Question

If $\omega$ is a complex cube root of unity, then $\sin \left(({\omega }^{10}+{\omega }^{23})\pi -\frac{\pi }{4}\right)$ is equal to

• A. $\frac{1}{\sqrt{2}}$
• B. $\frac{1}{2}$
• C. $1$
• D. $\frac{\sqrt{3}}{2}$

Answer 1

The correct option is A

$\frac{1}{\sqrt{2}}$

Explanation for the correct option

Step 1: Information required for the solution

Since, $\omega$ is a complex cube root of unity then we know that

$1+\omega +{\omega }^2=0\text{and}{\omega }^3=1$

Step 2: Simplification and calculation of the given expression

We can write the expression as,

$\begin{array}{rcl}\therefore \sin \left(\left({\omega }^{10}+{\omega }^{23}\right)\pi -\frac{\pi }{4}\right)& =& \sin \left(\left({\left({\omega }^3\right)}^3·\omega +{\left({\omega }^3\right)}^7·{\omega }^2\right)\pi -\frac{\pi }{4}\right)\\ & =& \sin \left(\left({\left(1\right)}^3·\omega +{\left(1\right)}^7·{\omega }^2\right)\pi -\frac{\pi }{4}\right)\\ & =& \sin \left(\left(\omega +{\omega }^2\right)\pi -\frac{\pi }{4}\right)\end{array}$

As $1+\omega +{\omega }^2=0$ then $\omega +{\omega }^2=-1$, we have

$\begin{array}{rcl}\therefore \sin \left(\left({\omega }^{10}+{\omega }^{23}\right)\pi -\frac{\pi }{4}\right)& =& \sin \left(-\pi -\frac{\pi }{4}\right)\\ & =& \sin \left(-\frac{5\pi }{4}\right)\\ & =& \frac{1}{\sqrt{2}}\end{array}$

Hence, the correct option is (A).