How do I find the value of$\sin \left( {\dfrac{{11\pi }}{6}} \right)$?

Hint:
Here we need to know that whenever we need to calculate such values we need to notice the quadrant in which angle is lying. So when we calculate we come to know that the angle is in the fourth quadrant and here the value of $\sin $ is negative. Hence we can write $\sin \left( {\dfrac{{11\pi }}{6}} \right)$ in the form of$\sin \left( {2\pi - \theta } \right)$ and then apply $\sin \left( {2\pi - \theta } \right) = - \sin \theta $.

Complete step by step solution:
Here we are given to find the value of $\sin \left( {\dfrac{{11\pi }}{6}} \right)$
We must know that we have four quadrants in which:
1) The $1{\text{st}}$quadrant lies between $0^\circ {\text{ and 90}}^\circ $
2) The ${\text{2nd}}$quadrant lies between $90^\circ {\text{ and 180}}^\circ $
3) The ${\text{3rd}}$quadrant lies between $180^\circ {\text{ and 270}}^\circ $
4) The $4{\text{th}}$quadrant lies between $270^\circ {\text{ and 360}}^\circ $
We also know that $\pi {\text{ radians}} = 180^\circ $
Now if we will calculate the given angle which is given in radian in degrees we will get:
$\pi {\text{ radians}} = 180^\circ $
\[\left( {\dfrac{{11\pi }}{6}} \right){\text{radian}} = \dfrac{{180}}{\pi } \times \dfrac{{11\pi }}{6} = 330^\circ \]
So we come to know that this angle lies in the fourth quadrant; the value of the trigonometric function $\sin $ is negative in the fourth quadrant. Hence our answer will also be negative.
We must know that $\sin {\text{ and csc}}$ are positive only in the first and second quadrant. In others they are negative.
Now we can write \[\left( {\dfrac{{11\pi }}{6}} \right) = 2\pi - \dfrac{\pi }{6}\]
We know that we have the formula:
$\sin \left( {2\pi - \theta } \right) = - \sin \theta $
Now we can compare \[\left( {\dfrac{{11\pi }}{6}} \right) = 2\pi - \dfrac{\pi }{6}\] with $\left( {2\pi - \theta } \right)$ we will get $\theta = \dfrac{\pi }{6}$ and we will get:
$\sin \left( {\dfrac{{11\pi }}{6}} \right) = \sin \left( {2\pi - \dfrac{\pi }{6}} \right)$
Now applying the above formula $\sin \left( {2\pi - \theta } \right) = - \sin \theta $, we will get:

$\sin \left( {\dfrac{{11\pi }}{6}} \right) = \sin \left( {2\pi - \dfrac{\pi }{6}} \right) = - \sin \left( {\dfrac{\pi }{6}} \right) = - \dfrac{1}{2}$

Note:
Here the student must know the trigonometric functions that are positive or negative in each quadrant. For example: If we would have been told to find $\cos \left( {\dfrac{{11\pi }}{6}} \right)$ then we must know that cosine function is positive in the fourth quadrant. Hence its value will come out to be positive. All the properties must be known.