Let $x=\cos \theta + i \sin \theta$

$\therefore \dfrac {1}{x} = \cos \theta - i \sin \theta$

$\therefore x + \dfrac {1}{x} = 2 \cos \theta$ and $x - \dfrac {1}{x} = 2i \sin \theta$

$x^n =( \cos \theta + i \sin \theta )^n = (\text{cosn} \theta + i \sin n\theta)$

$\therefore \dfrac {1}{x^n} = (\cos \theta - i \sin \theta )^n = (\text{cosn} \theta - i \sin n\theta)$

$x^n + \dfrac {1}{x^n} = 2 \text{ cosn and }x^n- \dfrac {1}{x^n} = 2i \sin n\theta$

Now, by binomial theorem,

$(2i \sin \theta )^7 = \left ( x - \dfrac {1}{x} \right )^7$

$= x^7 - 7x^6 \dfrac {1}{x} + 21x^5 \dfrac {1}{x^2} - 35 x^4 \dfrac {1}{x^3} - 35 x^3 \dfrac {1}{x^4} - 21 x^2 \dfrac {1}{x^5} + 7x\dfrac {1}{x^6} - \dfrac{1}{x^7}$

$= x^7 - 7x^5 + 21x^3 - 35x + \dfrac {35}{x} - \dfrac {21}{x^3} + \dfrac {7}{x^5} - \dfrac {1}{x^7}$

$\left ( x^7 - \dfrac {1}{x^7} \right ) - 7 \left(x^5 - \dfrac {1}{x^5} \right) + 21 \left ( x^3 - \dfrac {1}{x^3} \right ) - 35 \left ( x - \dfrac {35}{x} \right )$

$ = (2i\sin 7\theta) - 7 (2i \sin 5 \theta) + 21 (2i \sin 3\theta) - 35 (2i \sin \theta) - 2^6 \sin^7 \theta$

$= \sin 7 \theta - 7 \sin 5 \theta + 21 \sin 3\theta - 35 \sin \theta $

$\therefore \sin^7 \theta = \dfrac {-1}{2^6} \left ( \sin 7 \theta - 7 \sin 5\theta + 21 \sin 3\theta - 35 \sin \theta \right )$