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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..

Michael Rozenberg
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pi-π
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4 Answers4

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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
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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.

dxiv
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  • @ dxiv, Could you please explain your solution to me? – pi-π Feb 10 '17 at 05:47
  • @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. – dxiv Feb 10 '17 at 05:50
  • @ dxiv, I am satisfied with your answer above this ( Then the equality to prove becomes: ). But from this point downward, I could not understand. – pi-π Feb 10 '17 at 05:53
  • @ dxiv, If possible, Could you please explain me here at :: http://chat.stackexchange.com/rooms/53289/ keep-typing – pi-π Feb 10 '17 at 05:55
  • @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). – dxiv Feb 10 '17 at 06:01
  • @ dxiv, I could not understand, this :: Euclidian division? – pi-π Feb 10 '17 at 06:03
  • @Euler That's just another name for polynomial 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. – dxiv Feb 10 '17 at 06:05
  • @ dxiv, what is this? How it happened?? " Therefore the equality holds, which completes the proof."" – pi-π Feb 10 '17 at 06:09
  • 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. – dxiv Feb 10 '17 at 06:12
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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
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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.

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Will Jagy
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