Question

which of the following statements about reflecting periodic functions is correct? (1 point)

• if $f(x)=\sin x$ and $g(x)=\cos(-x)$, then $g(x)$ is a reflection of $f(x)$ across the $y$-axis.
• if $f(x)=\cos x$ and $g(x)=\cos(-x)$, then $g(x)$ is a reflection of $f(x)$ across the $y$-axis.
• if $f(x)=\sin x$ and $g(x)=\sin x$, then $g(x)$ is a reflection of $f(x)$ across the $y$-axis.
• if $f(x)=\sin x$ and $g(x)=\sin(-x)$, then $g(x)$ is a reflection of $f(x)$ across the $x$-axis.

Explanation:

Step1: Recall reflection rules

A reflection of \( f(x) \) across the \( y \)-axis is \( f(-x) \), and across the \( x \)-axis is \( -f(x) \). Also, \( \cos(-x)=\cos x \) (even function), \( \sin(-x)=-\sin x \) (odd function).

Step2: Analyze each option

• Option 1: \( f(x)=\sin x \), \( g(x)=\cos(-x)=\cos x \). \( \sin x \) and \( \cos x \) are different functions, not reflections. Eliminate.
• Option 2: \( f(x)=\cos x \), \( g(x)=\cos(-x)=\cos x \). Since \( \cos(-x)=\cos x = f(-x) \) (as \( \cos x \) is even), \( g(x)=f(-x) \), so it's a reflection across \( y \)-axis (since \( f(-x) \) is \( y \)-axis reflection). Check others.
• Option 3: \( f(x)=\sin x \), \( g(x)=\sin x \). No reflection (same function). Eliminate.
• Option 4: \( g(x)=\sin(-x)=-\sin x \), which is \( -f(x) \), a reflection across \( x \)-axis. But let's confirm Option 2 first. Wait, Option 2: \( f(x)=\cos x \), \( g(x)=\cos(-x)=f(-x) \), so it's \( y \)-axis reflection. Option 4: \( g(x)=-f(x) \), \( x \)-axis reflection, but let's check again. Wait, the fourth option has a typo? " \( g(x) - \sin(-x) \)" should be \( g(x)=\sin(-x) \). Then \( \sin(-x)=-\sin x=-f(x) \), so reflection over \( x \)-axis. But Option 2: \( f(x)=\cos x \), \( g(x)=\cos(-x)=f(-x) \), so \( y \)-axis reflection (since \( f(-x) \) is \( y \)-axis reflection, and \( \cos(-x)=\cos x = f(x) \)? Wait, no: \( f(x)=\cos x \), so \( f(-x)=\cos(-x)=\cos x = g(x) \). So \( g(x)=f(-x) \), which is reflection over \( y \)-axis. Wait, but \( \cos x \) is even, so \( f(-x)=f(x) \). So this is correct? Wait, let's re - check.

Wait, for a function \( f(x) \), reflection over \( y \)-axis is \( f(-x) \). If \( f(x) \) is even, \( f(-x)=f(x) \). So for \( f(x)=\cos x \), \( f(-x)=\cos(-x)=\cos x = g(x) \), so \( g(x) \) is \( f(-x) \), which is reflection over \( y \)-axis (even though it's the same as \( f(x) \) because \( \cos x \) is even). Now check Option 4: \( g(x)=\sin(-x)=-\sin x=-f(x) \), which is reflection over \( x \)-axis. But let's check the options again. Wait, the first option: \( f(x)=\sin x \), \( g(x)=\cos x \), not reflection. Third option: same function. So between Option 2 and 4. Wait, Option 2: \( f(x)=\cos x \), \( g(x)=\cos(-x)=f(-x) \), so it's \( y \)-axis reflection (since \( f(-x) \) is \( y \)-axis reflection). Option 4: \( g(x)=-f(x) \), \( x \)-axis reflection. But let's check the functions. For Option 2, since \( \cos(-x)=\cos x \), and \( f(x)=\cos x \), so \( g(x)=f(-x) \), which is the definition of \( y \)-axis reflection. So Option 2 is correct.

Answer:

B. If \( f(x)=\cos x \) and \( g(x)=\cos(-x) \), then \( g(x) \) is a reflection of \( f(x) \) across the \( y \)-axis. (Assuming the options are labeled as A, B, C, D with the second option as B)