2

Question

From the equation who's roots are $\cos\frac{\pi}9,\cos\frac{3\pi}9,\cos\frac{5\pi}9,\cos\frac{7\pi}9$ and hence prove
a) $8\cos\frac{\pi}9\cos\frac{5\pi}9\cos\frac{7\pi}9=1=8\cos\frac{\pi}9\cos\frac{2\pi}9\cos\frac{4\pi}9$
b) $\sec^4\frac{\pi}9+\sec^4\frac{2\pi}9+\sec^4\frac{4\pi}9=1104$

My Attempt:
Say $y=e^{i\pi(2k+1)/9}$

Thus $y^9+1=0$ has solutions $e^{i\pi(2k+1)/9}$ for $k\in\{0,1,\dots,8\}$
Let $y+1/y=2x\implies y=x\pm\sqrt{x^2-1}$

Thus $(x\pm\sqrt{x^2-1})^9+1=0$ has solutions $\cos\frac{(2k+1)\pi}{9}$ for $k\in\{0,1\dots8\}$

Thus $(x\pm\sqrt{x^2-1})^9+1=\prod_{k=0}^8(x-\cos\frac{(2k+1)\pi}{9})$

But if we put $x=0$, LHS is complex while RHS is Real

What am i doing wrong?

Anvit
  • 3,449

1 Answers1

0

$$\implies x^4-3x^2+1=x(x^2-2)$$

$$\implies(x^4-3x^2+1)^2=x^2(x^2-2)^2$$

Set $\sec^2\dfrac{(2k+1)\pi}9=\dfrac4{x^2}=v$(say)

On replacement & simplification

$$ v^4-v^3(24+16)+v^2(144+32+64)+v(\cdots)+1=0$$

whose roots are $\sec^2\dfrac{(2k+1)\pi}9,k=0,1,2,3$

$$\sum_{k=1}^4\left(\sec^2\dfrac{(2k+1)\pi}9\right)^2$$

$$=\left(\sum_{k=1}^4\sec^2\dfrac{(2k+1)\pi}9\right)^2-2\sum_{k_1,k_2=0,1,2,3, k_1>k_2}\sec^2\dfrac{(2k_1+1)\pi}9\cdot\sec^2\dfrac{(2k_2+1)\pi}9$$

$$=40^2-2(144+32+64)=?$$

Now for $k=1,\sec^2\dfrac{(2k+1)\pi}9=?$

Finally $\sec(\pi-t)=-\sec t,\sec^2(\pi-t)=\sec^2t$

Set $t=\dfrac{2\pi}9,\dfrac{4\pi}9$

lab bhattacharjee
  • 279,016
  • on that note, can you tell why exactly in my method LHS is complex but rhs is purely real? – Anvit Jun 06 '18 at 09:21
  • @AFalseName,$$(x\pm\sqrt{x^2-1})^9+1=\prod_{k=0}^8(x-\cos\frac{(2k+1)\pi}{9})$$ is not an identity. Observe that $x=0\implies y^2=-1$ which does not satisfy $$y^9+1=0$$ – lab bhattacharjee Jun 06 '18 at 09:33