Sequences and series (4)

The above infinite series computes to:

$sin^{2}x + sin^{4}x + sin^{6}x + ... = \frac{sin^{2}x}{1 - sin^{2}x} = \frac{sin^{2}x}{cos^{2}x} = tan^{2}x$

which now gives us the exponential expression $e^{tan^{2}x \cdot ln(2)}.$ The quadratic equation $x^2 - 9x + 8 = 0$ has roots at $x = 1,8$ , which requires $e^{tan^{2}x \cdot ln(2)} = e^{ln(2^{tan^{2}x})} = 2^{tan^{2}x} = 1$ or $8.$ This mandates $tan^{2}x = 0$ or $3$ , or:

$x = arctan(0) = 0$ or $x = arctan(\pm \sqrt{3}) = \pm \frac{\pi}{3}.$

of which $0 < \frac{\pi}{3} < \frac{\pi}{2}$ is satisfied. Finally, the expression $\frac{sin(x) - cos(x)}{sin(x) + cos(x)}$ computes to $\frac{sin(\frac{\pi}{3}) - cos(\frac{\pi}{3})}{sin(\frac{\pi}{3}) + cos(\frac{\pi}{3})}$ ,

or $\frac{\frac{\sqrt{3}}{2} - \frac{1}{2}}{\frac{\sqrt{3}}{2} + \frac{1}{2}} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \cdot \frac{\sqrt{3} - 1}{\sqrt{3} - 1} = \frac{4 - 2\sqrt{3}}{2} = \boxed{2 - \sqrt{3}}.$