Verifying identities | Year 11

01 / LEARN

Understand the idea.

Prove a statement on its domain

An identity asserts equality for every input in the common domain of its original sides. A numerical failure disproves it, but finitely many successful substitutions do not prove it over an infinite domain. A proof needs valid symbolic transformations.

Choose one side, often the more complicated one, and transform it into the other. Writing ratios in sine and cosine, using a common denominator, factorising and applying ${sin}^{2} x + {cos}^{2} x = 1$ are useful strategies. Give a reason for each equality.

{sin}^{2} x + {cos}^{2} x = 1

Do not lose the exclusions

Before simplifying, list the original denominator restrictions. Cancelling a factor is valid only where that factor is nonzero, and the simplified expression does not regain the excluded inputs. Its natural domain may be larger than the original domain.

Assuming the desired equality and reaching a true statement does not prove the claim unless every step is reversible. Starting from one side and reaching the other makes the logical direction and all cancellation restrictions easier to inspect.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Example 1

Prove $tan x + cot x = 1 / (sin x cos x)$ where both sides are defined.

Worked solution

1
State the common domain.
$sin x \neq 0, cos x \neq 0$
Tangent, cotangent and the proposed right side all require these denominator restrictions.
2
Rewrite the left side using sine and cosine.
$tan x + cot x = \frac{sin x}{cos x} + \frac{cos x}{sin x}$
These are the definitions of tangent and cotangent on the stated domain.
3
Combine the fractions.
$tan x + cot x = \frac{{sin}^{2} x + {cos}^{2} x}{sin x cos x}$
The common denominator is nonzero throughout the original domain.
4
Use the Pythagorean identity.
$tan x + cot x = \frac{1}{sin x cos x}$
The numerator is one for every real angle, completing the identity proof on its common domain.

Answer

The identity holds wherever neither sine nor cosine is zero.

EXAMPLE 02

Example 2

Simplify $(1 - cos 2 x) / sin 2 x$ , with the original restrictions.

Worked solution

1
Find the original exclusions.
$sin 2 x \neq 0 \Rightarrow x \neq \frac{k π}{2}, k \in ℤ$
The denominator of the given expression must be nonzero.
2
Apply the double-angle identities.
$\frac{1 - cos 2 x}{sin 2 x} = \frac{2 {sin}^{2} x}{2 sin x cos x}$
Use $1 - cos 2 x = 2 {sin}^{2} x$ and $sin 2 x = 2 sin x cos x$ .
3
Cancel only on the original domain.
$\frac{2 {sin}^{2} x}{2 sin x cos x} = \frac{sin x}{cos x} = tan x$
The original exclusions ensure both sine and cosine are nonzero, so this cancellation is valid.
4
Retain every original restriction.
$tan x, x \neq \frac{k π}{2}, k \in ℤ$
Tangent alone exists at some excluded inputs, but the original fraction remains undefined there.

Answer

$tan x$ , with $x \neq k π / 2$ , $k \in ℤ$ .

03 / YOUR TURN

Try it yourself.

Check your understanding

Is $sin x = cos x$ an identity?

Show answer & reasoning

No.

At $x = 0$ , sine is $0$ and cosine is $1$ . This single permitted failure shows that $sin x = cos x$ is an equation true at selected angles, not an identity.

Keep practising

6 worksheet questions · 12 marks

Open worksheet Worked solutions

Read the questions online

Prove ${sec}^{2} x - {tan}^{2} x = 1$ wherever defined.
2 marks
Show $\frac{sin x}{1 + cos x} = \frac{1 - cos x}{sin x}$ where both sides are defined.
2 marks
Prove ${(sin x + cos x)}^{2} = 1 + sin 2 x$ .
2 marks
Simplify $\frac{{cos}^{2} x}{1 - sin x}$ , preserving restrictions.
2 marks
Disprove $tan (A + B) = tan A + tan B$ using $A = B = \frac{π}{6}$ .
2 marks
Prove $(sec x - tan x) (sec x + tan x) = 1$ .
2 marks

Read the worked solutions

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

Prove ${sec}^{2} x - {tan}^{2} x = 1$ wherever defined.
2 marks
For $cos x \neq 0$ , ${tan}^{2} x + 1 = {sec}^{2} x$ .
Divide the Pythagorean identity by ${cos}^{2} x$ and rearrange.
Show $\frac{sin x}{1 + cos x} = \frac{1 - cos x}{sin x}$ where both sides are defined.
2 marks
On the common domain $sin x \neq 0$ , multiply numerator/denominator by $1 - cos x$ .
Use $1 - {cos}^{2} x = {sin}^{2} x$ .
The original left side alone is also defined at some additional points, such as 0.
It follows by multiplying by the conjugate and cancelling.
Prove ${(sin x + cos x)}^{2} = 1 + sin 2 x$ .
2 marks
The left side is ${sin}^{2} x + {cos}^{2} x + 2 sin x cos x = 1 + sin 2 x$ .
Expand, combine squares and use double-angle sine.
Simplify $\frac{{cos}^{2} x}{1 - sin x}$ , preserving restrictions.
2 marks
Factor the numerator as $(1 - sin x) (1 + sin x)$ and cancel the nonzero common factor.
$1 + sin x$ , for $sin x \neq 1$ .
Disprove $tan (A + B) = tan A + tan B$ using $A = B = \frac{π}{6}$ .
2 marks
Every relevant tangent exists at this pair, so it is a valid counterexample.
The sides are $\sqrt{3}$ and $\frac{2}{\sqrt{3}}$ , which differ.
Prove $(sec x - tan x) (sec x + tan x) = 1$ .
2 marks
For $cos x \neq 0$ , the product equals ${sec}^{2} x - {tan}^{2} x = 1$ .
Apply difference of squares and the secant identity.

Checking saved progress…

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 ↗

In this lesson

Understand the idea.

Prove a statement on its domain

Do not lose the exclusions

See the method in action.

Example 1

State the common domain.

Rewrite the left side using sine and cosine.

Combine the fractions.

Use the Pythagorean identity.

Example 2

Find the original exclusions.

Apply the double-angle identities.

Cancel only on the original domain.

Retain every original restriction.

Try it yourself.

Check your understanding

Keep practising