Use graphs and standard triangles to evaluate $\sin(\frac{11}{6}\pi)$ 
Use graphs and standard triangles to evaluate $\sin(\frac{11}{6}\pi)$.

I end up with $\sin(\pi + \frac{5}{6}\pi)$ which I can't use standard triangles on.
 A: Hint: Try $\sin\left(2\pi-\dfrac\pi6\right)$ instead.
To be more explicit: $\dfrac{11\pi}{6}=2\pi-\dfrac\pi6$ and $\sin(2\pi-x)=-\sin(x)$.
A: Actually I don't know exactly what you mean by "standard triangles", but I feel this can help you (from Wikipedia):

If you still don't understand, follow my explaination:
$$\sin\left(\frac{11}{6}\pi\right) = \sin\left(2\pi-\frac{1}{6}\pi\right) = -\sin\left(\frac{1}{6}\pi\right)$$
Now can't you imagine what value might $-\sin(\frac{1}{6}\pi)$ be? Don't worry, look at the picture above: do you see that the angle $\frac{\pi}{6} = 30^\circ$? Now draw a triangle right there, with the radius of the circle as one side, the sinus of $\frac{\pi}{6}$ as the other side, and you've already got a triangle since the remaining side is a part of the diameter (it's very simple: can you see it?) Now, the triangle you've just drawn is indeed half an equilateral triangle! So, if the radius of the circle is one, the side corresponding to the sinus of $\frac{\pi}{6}$ is exactly $\frac{1}{2}$ :)
Therefore $$-\sin\left(\frac{1}{6}\pi\right) = -\frac{1}{2}$$ which is your final result.
