swyk u6 l11/12 - domain and tangent functions\nhere is a graph of f given by f(θ)=tan(θ).\na. are π/2 and…

Explanation:

Step1: Recall tangent - function definition

The tangent function is defined as $f(\theta)=\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}$. The function is undefined when $\cos(\theta) = 0$.

Step2: Check $\frac{\pi}{2}$ and $\frac{3\pi}{2}$ for domain

We know that $\cos(\frac{\pi}{2})=0$ and $\cos(\frac{3\pi}{2}) = 0$. Since $\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}$, when $\cos(\theta)=0$, the function is undefined. So, $\frac{\pi}{2}$ and $\frac{3\pi}{2}$ are not in the domain of $f$.

Step3: Recall $\theta$-intercept definition

The $\theta$-intercepts of the graph of $y = f(\theta)$ are the values of $\theta$ for which $f(\theta)=0$. Since $f(\theta)=\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}$, $f(\theta) = 0$ when $\sin(\theta)=0$ and $\cos(\theta)\neq0$.

Step4: Find $\theta$-intercepts

We know that $\sin(\theta)=0$ when $\theta = k\pi$, where $k\in\mathbb{Z}$. And $\cos(k\pi)\neq0$ for $k\in\mathbb{Z}$. So the $\theta$-intercepts of the graph of $f$ are $\theta = k\pi$, where $k\in\mathbb{Z}$.

Answer:

a. No. Because $\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}$ and $\cos(\frac{\pi}{2}) = 0$, $\cos(\frac{3\pi}{2})=0$, so $\frac{\pi}{2}$ and $\frac{3\pi}{2}$ make the function undefined. b. $\theta = k\pi$, $k\in\mathbb{Z}$. Because $\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}$ and $\tan(\theta) = 0$ when $\sin(\theta)=0$ and $\cos(\theta)\neq0$, and $\sin(\theta)=0$ at $\theta = k\pi$, $k\in\mathbb{Z}$ with $\cos(k\pi)\neq0$.