Question

Whether the given equation $\cos \theta .\cos 2\theta .\cos 3\theta .\cos 4\theta .\cos 6\theta .\cos 7\theta =\frac{1}{68},15\theta =\pi$ is ?

• A. True
• B. False

Answer 1

The correct option is B False

$15\theta =\pi ={180}^{\circ }$

$\Rightarrow \theta =\frac{{180}^{\circ }}{15}={12}^{\circ }$

$\cos \theta .\cos 2\theta .\cos 3\theta .\cos 4\theta .\cos 6\theta .\cos 7\theta$

$=\frac{1}{2\sin \theta }\times 2\sin \theta \cos \theta .\cos 2\theta .\cos 3\theta .\cos 4\theta .\cos 6\theta .\cos 7\theta$

$=\frac{1}{4\sin \theta }\times 2\sin 2\theta \cos 2\theta .\cos 3\theta .\cos 4\theta .\cos 6\theta .\cos 7\theta$

$=\frac{1}{4\sin \theta }\times \sin 4\theta \cos 4\theta \cos 3\theta ..\cos 6\theta .cos7\theta$

$=\frac{1}{8\sin \theta }\times 2\sin 4\theta \cos 4\theta \cos 3\theta ..\cos 6\theta .cos7\theta$

$=\frac{1}{8\sin \frac{\pi }{15}}\times \sin \frac{8\pi }{15}\cos \frac{3\pi }{15}.\cos \frac{2\pi }{5}.cos\frac{7\pi }{15}$

$=\frac{1}{8\sin \frac{\pi }{15}}\times \sin (\pi -\frac{7\pi }{15})\cos \frac{3\pi }{15}.\cos \frac{2\pi }{5}.cos\frac{7\pi }{15}$

$=\frac{1}{8\sin \frac{\pi }{15}}\times \sin \frac{7\pi }{15}\cos \frac{3\pi }{15}.\cos \frac{2\pi }{5}.cos\frac{7\pi }{15}$

$=\frac{1}{16\sin \frac{\pi }{15}}\times 2\sin \frac{7\pi }{15}cos\frac{7\pi }{15}\cos \frac{3\pi }{15}.\cos \frac{2\pi }{5}$

$=\frac{1}{16\sin \frac{\pi }{15}}\times \sin \frac{14\pi }{15}\cos \frac{3\pi }{15}.\cos \frac{2\pi }{5}$

$=\frac{1}{16\sin \frac{\pi }{15}}\times \sin (\pi -\frac{\pi }{15})\cos \frac{3\pi }{15}.\cos \frac{2\pi }{5}$

$=\frac{1}{16\sin \frac{\pi }{15}}\times \sin \frac{\pi }{15}\cos \frac{3\pi }{15}.\cos \frac{2\pi }{5}$

$=\frac{1}{32}\times 2\cos \frac{3\pi }{15}.\cos \frac{6\pi }{15}$

$=\frac{1}{32}\times (\cos \left(\frac{9\pi }{15}\right)+\cos \left(\frac{-3\pi }{15}\right))$

$=\frac{1}{32}\times (\cos \left(\frac{3\pi }{5}\right)+\cos \left(\frac{\pi }{5}\right))$

$=\frac{1}{32}\times (\cos (\pi -\frac{2\pi }{5})+\cos \left(\frac{\pi }{5}\right))$

$=\frac{1}{32}\times (-\cos \frac{2\pi }{5}+\cos \left(\frac{\pi }{5}\right))$

$=\frac{1}{32}\left(2\sin \left(\frac{3\pi }{10}\right)\sin \left(\frac{2\pi }{10}\right)\right)$

$=\frac{1}{16}\left(\sin {54}^{\circ }\sin {36}^{\circ }\right)$

$=\frac{1}{16}\times \frac{\sqrt{5}+1}{4}\times \frac{\sqrt{10-2\sqrt{5}}}{4}$

$=\frac{(\sqrt{5}+1)\left(\sqrt{10-2\sqrt{5}}\right)}{256}\ne \frac{1}{68}$

Hence the given statement is false.