Graphing Cotangent

Answer: $\pi$

The period of $y = \cot(x)$ is $\pi$, unlike sine and cosine which have period $2\pi$. This is because cotangent completes one full cycle between consecutive asymptotes.

Answer: $x = n\pi$ for integer $n$

Since $\cot(x) = \frac{\cos(x)}{\sin(x)}$, it is undefined where $\sin(x) = 0$, which occurs at $x = n\pi$ for any integer $n$. This is different from tangent, which has asymptotes at $x = \frac{\pi}{2} + n\pi$.

Answer: 1

Since $\tan(\frac{\pi}{4}) = 1$, and $\cot(x) = \frac{1}{\tan(x)}$, we have $\cot(\frac{\pi}{4}) = \frac{1}{1} = 1$.

Answer: 0

$\cot(\frac{\pi}{2}) = \frac{\cos(\frac{\pi}{2})}{\sin(\frac{\pi}{2})} = \frac{0}{1} = 0$. This is the x-intercept of the cotangent function.

Answer: Decreases

Cotangent decreases from $+\infty$ to $-\infty$ within each period. This is opposite to tangent, which increases. At $x = 0^+$ the value approaches $+\infty$, passes through 0 at $x = \frac{\pi}{2}$, and approaches $-\infty$ as $x \to \pi^-$.

Answer: pi/3

For $y = \cot(Bx)$, the period is $\frac{\pi}{|B|}$. Here $B = 3$, so the period is $\frac{\pi}{3}$. The graph completes 3 cycles in the same interval where the basic cotangent completes 1.

Answer: $\frac{\pi}{3}$ to the right

In the form $y = \cot(x - C)$, the graph shifts right by $C$. Here $C = \frac{\pi}{3}$, so the graph shifts $\frac{\pi}{3}$ units to the right. All asymptotes move from $x = n\pi$ to $x = n\pi + \frac{\pi}{3}$.

Answer: Period 2π, first positive asymptote x = 2π

1. What is the coefficient $B$ of $x$ inside the cotangent? 0.5
2. Apply the period formula $\frac{\pi}{|B|}$. What is the period? 2*pi
3. The asymptotes are where $\frac{x}{2} = n\pi$, so $x = 2n\pi$. What is the first positive asymptote ($n = 1$)? 2*pi

Answer: Vertical stretch by factor 3, vertical shift down 2, period unchanged

1. What does the coefficient 3 in front of cot do? vertical stretch by 3
2. What does the -2 at the end do? vertical shift down 2
3. Does the period change? no

Answer: $x = \frac{\pi}{4}$

The phase shift of $\frac{\pi}{4}$ to the right moves all asymptotes from $x = n\pi$ to $x = n\pi + \frac{\pi}{4}$. For $n = 0$, the asymptote moves from $x = 0$ to $x = \frac{\pi}{4}$, which is the first positive asymptote.

Answer: Phase shift π/6 right, period π/3, first positive asymptote x = π/6

1. Rewrite as $2\cot(3(x - C)) + 1$. What is the phase shift $C$? pi/6
2. What is the period? (Use the formula $\frac{\pi}{|B|}$) pi/3
3. Basic cotangent has an asymptote at $x = 0$. After the phase shift to the right, where is the first positive asymptote? pi/6

Answer: $(-\infty, \infty)$

The range of $y = \cot(x)$ is all real numbers $(-\infty, \infty)$. Adding 5 shifts every output up by 5, but since the range was already all real numbers, adding 5 to every value still gives all real numbers. The range remains $(-\infty, \infty)$.

Answer: 4

The asymptotes are at $\frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}, ...$, so the period is $\frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$. Since Period = $\frac{\pi}{B}$, we have $\frac{\pi}{4} = \frac{\pi}{B}$, so $B = 4$.

Answer: $y = \cot(\frac{1}{2}x - \frac{\pi}{4}) + 3$

1. If period = $2\pi$ and Period = $\frac{\pi}{B}$, what is $B$? 0.5
2. For phase shift $\frac{\pi}{2}$ right with $B = \frac{1}{2}$, what is $c$ in $\cot(Bx - c)$? pi/4
3. What is added for a vertical shift of 3 up? 3

Answer: $y = \cot(6x)$

The asymptotes $\frac{\pi}{6}, \frac{\pi}{3}, \frac{\pi}{2}, ...$ are $\frac{\pi}{6}$ apart, so the period is $\frac{\pi}{6}$ and $B = 6$. For $y = \cot(6x)$ the asymptotes are where $6x = n\pi$, so $x = \frac{n\pi}{6}$: $n = 1, 2, 3$ give exactly $\frac{\pi}{6}, \frac{\pi}{3}, \frac{\pi}{2}$. $y = \cot(6x - \frac{\pi}{2})$ has the right period but is shifted by $\frac{\pi}{12}$, and $y = \cot(3x)$ has period $\frac{\pi}{3}$.