Mathematics - Complex Number Question with Solution | TestHub

$\cos 7\theta +i\sin 7\theta ={}^7{C}_0{\cos }^7\theta +{}^7{C}_1{\cos }^6\theta (i\sin \theta )++{}^7{C}_2{\cos }^5\theta (i\sin \theta {)}^2+\dots .$.\ $\sin 7\theta ={}^7{C}_1{\cos }^6\sin \theta -{}^7{C}_3{\cos }^4\theta {\sin }^3\theta +{}^7{C}_5{\cos }^2\theta {\sin }^5\theta -{}^7{C}_7{\sin }^7\theta$ $\sin 7\theta ={\sin }^7\theta \left[{}^7{C}_1{\cot }^6\theta -{}^7{C}_3{\cot }^4\theta +{}^7{C}_3{\cot }^2\theta -1\right]$ ${\cot }^2\theta =x\Rightarrow \sin 7\theta ={\sin }^7\theta \left({}^7{C}_1{x}^3-{}^7{C}_3{x}^2+{}^7{C}_{2}x-1\right)\theta =\frac{\pi }{7},\frac{2\pi }{7},\frac{3\pi }{7}$ are roots

$${\cot }^2\frac{\pi }{7}+{\cot }^2\frac{2\pi }{7}+{\cot }^2\frac{3\pi }{7}=\frac{{}^7{\mathrm{C}}_3}{7}=\frac{7\times 6\times 5}{3\times 2\times 7}=5$$