How do I find the value of $\sin (\dfrac{{5\pi }}{6})$?

Hint:
This sum can be solved using two methods. It is upto the students which one he/she prefers. First method is using the formula$\sin (180 - \theta ) = \sin \theta $. This formula is derived using the graphical properties of the sin function. Second method is to use the formula $\sin (A + B) = \sin A\cos B + \cos A\sin B$. Both methods would give the same answer, only difference is that the second method is a bit longer compared to the first one.

Complete step by step solution:
Since there are two methods, let's start with the shorter method.
We know that the value of $\pi is{180^ \circ }$, therefore $\dfrac{{5\pi }}{6} = \pi - \dfrac{\pi }{6}$.
We know that $\sin (180 - \theta ) = \sin \theta $, from above we can rearrange the value of $\dfrac{{5\pi }}{6}$ in the given question as $\sin (\dfrac{{5\pi }}{6}) = \sin (\pi - \dfrac{\pi }{6})$
Using the property of sin function, we can say that the above equation is equivalent to $\sin (\dfrac{\pi }{6})$
We know the value of $\sin ({30^ \circ })$.

Therefore the answer is $0.5$.

Second Method:
We can say that $\sin (\dfrac{{5\pi }}{6}) = \sin (150)$ since the value of $\pi is{180^ \circ }$.
Using the formula $\sin (A + B) = \sin A\cos B + \cos A\sin B$, we can input $A = {90^ \circ },B = {60^ \circ }$ . Substituting the value of A & B in the formula
$\sin (150) = \sin {90^ \circ }\cos {60^ \circ } + \cos {90^ \circ }\sin {60^ \circ }......(1)$
Since the value of $\cos {90^ \circ }$is $0$,we can say that
$\sin (150) = \sin {90^ \circ }\cos {60^ \circ }...........(2)$
Also we know that $\sin {90^ \circ } = 1\& \cos ({60^ \circ }) = \dfrac{1}{2}$,

Therefore value of $\sin ({150^ \circ })$is $1 \times \dfrac{1}{2} = 0.5$

Note:
Though there are two methods for this type of sums, it is always preferable to use the first method since it is shorter and much easier to comprehend. Second method is preferable when the values of A & B are given and the student has to find the answer to $\sin (A + B)$. Also the student should memorize the graphical properties of trigonometric functions for angles greater than ${90^ \circ },{180^ \circ },{270^ \circ }$ as there may be a change in the behavior, for example $\cos ({90^ \circ } + \theta ) = - \cos (\theta )$and so on for other trigonometric functions.