Trigonometric Functions

What you'll learn

How the reciprocal functions $sec⁡\sec$ , $csc⁡\csc$ and $cot⁡\cot$ connect to $sin⁡\sin$ , $cos⁡\cos$ and $tan⁡\tan$ .
How to sketch reciprocal trig graphs using asymptotes and key points.
How to prove and use the identities involving $tan^2 x$ , $sec^2 x$ , $cot^2 x$ and $csc^2 x$ .
How to solve trig equations over stated intervals in radians or degrees.

Foundations: units, signs and reciprocals

Trig questions may use radians or degrees. If the interval contains $π\pi$ , you are working in radians. If it uses 360° or 180°, you are working in degrees. A key conversion is:

radians180^\circ=\pi \text{ radians}

The unit circle is split into four quadrants. Moving anticlockwise from quadrant I, the positive functions are: All, Sin, Tan, Cos. This helps you find every solution, not just the calculator’s first answer.

A reference angle is the acute angle between the line and the nearest horizontal axis.

A unit-circle quadrant diagram showing the ASTC sign rule and how the reference angle is measured from the nearest horizontal axis.

Definition

Reciprocal trigonometric functions

The reciprocal trig functions are defined by

sec⁡x=1cos⁡x,csc⁡x=1sin⁡x,cot⁡x=1tan⁡x=cos⁡xsin⁡x\sec x=\frac{1}{\cos x},\qquad \csc x=\frac{1}{\sin x},\qquad \cot x=\frac{1}{\tan x}=\frac{\cos x}{\sin x}

You may see $csc⁡x\csc x$ written as cosec x; they mean the same thing.

Because these are reciprocals, they are undefined when the denominator is zero. For example, $sec⁡x\sec x$ is undefined when $cos⁡x=0\cos x=0$ .

Example

Solving a reciprocal equation in radians

Solve $3csc⁡θ=73\csc\theta=7$ for $0≤θ≤2π0\le \theta\le 2\pi$ , giving answers to 3 significant figures.

The two unit-circle positions where sine is positive for the same reference angle in the interval from 0 to 2π.

Rewrite using the reciprocal definition:
$3csc⁡θ=7⇒sin⁡θ=373\csc\theta=7 \Rightarrow \sin\theta=\frac{3}{7}$
Find the reference angle in radian mode:
$α=arcsin⁡(37)≈0.443\alpha=\arcsin\left(\frac{3}{7}\right)\approx 0.443$
Since sine is positive in quadrants I and II, use $θ=α\theta=\alpha$ and $θ=π−α\theta=\pi-\alpha$ :
$θ=0.443orθ≈2.70\theta=0.443 \quad \text{or} \quad \theta\approx 2.70$
The solutions are $2.70\theta=0.443,\ 2.70$ .

Graphs of reciprocal functions

The period of a trig graph is the horizontal distance before it repeats. The graphs of $sin⁡x\sin x$ , $cos⁡x\cos x$ , $sec⁡x\sec x$ and $csc⁡x\csc x$ have period $2π2\pi$ . The graphs of $tan⁡x\tan x$ and $cot⁡x\cot x$ have period $π\pi$ .

A vertical asymptote is a vertical line that a graph approaches but does not cross. Reciprocal graphs have vertical asymptotes where the original trig function is zero.

Reciprocal trig graphs have vertical asymptotes exactly where the denominator trig function equals zero.

For sketching:

$y=sec⁡xy=\sec x$ is the reciprocal of $y=cos⁡xy=\cos x$ .
$y=csc⁡xy=\csc x$ is the reciprocal of $y=sin⁡xy=\sin x$ .
$y=cot⁡xy=\cot x$ is the reciprocal of $y=tan⁡xy=\tan x$ , with asymptotes where $sin⁡x=0\sin x=0$ .

Common Mistake

Joining across asymptotes

Never connect branches through a vertical asymptote. The function is undefined there, so the graph must break.

Example

Sketching $y = cos θ$ and $y = sec θ$

Sketch the two graphs on the same axes for $0≤θ≤2π0\le \theta\le 2\pi$ .

The graph of sec θ is formed from the reciprocal of cos θ, with branches outside −1 ≤ y ≤ 1 and asymptotes where cos θ = 0.

Mark the key cosine points:
$(2π,1)(0,1),\ \left(\frac{\pi}{2},0\right),\ (\pi,-1),\ \left(\frac{3\pi}{2},0\right),\ (2\pi,1)$
Since $sec⁡θ=1cos⁡θ\sec\theta=\frac{1}{\cos\theta}$ , the sec graph passes through $y = 1$ when $cos⁡θ=1\cos\theta=1$ , and through $y = - 1$ when $cos⁡θ=−1\cos\theta=-1$ .
Put vertical asymptotes where $cos⁡θ=0\cos\theta=0$ :
$3π2\theta=\frac{\pi}{2},\ \frac{3\pi}{2}$
Draw the sec branches outside the band between $y = - 1$ and $y = 1$ : upper branches near $θ=0\theta=0$ and $θ=2π\theta=2\pi$ , and a lower branch centred at $θ=π\theta=\pi$ .

Identities you need all the time

Definition

Identity

An identity is an equation that is true for every allowed value of the variable. The symbol $≡\equiv$ means “identically equal to”.

Start from the fundamental identity:

The Pythagorean identity comes directly from the unit circle coordinates cos x and sin x.

sin^2 x+\cos^2 x=1

Dividing by $cos^2 x$ gives:

tan^2 x+1=\sec^2 x

Dividing by $sin^2 x$ gives:

1+\cot^2 x=\csc^2 x

Key Idea

Choose the identity that reduces the number of functions

If an equation contains $tan^2 x$ and $sec⁡x\sec x$ , use $tan^2 x=\sec^2 x-1$ so everything becomes a quadratic in $sec⁡x\sec x$ .

Example

A prove-and-hence style example

Show that $sin⁡x+cos⁡xcot⁡x≡csc⁡x\sin x+\cos x\cot x\equiv \csc x$ . Hence solve $sin⁡x+cos⁡xcot⁡x=2sin⁡x\sin x+\cos x\cot x=2\sin x$ for $0<x<2π0<x<2\pi$ .

Replace $cot⁡x\cot x$ with $cos⁡xsin⁡x\frac{\cos x}{\sin x}$ :
$sin⁡x+cos⁡xcot⁡x=sin⁡x+cos⁡2xsin⁡x\sin x+\cos x\cot x=\sin x+\frac{\cos^2 x}{\sin x}$
Put the terms over a common denominator:
$sin⁡x+cos⁡2xsin⁡x=sin⁡2x+cos⁡2xsin⁡x\sin x+\frac{\cos^2 x}{\sin x}=\frac{\sin^2 x+\cos^2 x}{\sin x}$
Use $sin^2 x+\cos^2 x=1$ :
$sin⁡2x+cos⁡2xsin⁡x=1sin⁡x=csc⁡x\frac{\sin^2 x+\cos^2 x}{\sin x}=\frac{1}{\sin x}=\csc x$
For the equation, replace the left-hand side with $csc⁡x\csc x$ :
$csc⁡x=2sin⁡x\csc x=2\sin x$
Multiply by $sin⁡x\sin x$ and solve:
$1=2sin⁡2x⇒sin⁡2x=121=2\sin^2 x \Rightarrow \sin^2 x=\frac{1}{2}$
Therefore $sin⁡x=±22\sin x=\pm \frac{\sqrt{2}}{2}$ , so
$7π4x=\frac{\pi}{4},\ \frac{3\pi}{4},\ \frac{5\pi}{4},\ \frac{7\pi}{4}$

Solving equations by turning trig into algebra

A substitution means temporarily replacing a trig expression with a letter, such as $u=sec⁡xu=\sec x$ , so the equation looks like a familiar quadratic.

Common Mistake

Range check

For real angles, $sin⁡x\sin x$ and $cos⁡x\cos x$ must lie between -1 and 1. Therefore $sec⁡x\sec x$ and $csc⁡x\csc x$ cannot lie between -1 and 1, except that they also cannot be zero.

The reciprocal transformation means sec x and csc x have values y ≤ −1 or y ≥ 1, never between −1 and 1.

Example

Using an identity to form a quadratic

Solve $tan^2 x+3\sec x-3=0$ for $0∘≤x≤360∘0^\circ\le x\le 360^\circ$ , giving answers to 1 decimal place where needed.

A unit-circle view shows why cos x = −1/4 gives two solutions in quadrants II and III, while cos x = 1 gives the endpoints 0° and 360°.

Use $tan^2 x=\sec^2 x-1$ :
$tan⁡2x+3sec⁡x−3=0⇒sec⁡2x+3sec⁡x−4=0\tan^2 x+3\sec x-3=0 \Rightarrow \sec^2 x+3\sec x-4=0$
Let $u=sec⁡xu=\sec x$ :
$u^2+3u-4=0$
Factor to get $u = - 4$ or $u = 1$ :
$u^2+3u-4=(u+4)(u-1)=0$
Convert back to cosine: $sec⁡x=−4\sec x=-4$ gives $cos⁡x=−14\cos x=-\frac{1}{4}$ , and $sec⁡x=1\sec x=1$ gives $cos⁡x=1\cos x=1$ .
For $cos⁡x=−14\cos x=-\frac{1}{4}$ , the reference angle is $arccos⁡(14)≈75.5∘\arccos\left(\frac{1}{4}\right)\approx 75.5^\circ$ , so the quadrant II and III angles are 104.5° and 255.5°.
For $cos⁡x=1\cos x=1$ , include both endpoints in the interval:
$360∘x=0^\circ,\ 104.5^\circ,\ 255.5^\circ,\ 360^\circ$

Transformed angles

Sometimes the angle inside the trig function is not just $x$ or $θ\theta$ . For example, $sec⁡(2θ−20∘)\sec(2\theta-20^\circ)$ has an inside angle of $2θ−20∘2\theta-20^\circ$ .

Tip

Inside angle first

Set the whole inside angle equal to a new letter, transform the interval, solve for that new letter, then convert back at the end.

Example

Solving with a transformed angle

Solve $sec⁡(2θ−20∘)=−1.25\sec(2\theta-20^\circ)=-1.25$ for $−180∘≤θ≤180∘-180^\circ\le \theta\le 180^\circ$ , giving answers to 1 decimal place.

Let $u=2θ−20∘u=2\theta-20^\circ$ . Transform the interval:

Transforming the θ-interval into a u-interval helps ensure all possible inside-angle solutions are found before converting back.

$$
-380^\circ\le u\le 340^\circ
$$

2. Convert from sec to cos:

$$
\cos u=-0.8
$$

3. The reference angle is $arccos⁡(0.8)≈36.9∘\arccos(0.8)\approx 36.9^\circ$ . Since cosine is negative in quadrants II and III, use the general form with $k$ an integer:

$$
u=180^\circ\pm 36.9^\circ+360^\circ k
$$

4. Choose the values in $−380∘≤u≤340∘-380^\circ\le u\le 340^\circ$ :

$$
u=-216.9^\circ,\ -143.1^\circ,\ 143.1^\circ,\ 216.9^\circ
$$

5. Convert back using $θ=u+20∘2\theta=\frac{u+20^\circ}{2}$ :

$$
\theta=-98.4^\circ,\ -61.6^\circ,\ 81.6^\circ,\ 118.4^\circ
$$

Exam technique

In the exam

Rewrite reciprocal functions in terms of $sin⁡\sin$ , $cos⁡\cos$ and $tan⁡\tan$ unless the question is asking for a sketch.
Check the angle unit before using your calculator: radians for $π\pi$ intervals, degrees for 180° or 360° intervals.
After solving algebraically, reject impossible values and check for undefined angles.

Self review

Check yourself

Where are the vertical asymptotes of $y=sec⁡xy=\sec x$ between $0$ and $2π2\pi$ ?
How can you derive $csc^2 x=1+\cot^2 x$ from $sin^2 x+\cos^2 x=1$ ?
If $sec⁡x=−2\sec x=-2$ for $0∘≤x≤360∘0^\circ\le x\le 360^\circ$ , which quadrants contain the solutions?

Trigonometric Functions

What you'll learn

How the reciprocal functions $sec⁡\sec$ , $csc⁡\csc$ and $cot⁡\cot$ connect to $sin⁡\sin$ , $cos⁡\cos$ and $tan⁡\tan$ .
How to sketch reciprocal trig graphs using asymptotes and key points.
How to prove and use the identities involving $tan^2 x$ , $sec^2 x$ , $cot^2 x$ and $csc^2 x$ .
How to solve trig equations over stated intervals in radians or degrees.

Foundations: units, signs and reciprocals

Trig questions may use radians or degrees. If the interval contains $π\pi$ , you are working in radians. If it uses 360° or 180°, you are working in degrees. A key conversion is:

radians180^\circ=\pi \text{ radians}

A reference angle is the acute angle between the line and the nearest horizontal axis.

A unit-circle quadrant diagram showing the ASTC sign rule and how the reference angle is measured from the nearest horizontal axis.

Definition

Reciprocal trigonometric functions

The reciprocal trig functions are defined by

sec⁡x=1cos⁡x,csc⁡x=1sin⁡x,cot⁡x=1tan⁡x=cos⁡xsin⁡x\sec x=\frac{1}{\cos x},\qquad \csc x=\frac{1}{\sin x},\qquad \cot x=\frac{1}{\tan x}=\frac{\cos x}{\sin x}

You may see $csc⁡x\csc x$ written as cosec x; they mean the same thing.

Because these are reciprocals, they are undefined when the denominator is zero. For example, $sec⁡x\sec x$ is undefined when $cos⁡x=0\cos x=0$ .

Example

Solving a reciprocal equation in radians

Solve $3csc⁡θ=73\csc\theta=7$ for $0≤θ≤2π0\le \theta\le 2\pi$ , giving answers to 3 significant figures.

The two unit-circle positions where sine is positive for the same reference angle in the interval from 0 to 2π.

Rewrite using the reciprocal definition:
$3csc⁡θ=7⇒sin⁡θ=373\csc\theta=7 \Rightarrow \sin\theta=\frac{3}{7}$
Find the reference angle in radian mode:
$α=arcsin⁡(37)≈0.443\alpha=\arcsin\left(\frac{3}{7}\right)\approx 0.443$
Since sine is positive in quadrants I and II, use $θ=α\theta=\alpha$ and $θ=π−α\theta=\pi-\alpha$ :
$θ=0.443orθ≈2.70\theta=0.443 \quad \text{or} \quad \theta\approx 2.70$
The solutions are $2.70\theta=0.443,\ 2.70$ .

Graphs of reciprocal functions

A vertical asymptote is a vertical line that a graph approaches but does not cross. Reciprocal graphs have vertical asymptotes where the original trig function is zero.

Reciprocal trig graphs have vertical asymptotes exactly where the denominator trig function equals zero.

For sketching:

$y=sec⁡xy=\sec x$ is the reciprocal of $y=cos⁡xy=\cos x$ .
$y=csc⁡xy=\csc x$ is the reciprocal of $y=sin⁡xy=\sin x$ .
$y=cot⁡xy=\cot x$ is the reciprocal of $y=tan⁡xy=\tan x$ , with asymptotes where $sin⁡x=0\sin x=0$ .

Common Mistake

Joining across asymptotes

Never connect branches through a vertical asymptote. The function is undefined there, so the graph must break.

Example

Sketching $y = cos θ$ and $y = sec θ$

Sketch the two graphs on the same axes for $0≤θ≤2π0\le \theta\le 2\pi$ .

The graph of sec θ is formed from the reciprocal of cos θ, with branches outside −1 ≤ y ≤ 1 and asymptotes where cos θ = 0.

Mark the key cosine points:
$(2π,1)(0,1),\ \left(\frac{\pi}{2},0\right),\ (\pi,-1),\ \left(\frac{3\pi}{2},0\right),\ (2\pi,1)$
Since $sec⁡θ=1cos⁡θ\sec\theta=\frac{1}{\cos\theta}$ , the sec graph passes through $y = 1$ when $cos⁡θ=1\cos\theta=1$ , and through $y = - 1$ when $cos⁡θ=−1\cos\theta=-1$ .
Put vertical asymptotes where $cos⁡θ=0\cos\theta=0$ :
$3π2\theta=\frac{\pi}{2},\ \frac{3\pi}{2}$
Draw the sec branches outside the band between $y = - 1$ and $y = 1$ : upper branches near $θ=0\theta=0$ and $θ=2π\theta=2\pi$ , and a lower branch centred at $θ=π\theta=\pi$ .

Identities you need all the time

Definition

Identity

An identity is an equation that is true for every allowed value of the variable. The symbol $≡\equiv$ means “identically equal to”.

Start from the fundamental identity:

The Pythagorean identity comes directly from the unit circle coordinates cos x and sin x.

sin^2 x+\cos^2 x=1

Dividing by $cos^2 x$ gives:

tan^2 x+1=\sec^2 x

Dividing by $sin^2 x$ gives:

1+\cot^2 x=\csc^2 x

Key Idea

Choose the identity that reduces the number of functions

If an equation contains $tan^2 x$ and $sec⁡x\sec x$ , use $tan^2 x=\sec^2 x-1$ so everything becomes a quadratic in $sec⁡x\sec x$ .

Example

A prove-and-hence style example

Show that $sin⁡x+cos⁡xcot⁡x≡csc⁡x\sin x+\cos x\cot x\equiv \csc x$ . Hence solve $sin⁡x+cos⁡xcot⁡x=2sin⁡x\sin x+\cos x\cot x=2\sin x$ for $0<x<2π0<x<2\pi$ .

Replace $cot⁡x\cot x$ with $cos⁡xsin⁡x\frac{\cos x}{\sin x}$ :
$sin⁡x+cos⁡xcot⁡x=sin⁡x+cos⁡2xsin⁡x\sin x+\cos x\cot x=\sin x+\frac{\cos^2 x}{\sin x}$
Put the terms over a common denominator:
$sin⁡x+cos⁡2xsin⁡x=sin⁡2x+cos⁡2xsin⁡x\sin x+\frac{\cos^2 x}{\sin x}=\frac{\sin^2 x+\cos^2 x}{\sin x}$
Use $sin^2 x+\cos^2 x=1$ :
$sin⁡2x+cos⁡2xsin⁡x=1sin⁡x=csc⁡x\frac{\sin^2 x+\cos^2 x}{\sin x}=\frac{1}{\sin x}=\csc x$
For the equation, replace the left-hand side with $csc⁡x\csc x$ :
$csc⁡x=2sin⁡x\csc x=2\sin x$
Multiply by $sin⁡x\sin x$ and solve:
$1=2sin⁡2x⇒sin⁡2x=121=2\sin^2 x \Rightarrow \sin^2 x=\frac{1}{2}$
Therefore $sin⁡x=±22\sin x=\pm \frac{\sqrt{2}}{2}$ , so
$7π4x=\frac{\pi}{4},\ \frac{3\pi}{4},\ \frac{5\pi}{4},\ \frac{7\pi}{4}$

Solving equations by turning trig into algebra

A substitution means temporarily replacing a trig expression with a letter, such as $u=sec⁡xu=\sec x$ , so the equation looks like a familiar quadratic.

Common Mistake

Range check

For real angles, $sin⁡x\sin x$ and $cos⁡x\cos x$ must lie between -1 and 1. Therefore $sec⁡x\sec x$ and $csc⁡x\csc x$ cannot lie between -1 and 1, except that they also cannot be zero.

The reciprocal transformation means sec x and csc x have values y ≤ −1 or y ≥ 1, never between −1 and 1.

Example

Using an identity to form a quadratic

Solve $tan^2 x+3\sec x-3=0$ for $0∘≤x≤360∘0^\circ\le x\le 360^\circ$ , giving answers to 1 decimal place where needed.

A unit-circle view shows why cos x = −1/4 gives two solutions in quadrants II and III, while cos x = 1 gives the endpoints 0° and 360°.

Use $tan^2 x=\sec^2 x-1$ :
$tan⁡2x+3sec⁡x−3=0⇒sec⁡2x+3sec⁡x−4=0\tan^2 x+3\sec x-3=0 \Rightarrow \sec^2 x+3\sec x-4=0$
Let $u=sec⁡xu=\sec x$ :
$u^2+3u-4=0$
Factor to get $u = - 4$ or $u = 1$ :
$u^2+3u-4=(u+4)(u-1)=0$
Convert back to cosine: $sec⁡x=−4\sec x=-4$ gives $cos⁡x=−14\cos x=-\frac{1}{4}$ , and $sec⁡x=1\sec x=1$ gives $cos⁡x=1\cos x=1$ .
For $cos⁡x=−14\cos x=-\frac{1}{4}$ , the reference angle is $arccos⁡(14)≈75.5∘\arccos\left(\frac{1}{4}\right)\approx 75.5^\circ$ , so the quadrant II and III angles are 104.5° and 255.5°.
For $cos⁡x=1\cos x=1$ , include both endpoints in the interval:
$360∘x=0^\circ,\ 104.5^\circ,\ 255.5^\circ,\ 360^\circ$

Transformed angles

Sometimes the angle inside the trig function is not just $x$ or $θ\theta$ . For example, $sec⁡(2θ−20∘)\sec(2\theta-20^\circ)$ has an inside angle of $2θ−20∘2\theta-20^\circ$ .

Tip

Inside angle first

Set the whole inside angle equal to a new letter, transform the interval, solve for that new letter, then convert back at the end.

Example

Solving with a transformed angle

Solve $sec⁡(2θ−20∘)=−1.25\sec(2\theta-20^\circ)=-1.25$ for $−180∘≤θ≤180∘-180^\circ\le \theta\le 180^\circ$ , giving answers to 1 decimal place.

Let $u=2θ−20∘u=2\theta-20^\circ$ . Transform the interval:

Transforming the θ-interval into a u-interval helps ensure all possible inside-angle solutions are found before converting back.

$$
-380^\circ\le u\le 340^\circ
$$

2. Convert from sec to cos:

$$
\cos u=-0.8
$$

3. The reference angle is $arccos⁡(0.8)≈36.9∘\arccos(0.8)\approx 36.9^\circ$ . Since cosine is negative in quadrants II and III, use the general form with $k$ an integer:

$$
u=180^\circ\pm 36.9^\circ+360^\circ k
$$

4. Choose the values in $−380∘≤u≤340∘-380^\circ\le u\le 340^\circ$ :

$$
u=-216.9^\circ,\ -143.1^\circ,\ 143.1^\circ,\ 216.9^\circ
$$

5. Convert back using $θ=u+20∘2\theta=\frac{u+20^\circ}{2}$ :

$$
\theta=-98.4^\circ,\ -61.6^\circ,\ 81.6^\circ,\ 118.4^\circ
$$

Exam technique

In the exam

Rewrite reciprocal functions in terms of $sin⁡\sin$ , $cos⁡\cos$ and $tan⁡\tan$ unless the question is asking for a sketch.
Check the angle unit before using your calculator: radians for $π\pi$ intervals, degrees for 180° or 360° intervals.
After solving algebraically, reject impossible values and check for undefined angles.

Self review

Check yourself

Where are the vertical asymptotes of $y=sec⁡xy=\sec x$ between $0$ and $2π2\pi$ ?
How can you derive $csc^2 x=1+\cot^2 x$ from $sin^2 x+\cos^2 x=1$ ?
If $sec⁡x=−2\sec x=-2$ for $0∘≤x≤360∘0^\circ\le x\le 360^\circ$ , which quadrants contain the solutions?

Trigonometric Functions

What you'll learn

Foundations: units, signs and reciprocals

Graphs of reciprocal functions

Identities you need all the time

Solving equations by turning trig into algebra

Transformed angles

Recap questions

You've reached the end

Trigonometric Functions

What you'll learn

Foundations: units, signs and reciprocals

Graphs of reciprocal functions

Identities you need all the time

Solving equations by turning trig into algebra

Transformed angles

Recap questions

You've reached the end

Trigonometric Functions

What you'll learn

Foundations: units, signs and reciprocals

Graphs of reciprocal functions

Identities you need all the time

Solving equations by turning trig into algebra

Transformed angles

Recap questions

You've reached the end

Trigonometric Functions Revision Guide

Trigonometric Functions

What you'll learn

Foundations: units, signs and reciprocals

Graphs of reciprocal functions

Identities you need all the time

Solving equations by turning trig into algebra

Transformed angles

Recap questions

You've reached the end

Trigonometric Functions Revision Guide