# What is the equation of the tangent line to the polar curve  f(theta)=2thetasin(theta)+thetacot^2(2theta)  at theta = pi/8?

May 27, 2017

$y \approx - 0.650 x + 0.949$

#### Explanation:

$f \left(\frac{\pi}{8}\right) = 2 \left(\frac{\pi}{8}\right) \sin \left(\frac{\pi}{8}\right) + \frac{\pi}{8} {\cot}^{2} \left(2 \left(\frac{\pi}{8}\right)\right) = \frac{\pi}{4} \sin \left(\frac{\pi}{8}\right) + \frac{\pi}{8} {\cot}^{2} \left(\frac{\pi}{4}\right)$

By the half angle formulae,
$\sin \left(\frac{x}{2}\right) = \pm \sqrt{\frac{1 - \cos \left(x\right)}{2}}$
$\sin \left(\frac{\pi}{8}\right) = \pm \sqrt{\frac{1 - \cos \left(\frac{\pi}{4}\right)}{2}} = \pm \sqrt{\frac{1 - \frac{\sqrt{2}}{2}}{2}} = \pm \frac{\sqrt{2 - \sqrt{2}}}{2}$
Using the CAST rule, we can deduce that $\sin \left(\frac{\pi}{8}\right) = + \frac{\sqrt{2 - \sqrt{2}}}{2}$

Hence $f \left(\frac{\pi}{8}\right) = \frac{\pi}{4} \frac{\sqrt{2 - \sqrt{2}}}{2} + \frac{\pi}{8} {\left(1\right)}^{2} = \frac{\pi}{8} \left(\sqrt{2 - \sqrt{2}} + 1\right)$

$f \left(\theta\right) = 2 \theta \sin \left(\theta\right) + \theta {\cot}^{2} \left(2 \theta\right)$
$f ' \left(\theta\right) = 2 \sin \theta + 2 \theta \cos \theta + {\cot}^{2} \left(2 \theta\right) + \theta \left(2 \cot \left(2 \theta\right) \cdot \left(- {\csc}^{2} \left(2 \theta\right)\right) \cdot 2\right)$

$\therefore f ' \left(\frac{\pi}{8}\right) = 2 \sin \left(\frac{\pi}{8}\right) + \frac{\pi}{4} \cos \left(\frac{\pi}{8}\right) + {\cot}^{2} \left(\frac{\pi}{4}\right) + \frac{\pi}{2} \cot \left(\frac{\pi}{4}\right) \cdot \left(- {\csc}^{2} \left(\frac{\pi}{4}\right)\right) = 2 \sin \left(\frac{\pi}{8}\right) + \frac{\pi}{4} \cos \left(\frac{\pi}{8}\right) + 1 - \frac{\pi}{2} {\csc}^{2} \left(\frac{\pi}{4}\right)$

By the half angle formulae,
$\cos \left(\frac{x}{2}\right) = \pm \sqrt{\frac{\cos \left(x\right) + 1}{2}}$
$\cos \left(\frac{\pi}{8}\right) = \pm \sqrt{\frac{\cos \left(\frac{\pi}{4}\right) + 1}{2}} = \pm \sqrt{\frac{\frac{\sqrt{2}}{2} + 1}{2}} = \pm \frac{\sqrt{2 + \sqrt{2}}}{2}$
Using the CAST rule, we can deduce that $\cos \left(\frac{\pi}{8}\right) = + \frac{\sqrt{2 + \sqrt{2}}}{2}$

$\because \csc \left(x\right) = \frac{1}{\sin} \left(x\right)$
therefore csc(pi/8)=1/(sqrt(2-sqrt2)/2)=2/(sqrt(2-sqrt2)
$\therefore {\csc}^{2} \left(\frac{\pi}{8}\right) = {\left(\frac{2}{\sqrt{2 - \sqrt{2}}}\right)}^{2} = \frac{4}{2 - \sqrt{2}}$

$\therefore f ' \left(\frac{\pi}{8}\right) = 2 \left(\frac{\sqrt{2 - \sqrt{2}}}{2}\right) + \frac{\pi}{4} \frac{\sqrt{2 + \sqrt{2}}}{2} + 1 - \frac{\pi}{2} \left(\frac{4}{2 - \sqrt{2}}\right) = \sqrt{2 - \sqrt{2}} + \frac{\pi \sqrt{2 + \sqrt{2}}}{8} + 1 - \frac{2 \pi}{2 - \sqrt{2}}$

$\therefore \frac{y - {y}_{1}}{x - {x}_{1}} = y '$
$\frac{y - f \left(\frac{\pi}{8}\right)}{x - \frac{\pi}{8}} = f ' \left(\frac{\pi}{8}\right)$
$y = f ' \left(\frac{\pi}{8}\right) x - \frac{\pi}{8} f ' \left(\frac{\pi}{8}\right) + f \left(\frac{\pi}{8}\right)$
$\therefore y = - 0.65013 x + 0.94875 \approx - 0.650 x + 0.949$