How to evaluate $\sin (\dfrac{{7\pi }}{8})$ using half angle formula.

Hint:In order to solve this question we need the half angle then with the help of the trigonometric identities we will have to transform it up till we get any exact value of it.

Complete step-by-step solution
For solving this we will have to convert it in the form of $\pi $ so that we will apply some identity there:
So,
$\sin \left( {\dfrac{{7\pi }}{6}} \right) = \sin \left( {\pi - \dfrac{\pi }{8}} \right)$
Now we know that the $\sin (\pi - \theta ) = \sin \theta $ so,
$\sin \left( {\pi - \dfrac{\pi }{8}} \right) = \sin \left( {\dfrac{\pi }{8}} \right)$
Now we will apply a trigonometric identity here i.e.,
$\cos 2\theta = 1 - 2{\sin ^2}\theta $
Putting $\theta = \dfrac{\pi }{8}$
$\cos \left( {\dfrac{\pi }{4}} \right) = 1 - 2{\sin ^2}\left( {\dfrac{\pi }{8}} \right)$
As we all know that the value of $\cos \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }} = \dfrac{{\sqrt 2 }}{2}$
Putting this value in this equation;
$\dfrac{{\sqrt 2 }}{2} = 1 - 2{\sin ^2}\left( {\dfrac{\pi }{8}} \right)$
On further solving:
${\sin ^2}\left( {\dfrac{\pi }{8}} \right) = \dfrac{{2 - \sqrt 2 }}{4}$
Now taking square root on both side we will get;
$\sin \left( {\dfrac{\pi }{8}} \right) = \dfrac{{\sqrt {2 - \sqrt 2 } }}{2}$
Since we have already proven that the value of $\sin \left( {\dfrac{{7\pi }}{8}} \right) = \sin \left( {\dfrac{\pi }{8}} \right)$

So this is the value of $\sin \left( {\dfrac{{7\pi }}{8}} \right)$.

Note: Since in these types of questions we are not getting the exact value so these are not memorable and hence we need a particular logic of every question to have a solution to the improper angled questions. One special thing is that only which are findable by easy calculations are asked in exams, others are given in if those are required in it.