Question

How do you find the value of $\theta $ for which $\cos \left( \theta +\dfrac{\pi }{2} \right)$ does not equal to $\cos \theta +\cos \left( \dfrac{\pi }{2} \right)$?

Answer 1

Hint: we start solving the problem by considering $\cos \left( \theta +\dfrac{\pi }{2} \right)\ne \cos \theta +\cos \left( \dfrac{\pi }{2} \right)$. We then make use of the fact that $\cos \left( \dfrac{\pi }{2} \right)=0$ to proceed through the problem. We then make use of the fact that $\cos A-\cos B=-2\sin \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{A-B}{2} \right)$ to proceed further through the problem. We then make the necessary calculations and then make use of the fact that if $\sin x=0$, then the general solution is $x=n\pi $, $n\in Z$ to get the required answer.

Complete step by step answer:
According to the problem, we are asked to find the value of $\theta $ for which $\cos \left( \theta +\dfrac{\pi }{2} \right)$ does not equal to $\cos \theta +\cos \left( \dfrac{\pi }{2} \right)$.
Now, let us consider $\cos \left( \theta +\dfrac{\pi }{2} \right)\ne \cos \theta +\cos \left( \dfrac{\pi }{2} \right)$ ---(1).
We know that $\cos \left( \dfrac{\pi }{2} \right)=0$. Let us use this result in equation (1).
\[\Rightarrow \cos \left( \theta +\dfrac{\pi }{2} \right)\ne \cos \theta +0\].
\[\Rightarrow \cos \left( \theta +\dfrac{\pi }{2} \right)\ne \cos \theta \].
\[\Rightarrow \cos \left( \theta +\dfrac{\pi }{2} \right)-\cos \theta \ne 0\] ---(2).
We know that $\cos A-\cos B=-2\sin \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{A-B}{2} \right)$. Let us use this result in equation (2).
\[\Rightarrow -2\sin \left( \dfrac{\theta +\dfrac{\pi }{2}+\theta }{2} \right)\sin \left( \dfrac{\theta +\dfrac{\pi }{2}-\theta }{2} \right)\ne 0\].
\[\Rightarrow -2\sin \left( \dfrac{2\theta +\dfrac{\pi }{2}}{2} \right)\sin \left( \dfrac{\dfrac{\pi }{2}}{2} \right)\ne 0\].
\[\Rightarrow -2\sin \left( \theta +\dfrac{\pi }{4} \right)\sin \left( \dfrac{\pi }{4} \right)\ne 0\].
\[\Rightarrow -2\sin \left( \theta +\dfrac{\pi }{4} \right)\left( \dfrac{1}{\sqrt{2}} \right)\ne 0\].
\[\Rightarrow -\sqrt{2}\sin \left( \theta +\dfrac{\pi }{4} \right)\ne 0\] ---(3).
We know that $-\sqrt{2}\ne 0$. So, we need to find the values of $\theta $ for which \[\sin \left( \theta +\dfrac{\pi }{4} \right)\ne 0\].
We know that if $\sin x=0$, then the general solution is $x=n\pi $, $n\in Z$. Let us use this result in equation (3).
\[\Rightarrow \theta +\dfrac{\pi }{4}\ne n\pi \].
\[\Rightarrow \theta \ne n\pi -\dfrac{\pi }{4}\], $n\in Z$.
So, we have found the values of $\theta $ for which $\cos \left( \theta +\dfrac{\pi }{2} \right)$ does not equal to $\cos \theta +\cos \left( \dfrac{\pi }{2} \right)$ as \[\theta \ne n\pi -\dfrac{\pi }{4}\], $n\in Z$.
$\therefore $ The values of $\theta $ for which $\cos \left( \theta +\dfrac{\pi }{2} \right)$ does not equal to $\cos \theta +\cos \left( \dfrac{\pi }{2} \right)$ is \[\theta \ne n\pi -\dfrac{\pi }{4}\], $n\in Z$.

Note:
We should perform each step carefully in order to avoid confusion and calculation mistakes while solving this problem. Here, we have assumed that we need to find the general solution for the values of $\theta $. We can also find the principal value of the angle $\theta $ for which the given condition is satisfied. Similarly, we can expect problems to find the values of $\theta $ for which $\cos \left( \theta +\dfrac{\pi }{4} \right)$ does not equal to $\cos \theta +\cos \left( \dfrac{\pi }{4} \right)$.