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,

& \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{aligned}$$ So now we can write both these factors separately from equation (3) and hence we get, $$\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{aligned} & \Rightarrow x+y=(2n+1)\pi \\\ & \Rightarrow x=(2n+1)\pi -y......(7) \\\ \end{aligned}$$ Now we know that -1 to the power any odd number will be -1. Hence using this information in equation (7) we get, $$\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.