Question

The exact value of $\cos \frac{2\pi }{28}\csc \frac{3\pi }{28}+\cos \frac{6\pi }{28}\csc \frac{9\pi }{28}+\cos \frac{18\pi }{28}\csc \frac{27\pi }{28}$ is?

Answer 1

$\cos \frac{2\pi }{28}cosec\frac{3\pi }{28}+\cos \frac{6\pi }{28}cosec\frac{9\pi }{28}+\cos \frac{18\pi }{28}cosec\frac{27\pi }{28}$

$\Rightarrow$ $\frac{\cos \frac{2\pi }{28}}{\sin \frac{3\pi }{28}}+\frac{\cos \frac{6\pi }{28}}{\sin \frac{9\pi }{28}}+\frac{\cos \frac{18\pi }{28}}{\sin \frac{27\pi }{28}}$

Put $\frac{\pi }{28}=\theta$

$\Rightarrow$ $\frac{\cos 2\theta }{\sin 3\theta }+\frac{\cos 6\theta }{\sin 9\theta }+\frac{\cos 18\theta }{\sin 27\theta }$

$\Rightarrow$ $\frac{\cos 2\theta }{\sin 3\theta }+\frac{\cos 6\theta }{\sin (14\theta -5\theta )}+\frac{\cos (14\theta +4\theta )}{\sin (28\theta -\theta )}$

$\frac{\pi }{28}=\theta$

$\pi =28\theta$

and $\frac{\pi }{2}=14\theta$

$\Rightarrow$ $\frac{\cos 2\theta }{\sin 3\theta }+\frac{\cos 6\theta }{\cos 5\theta }+\frac{(-\sin 4\theta )}{\sin \theta }$

$\Rightarrow$ $\frac{\cos 2\theta \cos 5\theta \sin \theta +\cos 6\theta \sin 3\theta \sin \theta -\sin 4\theta \cos 5\theta \sin 3\theta }{\sin 3\theta \cos 5\theta \sin \theta }$ --- ( 1 )

Now, $\cos 2\theta \cos 5\theta \sin \theta +\cos 6\theta \sin 3\theta \sin \theta -\sin 4\theta \cos 5\theta \sin 3\theta$

$\Rightarrow$ $\frac{2}{2}\left[\cos 2\theta \cos 5\theta \right)\sin \theta +\frac{\cos 6\theta 2\left(\sin 3\theta \sin \theta \right)}{2}-\frac{\sin 2\theta 2\left(\sin 4\theta \sin 3\theta \right)}{2}$

$\Rightarrow$ $\frac{1}{2}\left[\right(\cos 7\theta +\cos 3\theta )\sin \theta +\cos 6\theta (-\cos 4\theta +\cos 2\theta )-\cos 5\theta (-\cos 7\theta +\cos \theta \left)\right]$

$\Rightarrow$ $\frac{1}{4}[2\sin \theta \cos 7\theta +2\sin \theta \cos 3\theta -2\cos 6\theta \cos 4\theta +2\cos 6\theta \cos 2\theta +2\cos 5\theta \cos 7\theta -2\cos 5\theta \cos theta]$

$\Rightarrow$ $\frac{1}{4}\left[\sin \right(4\theta )-\sin (2\theta )+\sin 8\theta -\sin 6\theta +\cos 8\theta +\cos 4\theta -\cos 10\theta -\cos 2\theta -\cos 6\theta -\cos 4\theta +\cos 12\theta +\cos 2\theta ]$

$\Rightarrow$ $\frac{1}{4}\left[\sin \right(14\theta -10\theta )-\sin (14\theta -12\theta )+\sin (14\theta -6\theta )-\sin (14\theta -8\theta )+\cos 8\theta +\cos 4\theta -\cos 10\theta -\cos 2\theta -\cos 6\theta -\cos 4\theta +\cos 12\theta +\cos 2\theta ]$

$\Rightarrow$ $\frac{1}{4}[\cos 10\theta -\cos 12\theta +\cos 6\theta -\cos 8\theta +\cos 8\theta -\cos 10\theta -\cos 2\theta -\cos 6\theta +\cos 12\theta +\cos 2\theta ]$

$\Rightarrow$ $0$ ----- ( 2 )

Substituting ( 2 ) in ( 1 ) we get,

$\Rightarrow$ $\frac{0}{\sin 3\theta \cos 5\theta \sin \theta }$

$\Rightarrow$ $0$