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

Question

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

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..

Answer 1

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

$a+b+c=\dfrac04$

$ab+bc+ca=-\dfrac34$

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

Answer 2

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

Answer 3

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.

Answer 4

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.