1. **State the problem:** Evaluate $\left( \sin \frac{\pi}{9} + i \sin \frac{7\pi}{18} \right)^{-6}$ using De Moivre's theorem. 2. **Recall De Moivre's theorem:** For a complex number in polar form $r(\cos \theta + i \sin \theta)$, we have: $$\left(r(\cos \theta + i \sin \theta)\right)^n = r^n (\cos n\theta + i \sin n\theta)$$ 3. **Rewrite the expression:** Notice the given expression is $\sin \frac{\pi}{9} + i \sin \frac{7\pi}{18}$, which is not immediately in the form $\cos \theta + i \sin \theta$. 4. **Check if it can be expressed as $\cos \alpha + i \sin \alpha$:** We want to find $\alpha$ such that: $$\cos \alpha = \sin \frac{\pi}{9}, \quad \sin \alpha = \sin \frac{7\pi}{18}$$ 5. **Evaluate the values:** - $\sin \frac{\pi}{9} = \sin 20^\circ \approx 0.3420$ - $\sin \frac{7\pi}{18} = \sin 70^\circ \approx 0.9397$ 6. **Find $\alpha$:** Since $\sin \alpha = 0.9397$, $\alpha \approx 70^\circ = \frac{7\pi}{18}$. Check $\cos \alpha = \cos 70^\circ \approx 0.3420$, which matches $\sin \frac{\pi}{9}$. 7. **Therefore:** $$\sin \frac{\pi}{9} + i \sin \frac{7\pi}{18} = \cos \frac{7\pi}{18} + i \sin \frac{7\pi}{18}$$ 8. **Apply De Moivre's theorem:** $$\left( \cos \frac{7\pi}{18} + i \sin \frac{7\pi}{18} \right)^{-6} = \cos \left(-6 \times \frac{7\pi}{18}\right) + i \sin \left(-6 \times \frac{7\pi}{18}\right)$$ 9. **Simplify the angle:** $$-6 \times \frac{7\pi}{18} = -\frac{42\pi}{18} = -\frac{7\pi}{3}$$ 10. **Reduce the angle modulo $2\pi$:** $$-\frac{7\pi}{3} + 4\pi = -\frac{7\pi}{3} + \frac{12\pi}{3} = \frac{5\pi}{3}$$ 11. **Final evaluation:** $$\cos \frac{5\pi}{3} + i \sin \frac{5\pi}{3} = \frac{1}{2} - i \frac{\sqrt{3}}{2}$$ **Answer:** $$\boxed{\frac{1}{2} - i \frac{\sqrt{3}}{2}}$$