Question

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

• A. $(2n+1)\frac{\pi }{2};n\in Z$
• B. $(2n+1)\frac{\pi }{4};n\in Z$
• C. $(2n+1)\frac{\pi }{8};n\in Z$
• D. $(2n+1)\frac{\pi }{16};n\in Z$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $(2n+1)\frac{\pi }{2};n\in Z$
B $(2n+1)\frac{\pi }{4};n\in Z$
C $(2n+1)\frac{\pi }{8};n\in Z$

Given that:-

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

To find:- $\theta =?$

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

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

$2\cos \left(\frac{\theta +7\theta }{2}\right)\cos \left(\frac{\theta -7\theta }{2}\right)+2\cos \left(\frac{3\theta +5\theta }{2}\right)\cos \left(\frac{3\theta -5\theta }{2}\right)=0[\because \cos A+\cos B=2\cos \frac{A+B}{2}\cos \frac{A-B}{2}]$

$2(\cos 4\theta \cos 3\theta +\cos 4\theta \cos \theta )=0[\because \cos (-\theta )=\cos \theta ]$

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

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

$\Rightarrow 4\cos \theta \cos 2\theta \cos 4\theta =0$

$\Rightarrow \cos \theta =0\text{or}\cos 2\theta =0\text{or}\cos 4\theta =0$

Case I:-

For $\cos \theta =0$

$\Rightarrow \theta =(2n+1)\frac{\pi }{2};n\in Z.....\left(1\right)$

Case II:-

For $\cos 2\theta =0$

$\Rightarrow 2\theta =(2n+1)\frac{\pi }{2}$

$\Rightarrow \theta =(2n+1)\frac{\pi }{4};n\in Z.....\left(2\right)$

Case III:-

For $\cos 4\theta =0$

$\Rightarrow 4\theta =(2n+1)\frac{\pi }{2}$

$\Rightarrow \theta =(2n+1)\frac{\pi }{8};n\in Z.....\left(3\right)$

Hence equation $\left(1\right),\left(2\right)\&\left(3\right)$ are the solution of $\cos \theta +\cos 7\theta +\cos 3\theta +\cos 5\theta =0$.