Cotangent Function

Approach:

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

Steps:

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

Answer:

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

Approach:

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

Steps:

1. The addition formula gives $\cot(x + \pi) = \cot x$, so $\pi$ is a period; the poles occur at $x = k\pi$, exactly $\pi$ apart, so no smaller positive number can be a period.
2. Computing $f(-x) = \cot(-x) = -\cot x = -f(x)$, 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:

Test each option with $f(-x) = \pm f(x)$: the sine and the cotangent 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) $f(-x) = \cot(-x) = -\cot x = -f(x)$, so $y = \cot x$ 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.