Mathematics Advanced • Year 11 • Module 2 • Lesson 6
Reciprocal Trigonometric Functions
Build procedural fluency in evaluating csc, sec, and cot at standard reference angles and identifying where each reciprocal function is undefined.
1. Quick recall
Answer each question in the space provided. 1 mark each
Q1.1 Complete the three reciprocal definitions:
csc θ = 1 / __________ sec θ = 1 / __________ cot θ = __________ / sin θ
Q1.2 Match each reciprocal function with the angles at which it is undefined.
csc θ undefined when θ = ____________________
sec θ undefined when θ = ____________________
cot θ undefined when θ = ____________________
Q1.3 Why does $\csc \theta \cdot \sin \theta = 1$ wherever it is defined? Answer in one sentence using the word "reciprocal".
2. Worked example, evaluating sec(2π/3)
Every line of algebra is annotated. Follow it through, then attempt the faded version below.
Problem. Find the exact value of $\sec \frac{2\pi}{3}$.
Step 1, Rewrite using the reciprocal definition.
sec(2π/3) = 1 / cos(2π/3)
Reason: every reciprocal evaluation starts by writing the original function in the denominator.
Step 2, Locate 2π/3 on the unit circle and find the reference angle.
2π/3 is in Quadrant II. Reference angle = π − 2π/3 = π/3.
Reason: ASTC gives the sign; reference angle gives the magnitude.
Step 3, Apply ASTC and read the exact value.
cos is negative in QII. cos(π/3) = 1/2, so cos(2π/3) = −1/2.
Reason: in QII only sine (and its reciprocal) are positive.
Step 4, Take the reciprocal to finish.
sec(2π/3) = 1 / (−1/2) = −2.
Reason: reciprocals preserve the sign, cos negative ⇒ sec negative (Trap 1 in the lesson).
Conclusion. $\sec \frac{2\pi}{3} = \mathbf{-2}$.
3. Faded example, fill in the missing steps
Find the exact value of $\csc \frac{5\pi}{4}$. Fill in each blank line. 4 marks
Step 1, Reciprocal definition.
csc(5π/4) = 1 / ________
Step 2, Quadrant and reference angle.
5π/4 lies in Quadrant ________. Reference angle = 5π/4 − π = ________.
Step 3, Apply ASTC.
In QIII, sin is ________ (positive / negative). sin(π/4) = ________, so sin(5π/4) = ________.
Step 4, Take reciprocal and rationalise.
csc(5π/4) = 1 / ________ = ________ (rationalise the denominator).
Conclusion. $\csc \frac{5\pi}{4}$ = ____________.
4. Graduated practice
Leave all answers in exact form rationalise denominators where needed. Show one line of working at minimum.
Foundation, standard reference angles (4 questions)
| Q | Find | Working (1 line) | Answer |
|---|---|---|---|
| 4.1 1 | $\csc \frac{\pi}{6}$ | ||
| 4.2 1 | $\sec \frac{\pi}{4}$ | ||
| 4.3 1 | $\cot \frac{\pi}{4}$ | ||
| 4.4 1 | $\cot \frac{\pi}{3}$ |
Standard, ASTC + reciprocal (6 questions)
Each requires identifying the quadrant, applying ASTC to the base function, then taking the reciprocal.
4.5 Find $\sec \frac{5\pi}{6}$ exactly. 2 marks
4.6 Find $\csc \frac{7\pi}{6}$ exactly. 2 marks
4.7 Find $\cot \frac{5\pi}{3}$ exactly. 2 marks
4.8 Find $\sec \frac{4\pi}{3}$ exactly. 2 marks
4.9 State all values of $\theta$ in $[0, 2\pi]$ for which $\sec \theta$ is undefined. 2 marks
4.10 Simplify $\tan \theta \cdot \cot \theta$ (state any restrictions on $\theta$). 2 marks
Extension, combine identities (2 questions)
4.11 Simplify $\frac{\csc \theta}{\cot \theta}$ to a single trigonometric function. State the restrictions on $\theta$. 3 marks
4.12 Find the exact value of $\csc \frac{11\pi}{6} \cdot \sec \frac{11\pi}{6}$, leaving the answer rationalised. 3 marks
5. Self-check the easy 3
Tick the first three once you've verified your method works.
How did this worksheet feel?
What I'll revisit before next class:
Q1.1, Reciprocal definitions
$\csc \theta = \frac{1}{\sin \theta}$, $\sec \theta = \frac{1}{\cos \theta}$, $\cot \theta = \frac{\cos \theta}{\sin \theta}$ (also equal to $\frac{1}{\tan \theta}$).
Q1.2, Where each is undefined
$\csc \theta$ undefined when $\sin \theta = 0$, i.e. $\theta = n\pi$, $n \in \mathbb{Z}$.
$\sec \theta$ undefined when $\cos \theta = 0$, i.e. $\theta = \frac{\pi}{2} + n\pi$.
$\cot \theta$ undefined when $\sin \theta = 0$, i.e. $\theta = n\pi$ (same as csc).
Q1.3, Why the product is 1
By definition $\csc \theta = \frac{1}{\sin \theta}$. The product of any non-zero number with its reciprocal is 1, so $\csc \theta \cdot \sin \theta = \frac{\sin \theta}{\sin \theta} = 1$ (provided $\sin \theta \neq 0$).
Q3, Faded example: $\csc \frac{5\pi}{4}$
Step 1: $\csc \frac{5\pi}{4} = \frac{1}{\sin \frac{5\pi}{4}}$.
Step 2: QIII; reference angle = π/4.
Step 3: In QIII, sine is negative. $\sin \frac{\pi}{4} = \frac{\sqrt{2}}{2}$, so $\sin \frac{5\pi}{4} = -\frac{\sqrt{2}}{2}$.
Step 4: $\csc \frac{5\pi}{4} = \frac{1}{-\sqrt{2}/2} = -\frac{2}{\sqrt{2}} = -\sqrt{2}$ (after rationalising).
Conclusion: $\csc \frac{5\pi}{4} = \mathbf{-\sqrt{2}}$.
Q4.1, $\csc \frac{\pi}{6}$
$\csc \frac{\pi}{6} = \frac{1}{\sin \frac{\pi}{6}} = \frac{1}{1/2} = \mathbf{2}$.
Q4.2, $\sec \frac{\pi}{4}$
$\sec \frac{\pi}{4} = \frac{1}{\cos \frac{\pi}{4}} = \frac{1}{\sqrt{2}/2} = \frac{2}{\sqrt{2}} = \mathbf{\sqrt{2}}$.
Q4.3, $\cot \frac{\pi}{4}$
$\cot \frac{\pi}{4} = \frac{1}{\tan \frac{\pi}{4}} = \frac{1}{1} = \mathbf{1}$.
Q4.4, $\cot \frac{\pi}{3}$
$\cot \frac{\pi}{3} = \frac{1}{\tan \frac{\pi}{3}} = \frac{1}{\sqrt{3}} = \mathbf{\frac{\sqrt{3}}{3}}$ (rationalised).
Q4.5, $\sec \frac{5\pi}{6}$
QII, reference $\frac{\pi}{6}$. $\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}$; in QII cos negative, so $\cos \frac{5\pi}{6} = -\frac{\sqrt{3}}{2}$. $\sec \frac{5\pi}{6} = \frac{1}{-\sqrt{3}/2} = -\frac{2}{\sqrt{3}} = \mathbf{-\frac{2\sqrt{3}}{3}}$.
Q4.6, $\csc \frac{7\pi}{6}$
QIII, reference $\frac{\pi}{6}$. $\sin \frac{\pi}{6} = \frac{1}{2}$; in QIII sin negative, so $\sin \frac{7\pi}{6} = -\frac{1}{2}$. $\csc \frac{7\pi}{6} = \frac{1}{-1/2} = \mathbf{-2}$.
Q4.7, $\cot \frac{5\pi}{3}$
QIV, reference $\frac{\pi}{3}$. $\tan \frac{\pi}{3} = \sqrt{3}$; in QIV tan negative, so $\tan \frac{5\pi}{3} = -\sqrt{3}$. $\cot \frac{5\pi}{3} = \frac{1}{-\sqrt{3}} = \mathbf{-\frac{\sqrt{3}}{3}}$.
Q4.8, $\sec \frac{4\pi}{3}$
QIII, reference $\frac{\pi}{3}$. $\cos \frac{\pi}{3} = \frac{1}{2}$; in QIII cos negative, so $\cos \frac{4\pi}{3} = -\frac{1}{2}$. $\sec \frac{4\pi}{3} = \frac{1}{-1/2} = \mathbf{-2}$.
Q4.9, Where $\sec \theta$ is undefined on $[0, 2\pi]$
$\sec \theta$ is undefined where $\cos \theta = 0$. On $[0, 2\pi]$: $\theta = \mathbf{\frac{\pi}{2}, \frac{3\pi}{2}}$.
Q4.10, $\tan \theta \cdot \cot \theta$
$\tan \theta \cdot \cot \theta = \tan \theta \cdot \frac{1}{\tan \theta} = \mathbf{1}$. Restrictions: $\sin \theta \neq 0$ and $\cos \theta \neq 0$, i.e. $\theta \neq \frac{n\pi}{2}$ for any integer $n$ (both $\tan$ and $\cot$ must be defined).
Q4.11, $\frac{\csc \theta}{\cot \theta}$
$\frac{\csc \theta}{\cot \theta} = \frac{1/\sin \theta}{\cos \theta / \sin \theta} = \frac{1}{\sin \theta} \times \frac{\sin \theta}{\cos \theta} = \frac{1}{\cos \theta} = \mathbf{\sec \theta}$. Restrictions: $\sin \theta \neq 0$ and $\cos \theta \neq 0$, so $\theta \neq \frac{n\pi}{2}$.
Q4.12, $\csc \frac{11\pi}{6} \cdot \sec \frac{11\pi}{6}$
$\frac{11\pi}{6}$ is in QIV; reference $\frac{\pi}{6}$. $\sin \frac{11\pi}{6} = -\frac{1}{2}$, $\cos \frac{11\pi}{6} = \frac{\sqrt{3}}{2}$. So $\csc \frac{11\pi}{6} = -2$ and $\sec \frac{11\pi}{6} = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3}$. Product: $(-2) \cdot \frac{2\sqrt{3}}{3} = \mathbf{-\frac{4\sqrt{3}}{3}}$.