Question

which function has a graph that overlaps the graph of \\(\sin x\\)? (1 point)\\(\bigcirc f(x)=\sin(x-\pi)\\)\\(\bigcirc f(x)=\sin\left(x+\frac{\pi}{2}\
ight)\\)\\(\bigcirc f(x)=\sin(x+\pi)\\)\\(\bigcirc f(x)=\sin(x+2\pi)\\)

Explanation:

Step1: Use the periodicity of the sine function

The sine function \(y = \sin(x)\) has a period of \(2\pi\). That is, \(\sin(x + 2k\pi)=\sin(x)\) for all \(x\in R\) and \(k\in Z\).

Step2: Analyze each option

• For \(y = \sin(x-\pi)\), using the identity \(\sin(A - B)=\sin A\cos B-\cos A\sin B\), \(\sin(x-\pi)=-\sin x\).
• For \(y=\sin(x + \frac{\pi}{2})\), using the identity \(\sin(A + B)=\sin A\cos B+\cos A\sin B\), \(\sin(x+\frac{\pi}{2})=\cos x\).
• For \(y=\sin(x+\pi)\), using the identity \(\sin(A + B)=\sin A\cos B+\cos A\sin B\), \(\sin(x+\pi)=-\sin x\).
• For \(y=\sin(x + 2\pi)\), since the period of \(y = \sin(x)\) is \(T = 2\pi\), by the formula \(\sin(x+2k\pi)=\sin x\) (\(k = 1\) here), \(\sin(x + 2\pi)=\sin x\).

Answer:

\(f(x)=\sin(x + 2\pi)\)