Question

If a curve is represented parametrically by the equations $x =\sin \left( t +\frac{7 \pi}{12}\right)+\sin \left( t -\frac{\pi}{12}\right)+\sin \left( t +\frac{3 \pi}{12}\right), y =\cos \left( t +\frac{7 \pi}{12}\right)+\cos \left( t -\frac{\pi}{12}\right)+\cos \left( t +\frac{3 \pi}{12}\right)$ then find the value of $\frac{d}{d t}\left(\frac{x}{y}-\frac{y}{x}\right)$ at $t=\frac{\pi}{8}$.

Answer 1

We have
$x =\sin \left( t +\frac{7 \pi}{12}\right)+\sin \left( t -\frac{\pi}{12}\right)+\sin \left( t +\frac{3 \pi}{12}\right)=2 \sin \left( t +\frac{\pi}{4}\right) \cos \left(\frac{\pi}{3}\right)+\sin \left( t +\frac{\pi}{4}\right)=2 \sin \left( t +\frac{\pi}{4}\right)$
|||ly
$y=\cos \left(t+\frac{7 \pi}{12}\right)+\cos \left(t-\frac{\pi}{12}\right)+\cos \left(t+\frac{3 \pi}{12}\right)=2 \cos \left(t+\frac{\pi}{4}\right) \cos \left(\frac{\pi}{3}\right)+\cos \left(t+\frac{\pi}{4}\right)=2 \cos \left(t+\frac{\pi}{4}\right)$
Now, $\frac{x}{y}=\tan \left(t+\frac{\pi}{4}\right)=\frac{1+\tan t}{1-\tan t}$ and $\frac{y}{x}=\frac{1}{\tan \left(t+\frac{\pi}{4}\right)}=\frac{1-\tan t}{1+\tan t}$
$\therefore \left(\frac{x}{y}-\frac{y}{x}\right)=\left(\frac{1+\tan t}{1-\tan t}\right)-\left(\frac{1-\tan t}{1+\tan t}\right)=\frac{(1+\tan t)^2-(1-\tan t)^2}{1-\tan ^2 t}=\frac{4 \tan t}{1-\tan ^2 t}=2 \tan 2 t$
Hence $\left.\frac{ d }{ dt }\left(\frac{ x }{ y }-\frac{ y }{ x }\right)=\frac{ d }{ dt }(2 \tan 2 t )=4 \sec ^2 2 t \right]_{ t =\frac{\pi}{8}}=4 \sec ^2 \frac{\pi}{4}=8$