Topic 3.11 Notes – The Secant, Cosecant, and Cotangent Functions

What Secant, Cosecant, and Cotangent Are

These are reciprocal trig functions.

$$\sec \theta=\frac{1}{\cos \theta}, \qquad \cos \theta \ne 0$$

$$\csc \theta=\frac{1}{\sin \theta}, \qquad \sin \theta \ne 0$$

$$\cot \theta=\frac{\cos \theta}{\sin \theta}, \qquad \sin \theta \ne 0$$

Also,

$$\cot \theta=\frac{1}{\tan \theta}$$

but only where tangent exists and is not zero.

That restriction matters. If a denominator is $0$, the function is undefined, and on the graph that usually means a vertical asymptote.

A very common confusion is reciprocal vs inverse.

• $\sec \theta$, $\csc \theta$, and $\cot \theta$ are reciprocals
• $\cos^{-1}x$ means arccos, the inverse trig function
• $\sec^2\theta$ means $(\sec\theta)^2$

Key Characteristics of the Three Graphs

Secant

Because $\sec\theta=\frac{1}{\cos\theta}$, it is undefined where $\cos\theta=0$, so at $\theta=\frac{\pi}{2}+k\pi$.

• The graph stays outside the strip $-1<y<1$
• Key points where cosine is $\pm1$ stay fixed
• Examples: $(0,1)$, $(\pi,-1)$
• No zeros, because a reciprocal can’t equal $0$

Cosecant

Because $\csc\theta=\frac{1}{\sin\theta}$, it is undefined where $\sin\theta=0$, so at $\theta=k\pi$.

• Also stays outside $-1<y<1$
• Fixed points happen where sine is $\pm1$
• Examples: $(\pi/2,1)$, $(3\pi/2,-1)$
• No zeros

Cotangent

Because $\cot\theta=\frac{\cos\theta}{\sin\theta}$, it is undefined where $\sin\theta=0$, so again at $\theta=k\pi$.

• Zeros happen where $\cos\theta=0$, so $\theta=\frac{\pi}{2}+k\pi$
• On $(0,\pi)$, it decreases from $+\infty$ through $(\pi/2,0)$ to $-\infty$

Finding Values and Graphing from Sine and Cosine

The fastest way to get values is to use sine and cosine first.

1. Find $\sin\theta$ and $\cos\theta$
2. Take reciprocals for secant or cosecant
3. Use $\frac{\cos\theta}{\sin\theta}$ for cotangent
4. Check the denominator before doing anything

Example with $\theta=\frac{5\pi}{6}$:

• $\sin\frac{5\pi}{6}=\frac12$
• $\cos\frac{5\pi}{6}=-\frac{\sqrt3}{2}$

So,

• $\csc\frac{5\pi}{6}=2$
• $\sec\frac{5\pi}{6}=-\frac{2\sqrt3}{3}$
• $\cot\frac{5\pi}{6}=-\sqrt3$

Sign habits help a lot:

• secant has the same sign as cosine
• cosecant has the same sign as sine
• cotangent matches $\frac{\cos\theta}{\sin\theta}$, so it is positive in Quadrants I and III, negative in II and IV

For graphing, think from the parent graph:

• secant from cosine or cosecant from sine
• denominator zeros become vertical asymptotes
• points where the denominator is $1$ or $-1$ stay put
• other nonzero values turn into reciprocals
• branches live above $y=1$ or below $y=-1$

For cotangent, use the quotient view:

• sine zeros give asymptotes
• cosine zeros give x-intercepts
• each branch decreases

Transformed Inputs and Function Identification

A transformed input changes where asymptotes and key points happen.

Example:

$$\sec\left(2\theta-\frac{\pi}{3}\right)$$

Asymptotes happen when the cosine denominator is zero:

$$2\theta-\frac{\pi}{3}=\frac{\pi}{2}+k\pi$$

Solve for $\theta$. That gives the actual asymptote locations. The coefficient $2$ also changes the period from $2\pi$ to $\pi$. The usual secant range still stays $(-∞,-1]\cup[1,∞)$.

To identify a graph:

• asymptotes at $\frac{\pi}{2}+k\pi$, no zeros, outside $-1<y<1$ → secant
• asymptotes at $k\pi$, no zeros, outside $-1<y<1$ → cosecant
• asymptotes at $k\pi$, zeros at $\frac{\pi}{2}+k\pi$, decreasing branches → cotangent

Common mistakes show up a lot on quizzes:

• treating reciprocals like reflections
• calling denominator zeros the zeros of secant or cosecant
• using $\cot\theta=\frac{1}{\tan\theta}$ when tangent is undefined
• copying parent asymptotes into transformed functions without solving for the new input variable