Question

If $\cos \theta +\cos 7\theta +\cos 3\theta +\cos 5\theta =0$ then $\theta =$

• A. $\frac{n\pi }{4};n\in I$
• B. $\frac{n\pi }{2};n\in I$
• C. $\frac{n\pi }{8};n\in I;n\ne 8k$
• D. $\frac{n\pi }{3};n\in I$

Answer 1

The correct option is C $\frac{n\pi }{8};n\in I;n\ne 8k$

Given that

$\cos \theta +\cos 7\theta +\cos 3\theta +\cos 5\theta =0$

$\Rightarrow (\cos \theta +\cos 7\theta )+(\cos 3\theta +\cos 5\theta )=0$

We know that $\cos C+\cos D=2\cos \left(\frac{C+D}{2}\right)\cos \left(\frac{C-D}{2}\right)$

Applying this result, we get

$\Rightarrow 2\cos \left(\frac{\theta +7\theta }{2}\right)\cos \left(\frac{\theta -7\theta }{2}\right)+2\cos \left(\frac{3\theta +5\theta }{2}\right)\cos \left(\frac{3\theta -5\theta }{2}\right)=0$

$\Rightarrow 2\cos \left(4\theta \right)\cos (-3\theta )+2\cos \left(4\theta \right)\cos (-\theta )=0$

$\Rightarrow 2\cos 4\theta (\cos 3\theta +\cos \theta )=0$

Therefore $\Rightarrow \cos 4\theta =0or\cos 3\theta +\cos \theta =0$

Case 1:

$\Rightarrow 4\theta =(2n\pi +\frac{\pi }{2})$

$\Rightarrow \theta =2n\frac{\pi }{4}+\frac{\pi }{8}$

Case2: $\cos 3\theta +\cos \theta =0$

$\Rightarrow \cos 3\theta =-\cos \theta$

$\Rightarrow \cos 3\theta =\cos \left(\right(2n+1)\pi +\theta )$

$\Rightarrow 3\theta =(2n+1)\pi +\theta$

$\Rightarrow \theta =n\pi +\frac{\pi }{2}$