0
3.9kviews
Prove that αn + βn = 2cos n θ cosecn θ if α and β roots of the equation z2sin2 θ - z sin2θ + 1 = 0 .
1 Answer
0
1.1kviews

$z=\dfrac {\sin 20 \pm \sqrt{\sin^2 2 \theta - 4 \sin^2 \theta}}{2\sin^2 \theta} = \dfrac {2 \sin \theta \cos \theta \pm \sqrt{4 \sin ^2 \theta \cos^2 \theta - 4 \sin^2 \theta}} {2 \sin^2 \theta}$ $= \dfrac {\cos \theta \pm \sqrt{\cos^2 \theta -1}}{\sin \theta} = \dfrac {\cos \theta \pm i\sin \theta}{\sin \theta} = (\cos \theta \pm i \sin \theta) \ csc \theta $ $ \text{Let }\alpha = (\cos \theta + i\sin \theta) \csc \text{ and }\beta = (\cos \theta - i\sin \theta )\csc \theta$ $Hence, \alpha^n = (\cos n\theta + i \sin n\theta)\csc \ ^n\theta; \ \beta^n = (\cos n\theta - i \sin n\theta)\csc \ ^n\theta$ $\therefore a^n + \beta^n = 2 \cos \theta \csc \ ^n \theta$

Please log in to add an answer.