Question

Prove the theorem that: For any real numbers x and y, $\cos x=\cos y$, implies $x=2n\pi \pm y$, where $n\in Z$.

Answer 1

Hint: To prove the above theorem first of all write the equation $\cos x=\cos y$ as $\cos x-\cos y=0$ then as you can see that $\cos x-\cos y$ in the form of $\cos C-\cos D$identity then apply the trigonometric identity of $\cos C-\cos D$ which is equal to $\cos C-\cos D=-2\sin \dfrac{C+D}{2}\sin \dfrac{C-D}{2}$ and then solve for x and y. After solving, you will get the relation between x and y as $x=2n\pi \pm y$.


Complete step by step solution:
The trigonometric equation given in the question is:
$\cos x=\cos y$
Rearranging the above equation we get,
$\cos x-\cos y=0$
Now, as you can see that $\cos x-\cos y$ is in the form of $\cos C-\cos D$ trigonometric identity and we know that,
$\cos C-\cos D=-2\sin \dfrac{C+D}{2}\sin \dfrac{C-D}{2}$
Applying the above identity in $\cos x-\cos y$ we get,
$-2\sin \dfrac{x+y}{2}\sin \dfrac{x-y}{2}=0$
The solutions of the above equation are:
$\sin \dfrac{x+y}{2}=0,\sin \dfrac{x-y}{2}=0$
We know from the trigonometric solutions that:
$\begin{align}
  & \sin \theta =0 \\
 & \Rightarrow \theta =n\pi \\
\end{align}$
In the above equation”n” corresponds to any integer.
 Using the above relation in sin in the solutions of the given equation we get,
$\sin \dfrac{x+y}{2}=0,\sin \dfrac{x-y}{2}=0$
$\begin{align}
  & \sin \dfrac{x+y}{2}=0 \\
 & \Rightarrow \dfrac{x+y}{2}=n\pi \\
 & \Rightarrow x+y=2n\pi \\
\end{align}$
Similarly for $\sin \dfrac{x-y}{2}=0$ we get,
$\begin{align}
  & \sin \dfrac{x-y}{2}=0 \\
 & \Rightarrow \dfrac{x-y}{2}=n\pi \\
 & \Rightarrow x-y=2n\pi \\
\end{align}$
From the above, we have got the two equations:
$\begin{align}
  & x+y=2n\pi ; \\
 & x-y=2n\pi \\
\end{align}$
We can write the first equation as:
$x=2n\pi -y$…………. Eq. (1)
We can write the second equation as:
$x=2n\pi +y$…………. Eq. (2)
From equation 1&2, we get the relation between x and y is:
$x=2n\pi \pm y$
In the above equation”n” can be any integer.
Hence, we have shown that for any real numbers x and y,$\cos x=\cos y$ this imply that $x=2n\pi \pm y$, where $n\in Z$.


Note: The possible blunder that could happen in proving the theorem is to wrongly write the trigonometric identity $\cos C-\cos D$.
You might confuse that two sine will come or one cosine or one sine and usually in haste of solving the questions in exam we forget to put negative signs in the expansion of the identity $\cos C-\cos D$.
So, be careful while writing the expansion of $\cos C-\cos D$.