Question

Prove that for any real numbers x and y, \[\sin x=\sin y\] implies \[x=n\pi +{{(-1)}^{n}}y\], where \[n\in Z\].

Answer 1

Hint: We will first convert the given expression in a recognizable formula form and then substitute using the formula \[\sin x-\sin y=2\cos \left( \dfrac{x+y}{2} \right)\sin \left( \dfrac{x-y}{2} \right)\]. After this we will solve both the factors separately and then will combine both the results to get our answer.

Complete step-by-step answer:
The trigonometric equation mentioned in the question is \[\sin x=\sin y.......(1)\]
Now bringing all the terms to the left hand side of the equation (1) we get,
\[\Rightarrow \sin x-\sin y=0.......(2)\]
Now we know the formula that \[\sin x-\sin y=2\cos \left( \dfrac{x+y}{2} \right)\sin \left( \dfrac{x-y}{2} \right)\]. So hence substituting this in equation (2) we get,
\[\begin{align}
  & \Rightarrow 2\cos \left( \dfrac{x+y}{2} \right)\sin \left( \dfrac{x-y}{2} \right)=0 \\
 & \Rightarrow \cos \left( \dfrac{x+y}{2} \right)\sin \left( \dfrac{x-y}{2} \right)=0...........(3) \\
\end{align}\]
So now we can write both these factors separately from equation (3) and hence we get,
\[\Rightarrow \cos \left( \dfrac{x+y}{2} \right)=0......(4)\]
Now solve equation (4) as we know that \[\cos \left( (2n+1)\dfrac{\pi }{2} \right)=0\]. So now substituting this in place of 0 we get,
\[\Rightarrow \cos \left( \dfrac{x+y}{2} \right)=\cos \left( (2n+1)\dfrac{\pi }{2} \right)......(5)\]
Now cancelling similar terms on both sides in equation (5) we get,
\[\Rightarrow \dfrac{x+y}{2}=(2n+1)\dfrac{\pi }{2}......(6)\]
Again simplifying and rearranging in equation (6) we get,
\[\begin{align}
  & \Rightarrow x+y=(2n+1)\pi \\
 & \Rightarrow x=(2n+1)\pi -y......(7) \\
\end{align}\]
Now we know that -1 to the power any odd number will be -1. Hence using this information in equation (7) we get,
\[\Rightarrow x=(2n+1)\pi +{{(-1)}^{2n+1}}y......(8)\]
Now solving the second factor from equation (3) we get,
\[\Rightarrow \sin \left( \dfrac{x-y}{2} \right)=0......(9)\]
Now solving equation (9) as we know that \[\sin n\pi =0\]. So now substituting this in place of 0 we get,
\[\Rightarrow \sin \left( \dfrac{x-y}{2} \right)=\sin n\pi ......(10)\]
Now cancelling the similar terms on both sides in equation (10) we get,
\[\Rightarrow \dfrac{x-y}{2}=n\pi ......(11)\]
Now cross multiplying and rearranging in equation (11) we get,
\[\Rightarrow x=2n\pi +y......(12)\]
Now we know that -1 to the power any even number is 1. Hence using this information in equation (12) we get,
\[\Rightarrow x=2n\pi +{{(-1)}^{2n}}y......(13)\]
Combining equation (8) and equation (13) we get,
\[\Rightarrow x=n\pi +{{(-1)}^{n}}y\] where \[n\in Z\]. Hence proved.

Note: In trigonometry remembering the formulas and the identities is very important because then it becomes easy. Here we may get confused about powers of 1 but we have to remember that -1 to the power any odd number will yield -1 and -1 to the power any even number will yield 1.