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Question: If the solutions for \(\theta\)of \(\cos p\theta + \cos q\theta = 0,p > 0,q > 0\) are in A.P., then ...

If the solutions for θ\thetaof cos⁡pθ+cos⁡qθ=0,p>0,q>0\cos p\theta + \cos q\theta = 0,p > 0,q > 0 are in A.P., then the numerically smallest common difference of

A.P. is

A

πp+q\frac{\pi}{p + q}

B

2πp+q\frac{2\pi}{p + q}

C

π2(p+q)\frac{\pi}{2(p + q)}

D

1p+q\frac{1}{p + q}

Answer

2πp+q\frac{2\pi}{p + q}

Explanation

Solution

Given, cos⁡pθ=−cos⁡qθ=cos⁡(π+qθ)\cos p\theta = - \cos q\theta = \cos(\pi + q\theta)

⇒\mathbf{\Rightarrow} pθ=2nπ±(π+qθ),n∈I\mathbf{p\theta = 2n\pi \pm (\pi + q\theta),n \in I} ⇒\mathbf{\Rightarrow} θ=(2n+1)πp−q\mathbf{\theta =}\frac{\mathbf{(2n + 1)\pi}}{\mathbf{p - q}} or (2n−1)πp+q\frac{(2n - 1)\pi}{p + q},

n∈In \in I. Both the solutions form an A.P. θ=(2n+1)πp−q\theta = \frac{(2n + 1)\pi}{p - q} gives us an A.P. with common difference =2πp−q= \frac{2\pi}{p - q} and θ=(2n−1)πp+q\theta = \frac{(2n - 1)\pi}{p + q} gives us an A.P. with common difference = 2πp+q\frac{2\pi}{p + q}. Certainly, 2πp+q<∣2πp−q∣\frac{2\pi}{p + q} < \left| \frac{2\pi}{p - q} \right|.