Finding the shortest distance from the origin to this curve

Question

Find the shortest distance between the origin point and the curve \begin{align*} x&=2\sin t - \sin 2t\\ y&=2\cos t - \cos 2t \end{align*}

I don't even know how to draw this curve, help please.

Answer 1

Minimising the distance to a ppoint of the curve is also minimising the square of this distance: \begin{align} d^2(t)&=(2\sin t-\sin 2t)^2+(2\cos t-\cos 2t)^2=4+1-4(\sin t\sin 2t+\cos t\cos 2t)\\ &=5-4\cos(2t- t)=5-4\cos t. \end{align} So the minimum is $d=1$ (and the maximum is $d=9$).

Answer 2

HINT: the distance is given by $$\sqrt{(2\sin(t)-\sin(2t))^2+(2\cos(t)-\cos(2t))^2}$$ simplifying and differentiating with respect to $t$ gives $$\sqrt{5-4\cos(t)}$$ and the derivative is given by $$\frac{1}{2}(5-4\cos(t))^{-1/2}8\sin(t)$$