ}\right)=$$ $$=\frac{1}{2}\left(\cos120^{\circ}+\cos120^{\circ}+\cos240^{\circ}\right)=-\frac{3}{4}$$ and $$\cos40^{\circ}\cos80^{\circ}\cos160^{\circ}=\frac{8\sin40^{\circ}\cos40^{\circ}\cos80^{\circ}\cos160^{\circ}}{8\sin40^{\circ}}=$$ $$=\frac{4\sin80^{\circ}\cos80^{\circ}\cos160^{\circ}}{8\sin40^{\circ}}=\frac{2\sin160^{\circ}\cos160^{\circ}}{8\sin40^{\circ}}=\frac{\sin320^{\circ}}{8\sin40^{\circ}}=-\frac{1}{8}.$$

Let $\cos40^{\circ}=a$, $\cos80^{\circ}=b$ and $\cos160^{\circ}=c$.

Hence, $$\sec^2 20^\circ +\sec^2 40^\circ +\sec^2 80^\circ=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=$$ $$=\frac{a^2b^2+a^2c^2+b^2c^2}{a^2b^2c^2}=\frac{(ab+ac+bc)^2-2abc(a+b+c)}{a^2b^2c^2}=\frac{\left(-\frac{3}{4}\right)^2}{\left(-\frac{1}{8}\right)^2}=36$$

@ Michael Rozenberg, Could You please elaborate the upper portions of you solution.. — 2017-02-10
@ Michael Rozenberg, Why do we need to prove those 3 expressions prior to this? And, where are we expected to.use them? — 2017-02-10
@Euler, in case of interest, those first three proofs follow quickly from a method due to Gauss, from an 1875 book. Worth going through yourself. — 2017-02-10
@ Michael Rozenberg, Could you say me the difference in yours and lab's solution?? — 2017-02-10

user id:291201 58k · Accepted Answer

Let $z = \cos 20^\circ\,$ then by the triple angle formula $\frac{1}{2}=\cos 60^\circ = 4 z^3 - 3 z$ $\iff 8z^3-6z-1=0$.

By the double angle formula $\cos 40^\circ = 2 z^2 - 1\,$ and $\cos 80^\circ = 2(2z^2-1)^2-1=8z^4-8z^2+1\,$. But $8z^4=z\cdot8z^3=z(6z+1)$ per the previous equation, so $\cos 80^\circ = -2z^2+z+1\,$.

Then the equality to prove becomes:

$$ \frac{1}{z^2}+\frac{1}{(2z^2-1)^2}+\frac{1}{(2z^2-z-1)^2} = 36 $$

$$ \iff \quad (2z^2-1)^2 (2z^2-z-1)^2 + z^2 (2z^2-z-1)^2 + z^2 (2z^2-1)^2 - 36 z^2(2z^2-1)^2 (2z^2-z-1)^2 = 0 $$

After expanding and routine simplifications, the above reduces to:

$$ 576 z^{10} - 576 z^9 - 1024 z^8 + 880 z^7 + 740 z^6 - 452 z^5 - 265 z^4 + 82 z^3 + 41 z^2 - 2 z - 1 = 0 $$

It can be verified by Euclidian division that the latter polynomial has $8z^3-6z-1$ as a factor:

$$ (8 z^3 - 6 z - 1) \cdot (72 z^7 - 72 z^6 - 74 z^5 + 65 z^4 + 28 z^3 - 17 z^2 - 4 z + 1) $$

Therefore the equality holds, which completes the proof.

@Euler Essentially, the equality to be proven can be written in terms of $z = \cos 20^\circ\,$, and reduces to a polynomial equation in $z\,$. That expression turns out to be indeed $0$ because $z$ satisfies a simple cubic equation which is a factor of said polynomial. Please point out which steps are not clear, and I can elaborate. — 2017-02-10
@ dxiv, I am satisfied with your answer above this ( Then the equality to prove becomes: ). But from this point downward, I could not understand. — 2017-02-10
@ dxiv, If possible, Could you please explain me here at :: http://chat.stackexchange.com/rooms/53289/ keep-typing — 2017-02-10
@Euler The equality following that line is simply substituting the expressions in $z$ derived earlier for $\cos 20^\circ, \cos 40^\circ, \cos 80^\circ$. The next step is eliminating the denominators and bringing it to a polynomial equation (just edited the answer to make that more clear). — 2017-02-10
@Euler That's just another name for [polynomial division](https://en.wikipedia.org/wiki/Polynomial_long_division). Point (in this case) is that you can verify whether a polynomial divides another pretty much mechanically. You don't need to know their roots, or their factorizations, and it works no matter how high the degrees are or how "ugly" the two look. — 2017-02-10
@ dxiv, what is this? How it happened?? " Therefore the equality holds, which completes the proof."" — 2017-02-10
That last line is equivalent to the equality you had to prove. Turned out it's in the form $(8z^3-6z-1) \cdot \text{something} = 0\,$. But $8z^3-6z-1=0\,$, look back at the very *first* line of the answer. Therefore you have $0 \cdot \text{something} = 0\,$, which is obviously true, and does in fact complete the proof. — 2017-02-10

user id:33337 231k · Accepted Answer

Starting like dxiv,

let $a=\cos20^\circ,b=-\cos40^\circ,c=-\cos80^\circ,$

As $\cos(3\cdot20^\circ)=\dfrac12$

$\cos(3\cdot40^\circ)=-\dfrac12$

$\cos(3\cdot80^\circ)=-\dfrac12$

As $\cos3x=\cos60^\circ, 3x=360^\circ m\pm60^\circ$ where $m$ is any integer.

$\implies x=120^\circ m+20^\circ$ where $m\equiv-1,0,1\pmod3$

Now as $\cos3x=4\cos^3x-3\cos x$

The roots of $4t^3-3t-\dfrac12=0$ are $a,b,c$

and we need to find $\dfrac1{a^2}+\dfrac1{b^2}+\dfrac1{c^2}=\dfrac{a^2b^2+b^2c^2+c^2a^2}{(abc)^2}$ $=\dfrac{(ab+bc+ca)^2-2abc(a+b+c)}{(abc)^2}$

By Vieta's formula

$a+b+c=\dfrac04$

$ab+bc+ca=-\dfrac34$

$abc=-\dfrac1{2\cdot4}$

@ lab bhattacharjee, where does $\cos 3x=\textrm 120$ come from? — 2017-02-10
@ lab bhattacharjee, why did you let $\textrm b=- \cos 40^\circ$ and $\textrm c= (-)....$? — 2017-02-10
@ lab bhattacharjee, what is $\textrm m\equiv \textrm -1, \textrm 0, \textrm 1 (\textrm mod \textrm 3)$? — 2017-02-10
@ lab bhattacharjee, could you provide a bit more explanation on this solution? — 2017-02-10
@ lab bhattacharjee, How did you.get that $a,b,c$ are the roots of $4t^3-3t-1/2=0$? — 2017-02-11
Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/53454/discussion-between-lab-bhattacharjee-and-s-ramanujan). — 2017-02-11

Prove that: $\sec^2 20^\circ +\sec^2 40^\circ +\sec^2 80^\circ = \textrm 36$

4 Answers 4

Related Posts