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# If the solution for $\theta$ of $\cos p\theta +\cos q\theta =0,p>q>0$ are in A.P, then the numerically smallest common difference is A.P.(a) $\dfrac{\pi }{p+q}$(b) $\dfrac{2\pi }{p+q}$(c) $\dfrac{\pi }{2\left( p+q \right)}$ (d) $\dfrac{1}{p+q}$

Last updated date: 13th Jul 2024
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Hint: Use the formula of trigonometry $\operatorname{cosx}+cosy=2cos\left( \dfrac{x+y}{2} \right)cos\left( \dfrac{x-y}{2} \right)$. Apply the same to the expression given in the question. Then use the formula $\dfrac{\left( 2n+1 \right)\pi }{2}$ to find values where $\cos \theta =0$.

We are given an equation $\cos p\theta +\cos q\theta =0,p>q>0$.
We need to find its solution, so we will manipulate the given equation to find its solution.
We can use the general formula of trigonometry that is $\operatorname{cosx}+cosy=2cos\left( \dfrac{x+y}{2} \right)cos\left( \dfrac{x-y}{2} \right)$ .
$\cos p\theta +\cos q\theta =0,p>q>0$
Using the formula mentioned above we will get
$2\cos \left( \dfrac{p+q}{2} \right)\theta \cos \left( \dfrac{p-q}{2} \right)\theta =0.....\left( i \right)$
Now to get the solution, equate both its factors one by one with zero.
Now we will equate each factor to zero and we get,
$\cos \left( \dfrac{p+q}{2} \right)\theta =0$
We know the general solution when $\cos \theta =0$ then $\theta$ is equal to $\left( 2n+1 \right)\dfrac{\pi }{2}$ where n belongs to an integer.
So it’s clear from the general solution that,
$\left( \dfrac{p+q}{2} \right)\theta =\left( 2n+1 \right)\dfrac{\pi }{2}....\left( ii \right)$
where n belongs to integers.
And, similarly again we use the general solution formula, $\cos \theta =0$ then $\theta$ is equal to $\left( 2n+1 \right)\dfrac{\pi }{2}$ where n belongs to an integer for the second factor as well.
$\cos \left( \dfrac{p-q}{2} \right)\theta =0$
So,
$\left( \dfrac{p-q}{2} \right)\theta =\left( 2n+1 \right)\dfrac{\pi }{2}....\left( iii \right)$
where n belongs to integer
Now it is given in the question that solution of it would form an A.P so observing equation (ii) and equation (iii) we can observe that on putting the value of $n=1,2,3,,,,$ and so on we will get a common difference in each term i.e.
For equation (ii)
When $n=1$
$\left( \dfrac{p+q}{2} \right)\theta =\dfrac{3\pi }{2}$
When $n=2$
$\left( \dfrac{p+q}{2} \right)\theta =\dfrac{5\pi }{2}$
And so on.
We get a common difference $\pi$ for every successive value of n.
So,
\begin{align} & \left( \dfrac{p+q}{2} \right)\theta =\pi \\ & \theta =\dfrac{2\pi }{p+q} \\ \end{align}
We get one of the solutions of the given equation.
$\theta =\dfrac{2\pi }{p+q}.....\left( iv \right)$
Now for equation (iii)
When $n=1$
$\left( \dfrac{p-q}{2} \right)\theta =\dfrac{3\pi }{2}$
When $n=2$
$\left( \dfrac{p-q}{2} \right)\theta =\dfrac{5\pi }{2}$
And so on.
We get a common difference $\pi$ for every successive value of n.
So,
\begin{align} & \left( \dfrac{p-q}{2} \right)\theta =\pi \\ & \theta =\dfrac{2\pi }{p-q} \\ \end{align}
We get one of the solutions of the given equation.
$\theta =\dfrac{2\pi }{p-q}....\left( iv \right)$
Now we need to find the smallest solution for $\theta$,
So,
We will observe equation (iii) and (iv) and it is given in the question that $p>q>0$, so $p-q$ is less than $p+q$ and we know that if denominator value is greater, then fraction will be smaller. So, $\theta =\dfrac{2\pi }{p+q}$ the value of $\theta$ is much smaller than $\theta =\dfrac{2\pi }{p-q}$.
So the correct option is b.

Note: The possibility of mistake that could be done here is not considering both solutions of equation i.e. forgetting to evaluate the solution $\theta =\dfrac{2\pi }{p-q}$ although it will be rejected later on as it is not the smallest solution for $\theta$ . But we have to show that process in the solution.