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Question: Prove the theorem that: For any real numbers x and y, \(\cos x=\cos y\), implies \(x=2n\pi \pm y\), ...

Prove the theorem that: For any real numbers x and y, cos⁡x=cos⁡y\cos x=\cos y, implies x=2nπ±yx=2n\pi \pm y, where n∈Zn\in Z.

Explanation

Solution

Hint : To prove the above theorem first of all write the equation cos⁡x=cos⁡y\cos x=\cos y as cos⁡x−cos⁡y=0\cos x-\cos y=0 then as you can see that cos⁡x−cos⁡y\cos x-\cos y in the form of cos⁡C−cos⁡D\cos C-\cos Didentity then apply the trigonometric identity of cos⁡C−cos⁡D\cos C-\cos D which is equal to cos⁡C−cos⁡D=−2sin⁡C+D2sin⁡C−D2\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=2nπ±yx=2n\pi \pm y.

Complete step by step solution :
The trigonometric equation given in the question is:
cos⁡x=cos⁡y\cos x=\cos y
Rearranging the above equation we get,
cos⁡x−cos⁡y=0\cos x-\cos y=0
Now, as you can see that cos⁡x−cos⁡y\cos x-\cos y is in the form of cos⁡C−cos⁡D\cos C-\cos D trigonometric identity and we know that,
cos⁡C−cos⁡D=−2sin⁡C+D2sin⁡C−D2\cos C-\cos D=-2\sin \dfrac{C+D}{2}\sin \dfrac{C-D}{2}
Applying the above identity in cos⁡x−cos⁡y\cos x-\cos y we get,
−2sin⁡x+y2sin⁡x−y2=0-2\sin \dfrac{x+y}{2}\sin \dfrac{x-y}{2}=0
The solutions of the above equation are:
sin⁡x+y2=0,sin⁡x−y2=0\sin \dfrac{x+y}{2}=0,\sin \dfrac{x-y}{2}=0
We know from the trigonometric solutions that:
sin⁡θ=0 ⇒θ=nπ \begin{aligned} & \sin \theta =0 \\\ & \Rightarrow \theta =n\pi \\\ \end{aligned}
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⁡x+y2=0,sin⁡x−y2=0\sin \dfrac{x+y}{2}=0,\sin \dfrac{x-y}{2}=0
sin⁡x+y2=0 ⇒x+y2=nπ ⇒x+y=2nπ \begin{aligned} & \sin \dfrac{x+y}{2}=0 \\\ & \Rightarrow \dfrac{x+y}{2}=n\pi \\\ & \Rightarrow x+y=2n\pi \\\ \end{aligned}
Similarly for sin⁡x−y2=0\sin \dfrac{x-y}{2}=0 we get,
sin⁡x−y2=0 ⇒x−y2=nπ ⇒x−y=2nπ \begin{aligned} & \sin \dfrac{x-y}{2}=0 \\\ & \Rightarrow \dfrac{x-y}{2}=n\pi \\\ & \Rightarrow x-y=2n\pi \\\ \end{aligned}
From the above, we have got the two equations:
x+y=2nπ; x−y=2nπ \begin{aligned} & x+y=2n\pi ; \\\ & x-y=2n\pi \\\ \end{aligned}
We can write the first equation as:
x=2nπ−yx=2n\pi -y…………. Eq. (1)
We can write the second equation as:
x=2nπ+yx=2n\pi +y…………. Eq. (2)
From equation 1&2, we get the relation between x and y is:
x=2nπ±yx=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\cos x=\cos y this imply that x=2nπ±yx=2n\pi \pm y, where n∈Zn\in Z.

Note : The possible blunder that could happen in proving the theorem is to wrongly write the trigonometric identity cos⁡C−cos⁡D\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\cos C-\cos D.
So, be careful while writing the expansion of cos⁡C−cos⁡D\cos C-\cos D.