If $\omega $ is an imaginary cube root of unity, then the value of $\sin \left[ {\left( {{\omega ^{10}} + {\omega ^{23}}} \right)\pi - \dfrac{\pi }{4}} \right]$ is
$\left( a \right)\dfrac{{ - \sqrt 3 }}{2}$
\[\left( b \right)\dfrac{{ - 1}}{{\sqrt 2 }}\]
$\left( c \right)\dfrac{1}{{\sqrt 2 }}$
$\left( d \right)\dfrac{{\sqrt 3 }}{2}$

Hint: In this particular type of question use the concept that if $\omega $ is an imaginary cube root of unity then (\[{\omega ^3} = 1\]) and (${\omega ^2} + \omega + 1 = 0$) and use the concept that sin (-x) = -sin x and sine is negative in the third quadrant. So use these concepts to reach the solution of the question.

Complete step-by-step answer:
As it is given that $\omega $ is an imaginary cube root of unity, so it satisfies the following condition which is given as,
\[{\omega ^3} = 1\]......................... (1)
And
${\omega ^2} + \omega + 1 = 0$.................... (2)
Now given equation is
$\sin \left[ {\left( {{\omega ^{10}} + {\omega ^{23}}} \right)\pi - \dfrac{\pi }{4}} \right]$
Now first simplify the above equation we have,
$ \Rightarrow \sin \left[ {\left( {{{\left( {{\omega ^3}} \right)}^3}\omega + {{\left( {{\omega ^3}} \right)}^7}{\omega ^2}} \right)\pi - \dfrac{\pi }{4}} \right]$
Now substitute the value of \[{\omega ^3}\] from equation (1) in the above equation we have,
$ \Rightarrow \sin \left[ {\left( {{{\left( 1 \right)}^3}\omega + {{\left( 1 \right)}^7}{\omega ^2}} \right)\pi - \dfrac{\pi }{4}} \right]$
Now as we know any power of 1 is always remain 1 so use this property to simplify the above equation we have,
\[ \Rightarrow \sin \left[ {\left( {\omega + {\omega ^2}} \right)\pi - \dfrac{\pi }{4}} \right]\].......................... (3)
Now from equation (2) we have,
$ \Rightarrow {\omega ^2} + \omega + 1 = 0$
$ \Rightarrow {\omega ^2} + \omega = - 1$
So substitute this value in equation (3) we have,
\[ \Rightarrow \sin \left[ {\left( { - 1} \right)\pi - \dfrac{\pi }{4}} \right]\]
Now the above equation is also written as,
\[ \Rightarrow \sin \left[ { - \left( {\pi + \dfrac{\pi }{4}} \right)} \right]\]
Now as we know that sin (-x) = -sin x so use this property in the above equation we have,
\[ \Rightarrow \sin \left[ { - \left( {\pi + \dfrac{\pi }{4}} \right)} \right] = - \sin \left[ {\pi + \dfrac{\pi }{4}} \right]\]
Now as we know that sin in 3^rd quadrant is negative, so ($\sin \left( {\pi + \theta } \right) = - \sin \theta $), so use this property in the above equation we have,
\[ \Rightarrow \sin \left[ { - \left( {\pi + \dfrac{\pi }{4}} \right)} \right] = - \sin \left[ {\pi + \dfrac{\pi }{4}} \right] = - \left( { - \sin \dfrac{\pi }{4}} \right) = \sin \dfrac{\pi }{4}\]
Now the given equation is convert into simple standard value of sine i.e. \[\sin \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}\]
Therefore,
$\sin \left[ {\left( {{\omega ^{10}} + {\omega ^{23}}} \right)\pi - \dfrac{\pi }{4}} \right] = \dfrac{1}{{\sqrt 2 }}$
So this is the required answer.
Hence option (C) is the correct answer.

Note – Whenever we face such types of questions the key concept we have to remember is that always recall all the properties if $\omega $ is an imaginary cube root of unity which is all stated above, then first simplify the above equation using these properties as above, then apply the basic sine properties as above and simplify we will get the required answer.