Verified Answer :
- Two sine functions \(\sin(x-h_1)\) and \(\sin(x-h_2)\) are identical if \(\sin(x-h_1) = \sin(x-h_2)\) for all \(x\).
- Using the periodicity of the sine function, \(\sin(A) = \sin(B)\) implies \(A = B + 2k\pi\) or \(A = \pi - B + 2k\pi\).
- The second case fails because it requires \(x\) to be constant for all \(x\).
- The first case results in \(h_1 - h_2 = 2n\pi\), meaning the phase shifts must differ by an integer multiple of \(2\pi\).
- The final condition is \(h_1 - h_2 = 2n\pi\) for any non-zero integer \(n\), which is \(\boxed{h_1 - h_2 = 2n\pi}\).
Explanation
- Analyzing the condition for identical graphs
To determine when two sine functions \(y_1 = \sin(x - h_1)\) and \(y_2 = \sin(x - h_2)\) have the same graph, we require \(y_1 = y_2\) for all real values of \(x\). This is equivalent to solving the equation:
\[\sin(x - h_1) = \sin(x - h_2)\]
for all \(x \in \mathbb{R}\). The sine function is periodic with a period of \(2\pi\), meaning \(\sin(\theta) = \sin(\theta + 2k\pi)\) for any integer \(k\). Using the general properties of the sine function, the equality \(\sin(A) = \sin(B)\) holds if and only if \(A = B + 2k\pi\) or \(A = \pi - B + 2k\pi\) for some integer \(k\).
- Evaluating the cases for equality
We examine the two cases for the equality \(\sin(x - h_1) = \sin(x - h_2)\):
Case 1: \(x - h_1 = x - h_2 + 2k\pi\)
Subtracting \(x\) from both sides, we get:
\(-h_1 = -h_2 + 2k\pi\)
Rearranging terms, we find:
\(h_1 - h_2 = -2k\pi\)
Since \(k\) can be any integer, let \(n = -k\). Then \(h_1 - h_2 = 2n\pi\). This implies that the horizontal shift values must differ by an integer multiple of \(2\pi\). Since we are given \(h_1
eq h_2\), \(n\) must be a non-zero integer.
Case 2: \(x - h_1 = \pi - (x - h_2) + 2k\pi\)
Expanding the right side:
\(x - h_1 = \pi - x + h_2 + 2k\pi\)
Adding \(x\) to both sides:
\(2x - h_1 = \pi + h_2 + 2k\pi\)
This equation must hold for all real values of \(x\). However, the left side depends on \(x\), while the right side is constant. Therefore, this case cannot hold for all \(x\).
- Concluding the condition for identical graphs
From the analysis, we conclude that the only way for the two functions to have identical graphs is for the first case to hold. Specifically, the difference between the horizontal shifts \(h_1\) and \(h_2\) must be an integer multiple of \(2\pi\). Mathematically, this is expressed as:
\[h_1 - h_2 = 2n\pi\]
where \(n \in \{\dots, -2, -1, 1, 2, \dots\}\) is a non-zero integer. This ensures that the two waves are in perfect alignment due to the periodicity of the sine function. Thus, two sine functions with different phase shifts have the same graph if and only if the difference of their shifts is an integer multiple of \(2\pi\).
Examples
Consider a sound wave represented by a sine function. If you have two speakers playing the exact same tone, the sound waves they produce will overlap perfectly (constructive interference) if they are "in phase." In mathematical terms, this means their phase shifts differ by an integer multiple of their period, \(2\pi\). This principle is essential in audio engineering for creating sound systems where multiple speakers work together to produce a clear, coherent sound, or in noise-cancellation technology where waves are shifted by \(\pi\) to cancel each other out. Understanding this relationship helps in managing wave synchronization in various fields like telecommunications and acoustics.