Identify the Function

The given function is $f(x) = \sin x \cos x$.

Rewrite the Function using Trigonometric Identity

We can simplify this expression using the double angle identity for sine: $ \sin(2x) = 2 \sin x \cos x $ Rearranging this, we get: $ \sin x \cos x = \frac{1}{2} \sin(2x) $ Therefore, the function can be written as: $ f(x) = \frac{1}{2} \sin(2x) $

Analyze Key Features of $f(x) = \frac{1}{2} \sin(2x)$

• Amplitude: The amplitude is the coefficient of the sine function, which is $\frac{1}{2}$. The graph will oscillate between $-\frac{1}{2}$ and $\frac{1}{2}$.
• Period: The period of $\sin(kx)$ is $\frac{2\pi}{|k|}$. For $f(x) = \frac{1}{2} \sin(2x)$, $k=2$, so the period is $\frac{2\pi}{2} = \pi$. This means the function completes one full cycle over an interval of length $\pi$.
• Zeros: The function equals zero when $\sin(2x) = 0$. This occurs when $2x = n\pi$, where $n$ is an integer. Solving for $x$, we get $x = \frac{n\pi}{2}$. The zeros are located at $..., -\pi, -\frac{\pi}{2}, 0, \frac{\pi}{2}, \pi, ...$.

Compare Features with Graph Option B

Let's examine the graph corresponding to Option B:

• The graph's highest points are at $y = \frac{1}{2}$ and the lowest points are at $y = -\frac{1}{2}$, matching the amplitude of $\frac{1}{2}$.
• One complete cycle of the wave occurs between $x=0$ and $x=\pi$, confirming the period of $\pi$.
• The graph crosses the x-axis at $x=0, \frac{\pi}{2}, \pi$, and subsequent intervals, matching the zeros at $x = \frac{n\pi}{2}$.

Based on these characteristics, the graph in Option B accurately represents the function $f(x) = \sin x \cos x$.