Tangent Function

Approach:

The tangent is the ratio of the ordinate to the abscissa on the unit circle, so it is undefined exactly where the abscissa vanishes, that is at $x = \frac{\pi}{2} + k\pi$, and it takes every real value in between.

Steps:

1. By the unit-circle definition $\tan x = \frac{\sin x}{\cos x}$, so the expression is undefined when $\cos x = 0$, that is when $x = \frac{\pi}{2} + k\pi$.
2. Between two consecutive such points the function rises continuously from $-\infty$ to $+\infty$.
3. Therefore the range of $y = \tan x$ is all of $\mathbb{R}$.

Answer:

Domain: $x \neq \frac{\pi}{2} + k\pi$; Range: $\mathbb{R}$.

Approach:

Read the period off the addition formula for the tangent and check its parity by comparing $\tan(-x)$ with $\tan x$.

Steps:

1. The addition formula gives $\tan(x + \pi) = \tan x$, so $\pi$ is a period; the poles occur at $x = \frac{\pi}{2} + k\pi$, exactly $\pi$ apart, so no smaller positive number can be a period.
2. Computing $f(-x) = \tan(-x) = -\tan x = -f(x)$, the function is odd.
3. Hence the period is $\pi$ and the graph is symmetric about the origin.

Answer:

Period is $\pi$, it is an odd function.

Approach:

A trigonometric function is one whose argument is an angle, so check whether each expression is built from the sine, cosine or tangent of the variable $x$.

Steps:

1. $y = \sin x$ takes the angle $x$ and applies the sine function, so it is a trigonometric function.
2. $y = x^2$ is a polynomial in $x$ and has no angle argument, so it is not a trigonometric function.
3. $y = \tan x$ takes the angle $x$ and applies the tangent function, so it is a trigonometric function.
4. $y = e^x$ is an exponential function and has no angle argument, so it is not a trigonometric function.

Answer:

$y = \sin x$, $y = \tan x$ are trigonometric functions, $y = x^2$, $y = e^x$ are not.

Approach:

The tangent is the ratio of the ordinate to the abscissa on the unit circle, so it is undefined exactly where the abscissa vanishes, that is at $x = \frac{\pi}{2} + k\pi$, and it takes every real value in between.

Steps:

1. By the unit-circle definition $\tan x = \frac{\sin x}{\cos x}$, so the expression is undefined when $\cos x = 0$, that is when $x = \frac{\pi}{2} + k\pi$.
2. Between two consecutive such points the function rises continuously from $-\infty$ to $+\infty$.
3. Therefore the range of $y = \tan x$ is all of $\mathbb{R}$.

Answer:

Domain: $x \neq \frac{\pi}{2} + k\pi$; Range: $\mathbb{R}$.

Approach:

Test each option with $f(-x) = \pm f(x)$: the sine and the tangent change sign under the substitution, while the cosine and the square do not.

Steps:

1. (A) $f(-x) = \sin(-x) = -\sin x = -f(x)$, so $y = \sin x$ is an odd function.
2. (B) $f(-x) = \cos(-x) = \cos x = f(x)$, so $y = \cos x$ is an even function.
3. (C) $f(-x) = \tan(-x) = -\tan x = -f(x)$, so $y = \tan x$ is an odd function.
4. (D) $f(-x) = (-x)^2 = x^2 = f(x)$, so $y = x^2$ is an even function.

Answer:

(A), (C) are odd functions, (B), (D) are not.

Approach:

Read the period off the addition formula for the tangent and the domain off the places where the cosine vanishes.

Steps:

1. The addition formula gives $\tan(x + \pi) = \tan x$, so $\pi$ is a period, and the poles at $x = \frac{\pi}{2} + k\pi$ are exactly $\pi$ apart, so no smaller positive number can be a period.
2. The expression $\tan x = \frac{\sin x}{\cos x}$ is undefined when $\cos x = 0$, that is when $x = \frac{\pi}{2} + k\pi$.
3. Hence the period is $\pi$ and the domain excludes precisely those points.

Answer:

Period is $\pi$, domain is $x \neq \frac{\pi}{2} + k\pi$.