Question

Solve : $\frac{cos\theta +cos2\theta +cos3\theta +cos4\theta }{sin\theta +sin2\theta +sin3\theta +sin4\theta }=cot\frac{5\theta }{2}$

Answer 1

To prove that $\frac{\cos \theta +\cos 2\theta +\cos 3\theta +\cos 4\theta }{\sin \theta +\sin 2\theta +\sin 3\theta +\sin 4\theta }=\frac{\cot 5\theta }{2}$

$=\frac{\cos \theta +\cos 4\theta +\cos 2\theta +\cos 3\theta }{\sin \theta +\sin 4\theta +\sin 2\theta +\sin 3\theta }=\frac{2\cos \frac{5\theta }{2}\cos \frac{3\theta }{2}+2\cos \frac{5\theta }{2}\cos \frac{\theta }{2}}{2\sin \frac{5\theta }{2}.\sin \frac{\theta }{2}+2\sin \frac{5\theta }{2}.\sin \frac{\theta }{2}}$

$\frac{2\cos \frac{5\theta }{2}\left(\cos \frac{3\theta }{2}+\cos \frac{\theta }{2}\right)}{2\sin \frac{5\theta }{2}(\frac{\cos 3\theta }{2}+\frac{\cos \theta }{2})}=\cot \frac{5\theta }{2}$