Question

If m and n are positive
integers satisfying
$1+\cos 2\theta +\cos 4\theta +\cos 6\theta +\cos 8\theta +\cos 10\theta =\frac{\cos m\theta .\sin n\theta }{\sin \theta }$ then $(m+n)$ is equal to

• A. 9
• B. 10
• C. 11
• D. 12

Answer 1

The correct option is C 11

Given

$1+\cos 2\theta +\cos 4\theta +\cos 6\theta +\cos 8\theta +\cos 10\theta$

Multiplying and dividing by $2\sin \theta$

$⟹1+\frac{1}{2\sin \theta }(2\sin \theta \cos 2\theta +2\cos 4\theta \sin \theta +2\cos 6\theta \sin \theta +2\cos 8\theta \sin \theta +2\cos 10\theta \sin \theta )$

$⟹1+\frac{1}{2\sin \theta }(\sin 3\theta -\sin \theta +\sin 5\theta -\sin 3\theta +\sin 7\theta -\sin 5\theta +\sin 9\theta -\sin 7\theta +\sin 11\theta -\sin 9\theta )$

$(\sin 2\theta =2\sin \theta \cos \theta )$

$⟹1+\frac{1}{2\sin \theta }(\sin 11\theta -\sin \theta )$

$⟹\frac{2\sin \theta +\sin 11\theta -\sin \theta }{2\sin \theta }$

$⟹\frac{\sin 11\theta +\sin \theta }{2\sin \theta }$

$⟹\frac{\left(2\cos 5\theta \sin 6\theta \right)}{2\sin \theta }$

$(\sin A+\sin B=2\cos \frac{A-B}{2}\sin \frac{A+B}{2})$

$⟹\frac{\left(\cos 5\theta \sin 6\theta \right)}{\sin \theta }$

Here $m=5,n=6$

$⟹m+n=5+6=11$