$By\ De'Moivre\ Theorem:$ $(\cos \theta + i\sin \theta)^n= (\cos n\theta + i\sin n\theta)$ $(\cos \theta + i\sin \theta)^7= (\cos 7\theta + i\sin 7\theta) \cdots (1)$   By Binomial Theorem: $(\cos \theta\ +\ i\sin \theta)^7\ =\ (\cos^7\theta\ +\ 7 cos^6 \theta (i\sin \theta)\ +\ 21 \cos^5\theta.(i\sin \theta)^2\ +\ 35\cos^4\theta(i\sin \theta)^3\ +\ 35\cos^3\theta(i\sin \theta)^4\ +\ 21 \cos^2\theta.(i\sin \theta)^5\ +\ 7 \cos \theta.(i\sin \theta)^6\ +\ \cos \theta(i\sin \theta)^7 $ $= \cos ^7\theta + 7 cos^6 \theta (i\sin \theta) - 21 \cos^5\theta.sin^2 \theta-35\cos^4\theta. i\sin^3 \theta+35\cos^3\theta.\sin ^4\theta+ 21 \cos^2\theta.i\sin^5 \theta- 7 \cos \theta.\sin^6 \theta- i\sin^7 \theta $ $= \cos ^7\theta - 21 \cos^5\theta.sin^2 \theta+35\cos^3\theta. \sin^4 \theta-7\cos \theta \sin^6\theta+i(7\cos^6 \theta \sin\theta-35\cos^4\theta.\sin ^3\theta+ 21 \cos^2\theta.\sin^5 \theta- \sin^7 \theta \cdots (2) $   Equating imaginary parts, from 1 and 2 we get, $\sin 7 \theta= (7\cos^6 \theta \sin\theta-35\cos^4\theta.\sin ^3\theta+ 21 \cos^2\theta.\sin^5 \theta- \sin^7 \theta \cdots (3) $ $\dfrac {\sin 7 \theta} {sin \theta}= \dfrac {(7\cos^6 \theta \sin\theta-35\cos^4\theta.\sin ^3\theta+ 21 \cos^2\theta.\sin^5 \theta- \sin^7 \theta)} {\sin \theta}$ $= 7\cos^6 \theta -35\cos^4\theta.\sin ^2\theta+ 21 \cos^2\theta.\sin^4 \theta- \sin^6 \theta $ $= 7 (1-sin^2\theta)^3-35 (1-sin^2\theta)^2.\sin^2\theta+21 (1-sin^2\theta)sin^4 \theta-\sin^6 \theta$ $= 7 (1-3sin^2\theta +3 \sin ^4 \theta - \sin^6 \theta)-35 (1-2sin^2\theta +\sin ^4 \theta).\sin^2\theta+21 (1-sin^2\theta)sin^4 \theta-\sin^6 \theta$ $= 7 -21sin^2\theta +21 \sin ^4 \theta - 7\sin^6 \theta- 35 \sin^2 \theta +70\sin ^4 \theta-35\sin ^6 \theta +21sin^4\theta-21\sin^6 \theta-\sin^6 \theta$ $= 7 -56sin^2\theta +112 \sin ^4 \theta - 64\sin^6 \theta$