Question

If x= 2 cos t - cos 2t and y= 2 Sint - sin 2t then prove that dy/DX = tan (3t/2)

Answer 1

$\textbf{Given:}$

$x=2\,cos\,t-cos 2t\text{ and }y= 2\, Sin\,t-sin\,2t$

$\text{Then,}$

$\displaystyle\frac{dx}{dt}=-2\,sin\,t+2sin\,2t$

$\displaystyle\frac{dx}{dt}=2[sin\,2t-sin\,t]$

$\text{and}$

$\displaystyle\frac{dy}{dt}=2\,cos\,t-2cos\,2t$

$\displaystyle\frac{dy}{dt}=2[cos\,t-cos\,2t]$

$Now,\;\;\displaystyle\frac{dy}{dx}$

$=\displaystyle\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

$=\displaystyle\frac{2[cos\,t-cos\,2t]}{2[sin\,2t-sin\,t]}$

$=\displaystyle\frac{cos\,t-cos\,2t}{sin\,2t-sin\,t}$

$\text{Using}$

$\boxed{\begin{minipage}{7cm} \bf\,cosC-cosD=-2\,sin(\frac{C+D}{2})\,sin(\frac{C-D}{2})\\\\sinC-sinD=2\,cos(\frac{C+D}{2})\,sin(\frac{C-D}{2}) \end{minipage}}$

$=\displaystyle\frac{-2\,sin(\frac{t+2t}{2})\,sin(\frac{t-2t}{2})}{2\,cos(\frac{2t+t}{2})\,sin(\frac{2t-t}{2})}$

$=\displaystyle\frac{-\,sin\,\frac{3t}{2}\,sin\,\frac{-t}{2}}{cos\,\frac{3t}{2}\,sin\,\frac{t}{2}}$

$=\displaystyle\frac{sin\,\frac{3t}{2}\,sin\,\frac{t}{2}}{cos\,\frac{3t}{2}\,sin\,\frac{t}{2}}$

$=\displaystyle\frac{sin\,\frac{3t}{2}}{cos\,\frac{3t}{2}}$

$=\displaystyle\,tan\,\frac{3t}{2}$

$\implies\boxed{\bf\frac{dy}{dx}=tan\,\frac{3t}{2}}$