Question

If $x$ and $y$ are connected parametrically by the equation,without eliminating the parameter,find $\frac{dy}{dx}$.
$x=cos\,θ-cos2θ,y=sin\,θ-sin2θ$

Answer 1

The correct answer is $\frac{cos\,θ-2cos\,2θ}{2sin\,2θ-sin\,θ}$
The given equations are $x=cos\,θ-cos2θ,y=sin\,θ-sin2θ$
Then,$\frac{dx}{dθ}=\frac{d}{dθ}(cos\,θ-cos\,2θ)=\frac{d}{dθ}(cos\,θ)-\frac{d}{dθ}(cos\,2θ)$
$=-sinθ-(-2sin\,2θ)=2sin\,2θ-sin\,θ$
$\frac{dy}{dθ}=\frac{d}{dθ}(sinθ-sin\,2θ)=\frac{d}{dθ}(sin\,θ)-\frac{d}{dθ}(sin\,2θ)$
$=cos\,θ-2cos\,2θ$
$∴\frac{dy}{dx}=\frac{(\frac{dy}{dθ})}{(\frac{dx}{dθ})}=\frac{cos\,θ-2cos\,2θ}{2sin\,2θ-sin\,θ}$