Question

If a curve is represented parametrically by the equations.
$x=\sin (t+\frac{7\pi }{12})+$$\sin (t-\frac{\pi }{12})+\sin (t+\frac{3\pi }{12}),$

$y=\cos (t+\frac{7\pi }{12})+\cos (t-\frac{\pi }{12})+\cos (t+\frac{3\pi }{12})$ then find the value of $\frac{d}{dt}(\frac{x}{y}-\frac{y}{x})$ at $t=\frac{\pi }{8}$

Answer 1

$x=\sin \left(t+\frac{7\pi }{12}\right)+\sin \left(t-\frac{\pi }{12}\right)+\sin (t+\frac{3\pi }{12})$

$x=2\sin \left(\frac{t+\frac{7\pi }{12}+t-\frac{\pi }{12}}{2}\right)\cos \left(\frac{\frac{7\pi }{12}+\frac{\pi }{12}}{2}\right)+\sin (t+\frac{\pi }{4})$

$=2\sin (t+\frac{\pi }{4})\cos \frac{\pi }{3}+\sin (t+\frac{\pi }{4})$

$=\sin (t+\frac{\pi }{4})+\sin (t+\frac{\pi }{4})$

$=2\sin (t+\frac{\pi }{4})$

$y=2\cos \left(\frac{t+\frac{7\pi }{12}+t-\frac{\pi }{12}}{2}\right)\cos \left(\frac{8\pi }{12\ast 2}\right)+\cos (t+\frac{\pi }{4})$

$=2\cos (t+\frac{\pi }{4})\cos \frac{\pi }{3}+\cos (t+\frac{\pi }{4})$

$=2\cos (t+\frac{\pi }{4})$

$\frac{d}{dt}(\frac{x}{y}-\frac{y}{x})=\frac{d}{dt}(\tan (t+\frac{\pi }{4})-\cot (t+\frac{\pi }{4}))$
$={\sec }^2(t+\frac{\pi }{4})+{\csc }^2(t+\frac{\pi }{4})$
At $t=\frac{\pi }{8}$
${\sec }^2(\frac{\pi }{8}+\frac{\pi }{4})+{\csc }^2(\frac{\pi }{8}+\frac{\pi }{4})$
$={\sec }^2\frac{3\pi }{8}+{\csc }^2\frac{3\pi }{8}$
$=\frac{1}{{\cos }^2\frac{3\pi }{8}}+\frac{1}{{\sin }^2\frac{3\pi }{8}}$
$=\frac{2}{1+\cos \frac{3\pi }{4}}+\frac{2}{1-\cos \frac{3\pi }{4}}$
$=\frac{2}{1-\frac{1}{\sqrt{2}}}+\frac{2}{1+\frac{1}{\sqrt{2}}}$
$=\frac{2\sqrt{2}}{\sqrt{2}-1}+\frac{2\sqrt{2}}{\sqrt{2}+1}=2\sqrt{2}(\sqrt{2}+1)+2\sqrt{2}(\sqrt{2}-1)$
$=2\times 2+2\sqrt{2}+2\times 2-2\sqrt{2}$
$=8$