Find the value of $\sin({-11\pi}/{3})$?

Question

In the image the rotation starts from point 1 and rotates clockwise until 7 comes at point 3

Now in this question I know how to solve it by using $\sin(2n\pi-\theta)$ but if I try to solve it in this way

$$\sin(\frac{-11\pi}{3})= \sin(\frac{-11\pi}{3}\cdot\frac{180}{\pi})= -\sin(660^\circ)= -\sin(90^\circ\cdot7+30^\circ)= -\cos(30^\circ)= -\frac{\sqrt3}{2}$$

Now as there is negative outside the bracket we should move clockwise in the Cartesian plane, right? If I start from the $+x$ axis then after moving seven times I would arrive at $-x$ axis. As it is $+30^\circ$ the angle should lie in II quadrant and as it is sin the value should be +ve. So the answer should be $-\frac{\sqrt3}{2}$ but the answer is $+\frac{\sqrt3}{2}$. So my doubt is which axis should I start from in the Cartesian plane if the movement is clockwise and also if the movement is anti clockwise or counter clockwise? Which axis should my point 1 be in?Sorry I went through the question again and I saw that I had made a mistake in one of the steps.Thank you all for your time and effort.

Answer 1

Hint:

Since $\sin x$ has period $2\pi$, one has $$\sin\Bigl(-\dfrac{11\pi}3\Bigr)=\sin\Bigl(-\dfrac{11\pi}3+4\pi\Bigr).$$

Answer 2

You incorrectly concluded $\sin(7\pi/2+x)=\cos x$. Shifting by two periods, $\sin(7\pi/2+x)=\sin(x-\pi/2)=-\cos x$.