Question

$\frac{\cos 6\theta +6\cos 4\theta +15\cos 2\theta +10}{\cos 5\theta +5\cos 3\theta +10\cos \theta }=$

• A. $\sin \theta$
• B. $\cos \theta$
• C. $2\sin \theta$
• D. $2\cos \theta$

Answer 1

Answer: (D)

$2\cos \theta$

$\cos 6\theta +6\cos 4\theta +15\cos 2\theta +10$

$=(\cos 6\theta +\cos 4\theta )+5(\cos 4\theta +\cos 2\theta )+10(\cos 2\theta +1)$

$=\left(2\cos \left(\frac{6\theta +4\theta }{2}\right)\cos \left(\frac{6\theta -4\theta }{2}\right)\right)+5\left(2\cos \left(\frac{4\theta +2\theta }{2}\right)\cos \left(\frac{4\theta -2\theta }{2}\right)\right)+10\left(2{\cos }^2\theta \right)$

$=2\cos 5\theta \cos \theta +10\cos 3\theta \cos \theta +20{\cos }^2\theta$

$=2(\cos 5\theta +5\cos 3\theta +10\cos \theta )\cos \theta$

$\therefore \frac{\cos 6\theta +6\cos 4\theta +15\cos 2\theta +10}{\cos 5\theta +5\cos 3\theta +10\cos \theta }=2\cos \theta$