$\cos 6\theta + i \sin 6 \theta = (\cos \theta + i \sin \theta )^6 \ \cdots \cdots (De - Moivr's Theorem )$ $\cos 6 \theta + i \sin \theta = \cos^6 \theta + 6 i \cos^5 \theta \sin \theta + 15 \cos^4 \theta \ i^2 \sin^2 \theta + 20 \cos^3 \theta \ i^3\sin^3 \theta + 15 \cos^2 \theta \ i^4 \sin ^4 \theta + 6 \cos \theta \ i^5 \sin^5 \theta + i^6 \sin^6 \theta$ ${i^2 = -1, \ i^3 = -i, \ i^4=1, \ i^5=i, \ i^6=-1}$ Comparing imaginary parts, $\sin 6 \theta = 6 \cos^5 \theta \sin \theta - 20 \cos^3 \theta \sin^3 \theta + 6 \cos \theta \sin^5 \theta $ $\dfrac {\sin 6 \theta}{\sin \theta}$ $=6 \cos^5 \theta - 20 \cos^3 \theta \sin^2 \theta + 6 \cos \theta \sin^4 \theta \ = 6 \cos^5 \theta -20 \cos^3 \theta (1-\cos^2 \theta) + 6 \cos \theta (1-\cos^2 \theta)^2 \ = (6+20)\cos^5 \theta - 20 \cos^3 \theta + 6 \cos \theta (1-2 \cos^2 \theta + \cos^4 \theta) \ = (26+6)\cos^5 \theta - (20 +12) \cos^3 \theta +6 \cos \theta \ =32 \cos^5 \theta -32 \cos^3 \theta + 6 \cos \theta \= 2 \cos \theta (16 \cos^4 \theta -16 \cos^2 \theta +3) $ $\therefore \dfrac {\sin 6 \theta}{\sin \theta}= 2 \cos \theta (16 \cos^4 \theta - 16 \cos^2 \theta +3) $ $\dfrac {\sin 6 \theta} {2 \cos \theta \sin \theta } = (16 \cos ^4 \theta -16 \cos^2 \theta +3) $ $\dfrac {\sin 6 \theta}{\sin 2 \theta}= 16 \cos^4 \theta - 16 \cos^2 \theta +3$