Question 1170577: By solving  another way, show that the exact value of  is 
I know that to solve this, the equation needs to be divided by  specifically, but I don't know how to find the exact value.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let's solve the equation $x^4 + 4x^3 - 6x^2 - 4x + 1 = 0$ in a way that relates to trigonometric functions.
Divide the equation by $x^2$:
$$x^2 + 4x - 6 - \frac{4}{x} + \frac{1}{x^2} = 0$$
Rearrange the terms:
$$\left(x^2 + \frac{1}{x^2}\right) + 4\left(x - \frac{1}{x}\right) - 6 = 0$$
Let $y = x - \frac{1}{x}$. Then $y^2 = x^2 - 2 + \frac{1}{x^2}$, so $x^2 + \frac{1}{x^2} = y^2 + 2$.
Substitute into the equation:
$$(y^2 + 2) + 4y - 6 = 0$$
$$y^2 + 4y - 4 = 0$$
Solve for $y$ using the quadratic formula:
$$y = \frac{-4 \pm \sqrt{16 - 4(1)(-4)}}{2} = \frac{-4 \pm \sqrt{32}}{2} = \frac{-4 \pm 4\sqrt{2}}{2} = -2 \pm 2\sqrt{2}$$
So, $y = x - \frac{1}{x} = -2 \pm 2\sqrt{2}$.
Now, let $x = \tan(\theta)$. Then
$$\tan(\theta) - \frac{1}{\tan(\theta)} = \tan(\theta) - \cot(\theta) = -2 \pm 2\sqrt{2}$$
We are given that $\tan(\frac{\pi}{16}) - \cot(\frac{\pi}{16}) = -2 - 2\sqrt{2}$.
Let's show why it must be the negative root.
We know that $0 < \frac{\pi}{16} < \frac{\pi}{4}$. Thus, $0 < \tan(\frac{\pi}{16}) < 1$ and $\cot(\frac{\pi}{16}) > 1$.
Therefore, $\tan(\frac{\pi}{16}) - \cot(\frac{\pi}{16}) < 0$.
Also, since $\tan(\frac{\pi}{16})$ is small and $\cot(\frac{\pi}{16})$ is large, $\tan(\frac{\pi}{16}) - \cot(\frac{\pi}{16})$ will be a negative number with a large magnitude.
Since $-2 + 2\sqrt{2} \approx 0.828$ and $-2 - 2\sqrt{2} \approx -4.828$, the negative root is the correct one.
Therefore, $\tan(\frac{\pi}{16}) - \cot(\frac{\pi}{16}) = -2 - 2\sqrt{2}$.
**Conclusion:**
By solving the given equation by dividing by $x^2$ and substituting $y = x - \frac{1}{x}$, we obtained $y = -2 \pm 2\sqrt{2}$.
Letting $x = \tan(\theta)$, we get $\tan(\theta) - \cot(\theta) = -2 \pm 2\sqrt{2}$.
Since $\tan(\frac{\pi}{16}) - \cot(\frac{\pi}{16})$ is negative, we conclude that $\tan(\frac{\pi}{16}) - \cot(\frac{\pi}{16}) = -2 - 2\sqrt{2}$.
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