The value of $\sin {18^\circ }$ is:

Hint: In order to solve this question, to know the value of given trigonometric expression, first we will assume the given degree as a variable and then we will work on to solve for the value of $\sin {18^\circ }$.

Complete step-by-step solution:
The given expression is $\sin {18^\circ }$
Let $A = {18^\circ }$
Therefore, $5A = {90^\circ }$
We can also write the above equation as-
 $ \Rightarrow 2A + 3A = {90^\circ } $
  or, $ 2A = {90^\circ } - 3A $
Now, we will take sine on both sides, we get:
$\sin 2A = \sin ({90^\circ } - 3A) = \cos 3A $
$\Rightarrow 2\sin A\cos A = 4{\cos ^3}A - 3\cos A $
$\Rightarrow 2\sin A\cos A - 4{\cos ^3}A + 3\cos A = 0 $
Now, by dividing both sides by $\cos A$ and $\cos A = \cos {18^\circ } \ne 0$ we get:
$ \Rightarrow 2\sin A - 4(1 - {\sin ^2}A) + 3 = 0 $
$ \Rightarrow 4{\sin ^2}A + 2\sin A - 1 = 0 $
which is a quadratic in $\sin A$
If we compare the above equation with a quadratic equation $a{x^2}+bx+c=0$ we get the coefficients as $a=4, b=2 \text{ and }c=-1$.
Therefore,
We use the sridharacharya formula to solve the quadratic equation such as:
$ x = \dfrac{{ - 2 \pm \sqrt {b^2 - (4)a.(c)} }}{{2 \times a}} $
Here on putting the values of a,b and c in the above formula.
$ \sin A = \dfrac{{ - 2 \pm \sqrt {4 - (4)4.( - 1)} }}{{2 \times 4}} $
$ = \dfrac{{ - 2 \pm \sqrt {4 + 16} }}{8} $
$= \dfrac{{ - 2 \pm 2\sqrt 5 }}{8} $
$\therefore \sin A = \dfrac{{ - 1 \pm \sqrt 5 }}{4}$
Now, $\sin {18^\circ }$ is positive, as ${18^\circ }$ lies in the first quadrant.
Therefore, $\sin A = \dfrac{{ - 1 + \sqrt 5 }}{4}$.

Note: Only a right-angle triangle can be used with trigonometric ratios. A right-angle triangle is one in which one of the angles is 90 degrees and the other two are less than 90 degrees. Each side of the right angle triangle is also given a name.