question 3 (4 points)\n1. draw f(x) = sin x and g(x) = csc x on the same grid (for 0 ≤ x ≤ 2π)\n2. explain…

Explanation:

Step1: Recall reciprocal relationship

We know that $\csc x=\frac{1}{\sin x}$.

Step2: Analyze vertical - asymptote condition

A function $y = g(x)$ has a vertical asymptote where the denominator of a rational - form of the function is zero. For $g(x)=\csc x=\frac{1}{\sin x}$, vertical asymptotes occur when $\sin x = 0$.

Step3: Find values of $x$ for $\sin x = 0$

In the interval $0\leq x\leq2\pi$, $\sin x = 0$ when $x = 0,\pi,2\pi$. In general, $\sin x=0$ when $x = n\pi$, where $n\in\mathbb{Z}$ (the set of all integers).

1. To draw $y = \sin x$:
   • When $x = 0$, $\sin(0)=0$; when $x=\frac{\pi}{2}$, $\sin(\frac{\pi}{2}) = 1$; when $x=\pi$, $\sin(\pi)=0$; when $x=\frac{3\pi}{2}$, $\sin(\frac{3\pi}{2})=-1$; when $x = 2\pi$, $\sin(2\pi)=0$. The graph of $y = \sin x$ is a smooth wave - like curve.
   • For $y=\csc x=\frac{1}{\sin x}$, when $\sin x$ is close to $0$, $\csc x$ has large positive or negative values. At $x = 0,\pi,2\pi$, $\csc x$ has vertical asymptotes. When $\sin x = 1$ (i.e., $x=\frac{\pi}{2}$), $\csc x = 1$; when $\sin x=-1$ (i.e., $x=\frac{3\pi}{2}$), $\csc x=-1$.
2. $g(x)=\csc x$ has vertical asymptotes because $\csc x=\frac{1}{\sin x}$, and vertical asymptotes occur at the values of $x$ for which the denominator $\sin x = 0$.
3. The vertical asymptotes of $g(x)=\csc x$ are given by the expression $x = n\pi$, where $n\in\mathbb{Z}$.

Answer:

1. Graph $y = \sin x$ as a wave - like curve with key points $(0,0),(\frac{\pi}{2},1),(\pi,0),(\frac{3\pi}{2},-1),(2\pi,0)$ and graph $y=\csc x$ with vertical asymptotes at $x = 0,\pi,2\pi$ and passing through $(\frac{\pi}{2},1),(\frac{3\pi}{2},-1)$.
2. Because $\csc x=\frac{1}{\sin x}$ and vertical asymptotes occur when the denominator $\sin x = 0$.
3. $x = n\pi$, $n\in\mathbb{Z}$.