find the value of $\sec^4(\pi/9) + \sec^4( 2\pi/9) + \sec^4 (4\pi/9)$

Question

find the value of $\sec^4(\pi/9) + \sec^4( 2\pi/9) + \sec^4 (4\pi/9)$

I tried converting $\sec^2(\theta)$ to $\tan^2(\theta)$ but for no vain.

Answer 1

Like Evaluate $\tan^{2}(20^{\circ}) + \tan^{2}(40^{\circ}) + \tan^{2}(80^{\circ})$

$$\cos3t=4\cos^3t-3\cos t$$

Observe that $\sec\left(3\cdot\dfrac{\pi}9\right)=2,\sec\left(3\cdot\dfrac{2\pi}9\right)=-2, \sec\left(3\cdot\dfrac{4\pi}9\right)=-2$

Put $3t=\dfrac\pi3$

So, the roots of $4c^3-3c-\dfrac12=0\iff8c^3-6c-1=0$ are $\cos\dfrac{(6n+1)\pi}9$ where $n=0,1,2$

As $\cos(\pi\pm y)=-\cos y$

$\cos\dfrac{(6\cdot1+1)\pi}9=\cdots=-\cos\dfrac{2\pi}9$ and $\cos\dfrac{(6\cdot2+1)\pi}9=\cdots=-\cos\dfrac{4\pi}9$

Set $\dfrac1c=s$ to find the roots of $s^3+6s^2-8=0$ are $$ \sec\dfrac\pi9, -\sec\dfrac{2\pi}9, -\sec\dfrac{4\pi}9$$

Square both sides of $s^3=8-6s^2$ and replace $s^2=t$ to find the roots of $$t^3-36t^2+96t-64=0$$ are $$a=\sec^2\dfrac\pi9,b=\sec^2\dfrac{2\pi}9,c=\sec^2\dfrac{4\pi}9$$

We need $$a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)$$