Animated Solution for Mathematics - Complex Numbers: If $\omega$ is an imaginary cube root of unity then the value of $\sin \{({\omega }^{10}+{\omega }^{23})\pi -\frac{\pi }{4}\}$ is

The cube roots of unity—$1$, $\omega$, and ${\omega }^2$—reside on the unit circle, forming an equilateral triangle. The core of this problem relies on the cyclic nature of $\omega$, defined by the identity:

Because of this property, any power of $\omega$ functions as a loop. We can simplify higher powers by dividing the exponent by $3$ and observing the remainder.

For ${\omega }^{10}$, we divide $10$ by $3$: Since the remainder is $1$, we have ${\omega }^{10}={\omega }^1=\omega$.

For ${\omega }^{23}$, we divide $23$ by $3$: Since the remainder is $2$, we have ${\omega }^{23}={\omega }^2$.

Now, consider the sum ${\omega }^{10}+{\omega }^{23}=\omega +{\omega }^2$. We utilize the fundamental identity for the roots of unity:

This implies that $\omega +{\omega }^2=-1$. The complex expression has successfully collapsed into a simple real number.

We substitute our result into the original trigonometric expression:

To simplify, we factor out the negative sign:

Since sine is an odd function, we apply the property $\sin (-\theta )=-\sin (\theta )$:

The angle $\pi +\frac{\pi }{4}$ lies in the third quadrant, where the sine function is negative. Using the reduction formula $\sin (\pi +\theta )=-\sin (\theta )$, we obtain:

Evaluating this standard trigonometric value, we arrive at the final result:

The final answer is $\frac{1}{\sqrt{2}}$.