Question

determine the equation of the sine function shown in the graph. (1 point) \\( \bigcirc f(x) = \sin x - \frac{\pi}{2} \\) \\( \bigcirc f(x) = \sin\left(x - \frac{\pi}{2}\
ight) \\) \\( \bigcirc f(x) = \sin x + \frac{\pi}{2} \\) \\( \bigcirc f(x) = \sin\left(x + \frac{\pi}{2}\
ight) \\)

Explanation:

Step1: Recall the transformation rules of sine function

The general form of a sine function is \(y = A\sin(B(x - C))+D\). Here \(A = 1\), \(B = 1\), \(D=0\). The phase - shift formula is \(x = C\). The standard sine function \(y=\sin(x)\) has a minimum at \(x=\frac{3\pi}{2}+ 2k\pi,k\in\mathbb{Z}\), and the given function has a minimum at \(x = 0\).

Step2: Apply the phase - shift formula

For the sine function \(y=\sin(x - C)\), when \(x = C\), \(y=\sin(0)\). We know that \(\sin(x)\) shifted to get \(\sin(x +\frac{\pi}{2})\) because \(\sin(x+\frac{\pi}{2})=\cos(x)\). The standard \(\sin(x)\) graph: \(y = \sin(x)\) has key points \((0,0)\), \((\frac{\pi}{2},1)\), \((\pi,0)\), \((\frac{3\pi}{2},- 1)\), \((2\pi,0)\). Using the identity \(\sin(x+\frac{\pi}{2})=\cos(x)\), and the graph of \(y = \cos(x)\) has key points \((0,1)\), \((\frac{\pi}{2},0)\), \((\pi,-1)\), \((\frac{3\pi}{2},0)\), \((2\pi,1)\). If we consider the transformation of \(y = \sin(x)\) to \(y=\sin(x +\frac{\pi}{2})\), using the formula \(y=\sin(x - C)\) (where \(C=-\frac{\pi}{2}\)), the graph of \(y = \sin(x)\) is shifted to the left by \(\frac{\pi}{2}\) units.

Answer:

\(f(x)=\sin(x+\frac{\pi}{2})\)