Reciprocal Trig Functions and Graphs (Year 11)

01 / LEARN

Understand the idea.

Reciprocal is not inverse

A reciprocal trig function takes the reciprocal of a ratio: secant uses cosine, cosecant uses sine, and cotangent is cosine divided by sine. This differs from an inverse trig function, which returns an angle from a ratio.

Exclude every zero of the original denominator. Secant excludes $\frac{π}{2} + k π$ , while cosecant and cotangent exclude $k π$ , for integers $k$ . Near those excluded values the graphs have vertical asymptotes; the function has no value on an asymptote.

\begin{aligned} sec x & = \frac{1}{cos x} \\ csc x & = \frac{1}{sin x} \\ cot x & = \frac{cos x}{sin x} \end{aligned}

Sketch separate branches

Taking the reciprocal of a nonzero sine or cosine produces magnitude at least one. Thus secant and cosecant have range $(- \infty, - 1] \cup [1, \infty)$ and period $2 π$ . Their vertices occur where the original ratio equals $1$ or $- 1$ .

Cotangent has range $ℝ$ , period $π$ and zeros where cosine vanishes but sine does not. Sketch each branch separately across asymptotes. For equations, convert to sine or cosine while retaining the original nonzero-denominator restriction.

The graph y equals secant x from zero to two pi radians, with vertical window minus four to four and separate branches. Dashed asymptotes lie at pi divided by two and three pi divided by two.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Example 1

State the domain and range of $y = sec x$ .

Worked solution

1
Exclude the zeros of cosine.
$x \neq \frac{π}{2} + k π, k \in ℤ$
Secant equals $1 / cos x$ , so its denominator must be nonzero.
2
Bound the reciprocal magnitude.
$0 < ∣ cos x ∣ \leq 1 \Rightarrow ∣ sec x ∣ \geq 1$
Taking reciprocals of positive magnitudes reverses this inequality.
3
Check endpoints and state the range.
$sec 0 = 1, sec π = - 1, y \in (- \infty, - 1] \cup [1, \infty)$
The endpoint values occur, and the continuous branches grow without bound near cosine zeros.

Answer

Domain excludes $\frac{π}{2} + k π$ , $k \in ℤ$ ; range $(- \infty, - 1] \cup [1, \infty)$ .

EXAMPLE 02

Example 2

Solve $csc x = 2$ on $[0, 2 π)$ .

Worked solution

1
Record the original restriction.
$sin x \neq 0$
Cosecant is defined only when its sine denominator is nonzero.
2
Rewrite in terms of sine.
$\frac{1}{sin x} = 2 \Rightarrow sin x = \frac{1}{2}$
Multiplication by the nonzero sine is valid on the original domain.
3
Find and check both branches.
$x = \frac{π}{6}, \frac{5 π}{6}$
Both quadrant I and II values lie in $[0, 2 π)$ and have nonzero sine, so both solve the original equation.

Answer

$x = \frac{π}{6}, \frac{5 π}{6}$ .

03 / YOUR TURN

Try it yourself.

Check your understanding

Where is $cot x = 0$ ?

Show answer & reasoning

$x = \frac{π}{2} + k π$ , $k \in ℤ$ .

For $cot x = cos x / sin x$ to vanish, cosine must be zero and sine must be nonzero. This occurs at $x = \frac{π}{2} + k π$ , with $k \in ℤ$ .

Keep practising

6 worksheet questions · 12 marks

Open worksheet Worked solutions

Read the questions online

Find $sec (\frac{π}{3})$ .
2 marks
Find $csc (\frac{7 π}{6})$ .
2 marks
State the excluded values for $csc (2 x)$ .
2 marks
Solve $sec x = - 2$ on $[0, 2 π)$ .
2 marks
State the range of $2 sec x + 1$ .
2 marks
Find the period and zeros of $cot (2 x)$ .
2 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Find $sec (\frac{π}{3})$ .
2 marks
Take the reciprocal of $cos (\frac{π}{3}) = \frac{1}{2}$ .
2.
Find $csc (\frac{7 π}{6})$ .
2 marks
The sine is $- \frac{1}{2}$ ; its reciprocal is $- 2$ .
-2.
State the excluded values for $csc (2 x)$ .
2 marks
Exclude $sin (2 x) = 0$ , giving $2 x = k π$ .
$x = \frac{k π}{2}$ , $k \in ℤ$ .
Solve $sec x = - 2$ on $[0, 2 π)$ .
2 marks
Take reciprocals to solve $cos x = - \frac{1}{2}$ .
$x = \frac{2 π}{3}, \frac{4 π}{3}$ .
State the range of $2 sec x + 1$ .
2 marks
Scale secant by 2 and shift both branches up 1.
$(- \infty, - 1] \cup [3, \infty)$ .
Find the period and zeros of $cot (2 x)$ .
2 marks
Cotangent repeats when its argument increases by $π$ ; its zeros occur at argument $\frac{π}{2} + k π$ .
Period $\frac{π}{2}$ ; zeros $x = \frac{π}{4} + \frac{k π}{2}$ .

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 Mathematics, Further Trigonometry: periodic functions and models, trigonometric identities and reciprocal functions; the Library sequence follows the approved school-topic structure.

Official curriculum source ↗

Reciprocal trig functions and their graphs

In this lesson

Understand the idea.

Reciprocal is not inverse

Sketch separate branches

See the method in action.

Example 1

Exclude the zeros of cosine.

Bound the reciprocal magnitude.

Check endpoints and state the range.

Example 2

Record the original restriction.

Rewrite in terms of sine.

Find and check both branches.

Try it yourself.

Check your understanding

Keep practising