A partical moves in the x-y plane according to the scheme x = -8 sinπ

1.	A partical moves in the x-y plane according to the scheme x = -8 sinπt and y = -2 cos²πt, where t is time. Find the equation of path of the particle.DONT SPAM IF YOU SPAM I WILL REPORT YOUR 25 ANSWERS
Answer» $\blue{\Large{\boxed{\tt{ \red{Solution:- }}}}}$ $\dashrightarrow$ $\tt{y = -<klux>2</klux>\cos^{2}\pi t}$ $\dashrightarrow$ $\tt{y = -2(\cos^{2}\pi t-\sin^{2}\pi t)}$ $\dashrightarrow$ $\tt{y = 1-2\sin^{2}\pi t ---(1)}$ Now, $\dashrightarrow$ $\tt{x = -8\sin\pi t}$ $\dashrightarrow$ $\tt{\sin\pi t = \frac{-x}{8}---(2)}$ Putting the value of $\sf{eq^{n}}$ (2)in $\sf{eq^{n}}$ (1) we'll GET, $\green\rightarrowtail$ $\sf{y = - 2 \{1 - 2( \frac{ - x}{8})^{2} \}}$ $\green\rightarrowtail$ $\sf{y = - 2(1 - \frac{2x ^{2} }{64})}$ $\green\rightarrowtail$ $\sf{y = - 2 + \frac{4x^{2} }{64}}$ $\green\rightarrowtail$ $\sf{y = \frac{x^2}{16} -2}$ $\green\rightarrowtail$ $\sf{\frac{x^2}{16}-2 = 0}$ $\green\rightarrowtail$ $\sf{x^2 = 2×16}$ $\green\rightarrowtail$ $\sf{x =\sqrt{32}}$ $\green\rightarrowtail$ $\sf\boxed{x =\pm4\sqrt{2}}$ Identities used:- $\purple\mapsto$ $\rm{\cos2\theta= \cos^2- \sin^2\theta}$ $\purple\mapsto$ $\rm{\sin^2\theta+\cos^2\theta= 1}$ NOTE:- for the FIGURE you may REFER to the attachment. Hope it helps :)