1.8.8 Proofs with trigonometric identities

What you'll learn

How to distinguish an identity from an equation.
How to select and apply the main trigonometric identities.
How to construct a proof by simplifying one side of a statement.
How to use fractions, factorisation and conjugates in harder proofs.

What is a trigonometric identity?

A trigonometric function is a function such as sine, cosine or tangent. These are written as $sin⁡x\sin x$ , $cos⁡x\cos x$ and $tan⁡x\tan x$ .

An equation is true only for particular values of its variable. For example, $sin⁡x=0\sin x=0$ is true when $x$ is an integer multiple of $π\pi$ , but it is not true for every value of $x$ .

An identity is an equality that is true for every value of the variable for which both sides are defined. The identity symbol is $≡\equiv$ .

For example,

sin⁡2x+cos⁡2x≡1.\sin^2 x+\cos^2 x\equiv 1.

Here, $sin^2x$ means $sin x)^2$ .

Definition

Trigonometric identity

A trigonometric identity is a statement involving trigonometric functions that is true for every value in the common domain of its two sides.

Key Idea

What a proof must show

A few successful substitutions do not prove an identity. You must use accepted identities and valid algebra to show that one side is equal to the other for all permitted values of the variable.

The identities you need

The fundamental Pythagorean identity is

sin⁡2x+cos⁡2x≡1.\sin^2x+\cos^2x\equiv 1.

Rearranging it gives two useful forms:

1−sin⁡2x≡cos⁡2x1-\sin^2x\equiv\cos^2x

and

1−cos⁡2x≡sin⁡2x.1-\cos^2x\equiv\sin^2x.

The quotient identity is

tan⁡x≡sin⁡xcos⁡x.\tan x\equiv\frac{\sin x}{\cos x}.

The reciprocal functions are defined by

sec⁡x≡1cos⁡x,cosec⁡x≡1sin⁡x,cot⁡x≡cos⁡xsin⁡x.\sec x\equiv\frac{1}{\cos x}, \qquad \cosec x\equiv\frac{1}{\sin x}, \qquad \cot x\equiv\frac{\cos x}{\sin x}.

Dividing the Pythagorean identity by $cos^2x$ gives

1+tan⁡2x≡sec⁡2x.1+\tan^2x\equiv\sec^2x.

Dividing it by $sin^2x$ gives

1+cot⁡2x≡cosec⁡2x.1+\cot^2x\equiv\cosec^2x.

Tip

Choosing an identity

Look at the expression you want to remove. Expressions such as $1-\sin^2x$ , $1+\tan^2x$ and $sin⁡xcos⁡x\frac{\sin x}{\cos x}$ are strong signals for a standard identity.

The basic proof method

When asked to prove an identity, begin with the more complicated side and transform it into the simpler side.

Write either LHS for the left-hand side or RHS for the right-hand side. Work on that side only until you reach the other side.

Do not begin by writing the entire identity and then altering both sides simultaneously. That can hide circular reasoning, where you accidentally assume the result you are meant to prove.

Example

Using the Pythagorean identity

Prove that

1−sin⁡2xcos⁡x≡cos⁡x.\frac{1-\sin^2x}{\cos x}\equiv\cos x.

Start with the more complicated left-hand side:
$LHS=1−sin⁡2xcos⁡x.\text{LHS}=\frac{1-\sin^2x}{\cos x}.$
Use $1−sin⁡2x≡cos⁡2x1-\sin^2x\equiv\cos^2x$ :
$LHS=cos⁡2xcos⁡x.\text{LHS}=\frac{\cos^2x}{\cos x}.$
Cancel the common factor of $cos⁡x\cos x$ :
$LHS=cos⁡x=RHS.\text{LHS}=\cos x=\text{RHS}.$
Therefore, the identity is proved wherever the original expression is defined.

Common Mistake

Cancelling terms across subtraction

You can cancel common factors, not separate terms. For example, you cannot cancel $sin^2x$ from $1-\sin^2x$ . Factorise or apply an identity first.

Converting to sine and cosine

If an expression contains several different trigonometric functions, rewriting everything in terms of $sin⁡x\sin x$ and $cos⁡x\cos x$ often reveals the underlying algebra.

This is especially useful when $tan⁡x\tan x$ , $sec⁡x\sec x$ , $cosec⁡x\cosec x$ or $cot⁡x\cot x$ appears.

Example

Converting tangent and secant

Prove that

tan⁡xsec⁡x≡sin⁡x.\frac{\tan x}{\sec x}\equiv\sin x.

Start with the left-hand side and replace each function using its definition:
$LHS=sin⁡xcos⁡x1cos⁡x.\text{LHS} =\frac{\frac{\sin x}{\cos x}}{\frac{1}{\cos x}}.$
Dividing by a fraction is equivalent to multiplying by its reciprocal:
$LHS=sin⁡xcos⁡x×cos⁡x.\text{LHS} =\frac{\sin x}{\cos x}\times\cos x.$
Cancel the common factor of $cos⁡x\cos x$ :
$LHS=sin⁡x=RHS.\text{LHS}=\sin x=\text{RHS}.$

Common Mistake

Restrictions on the variable

A trigonometric fraction is defined only when its denominator is non-zero. For example, $tan⁡x\tan x$ and $sec⁡x\sec x$ are undefined when $cos⁡x=0\cos x=0$ . Cancelling may make a restriction disappear from the final expression, but the restriction from the original expression still applies.

Combining fractions

When the chosen side contains added or subtracted fractions, a common denominator is often the key step.

For fractions with denominators $a$ and $b$ ,

pa+qb=pb+qaab.\frac{p}{a}+\frac{q}{b} = \frac{pb+qa}{ab}.

The same algebra applies when the denominators contain trigonometric functions.

Example

Combining trigonometric fractions

Prove that

11−sin⁡x+11+sin⁡x≡2sec⁡2x.\frac{1}{1-\sin x}+\frac{1}{1+\sin x}\equiv 2\sec^2x.

Combine the fractions using the common denominator $(1−sin⁡x)(1+sin⁡x)(1-\sin x)(1+\sin x)$ :
$LHS=1+sin⁡x+1−sin⁡x(1−sin⁡x)(1+sin⁡x)=21−sin⁡2x.\begin{aligned} \text{LHS} &= \frac{1+\sin x+1-\sin x} {(1-\sin x)(1+\sin x)}\\ &= \frac{2}{1-\sin^2x}. \end{aligned}$
Apply the difference of two squares and then the Pythagorean identity:
$1−sin⁡2x≡cos⁡2x.1-\sin^2x\equiv\cos^2x.$
Therefore,
$LHS=2cos⁡2x.\text{LHS}=\frac{2}{\cos^2x}.$
Since $sec⁡x≡1cos⁡x\sec x\equiv\frac{1}{\cos x}$ ,
$2cos⁡2x=2sec⁡2x=RHS.\frac{2}{\cos^2x}=2\sec^2x=\text{RHS}.$

Tip

Useful denominator pattern

The product $(1−sin⁡x)(1+sin⁡x)(1-\sin x)(1+\sin x)$ is a difference of two squares, so it simplifies to $1-\sin^2x$ , and hence to $cos^2x$ . The corresponding cosine pattern simplifies to $sin^2x$ .

Multiplying by a conjugate

Expressions such as $1+sin⁡x1+\sin x$ and $1−sin⁡x1-\sin x$ form a conjugate pair: they have the same terms but the sign between them is reversed.

Multiplying conjugates removes the middle terms:

1+\sin x)(1-\sin x)=1-\sin^2x=\cos^2x.

This technique is useful when a denominator contains $1+sin⁡x1+\sin x$ , $1−sin⁡x1-\sin x$ , $1+cos⁡x1+\cos x$ or $1−cos⁡x1-\cos x$ .

Example

Using a conjugate

Prove that

cos⁡x1+sin⁡x≡1−sin⁡xcos⁡x.\frac{\cos x}{1+\sin x}\equiv\frac{1-\sin x}{\cos x}.

Start with the left-hand side and multiply the numerator and denominator by the conjugate $1−sin⁡x1-\sin x$ :
$LHS=cos⁡x1+sin⁡x×1−sin⁡x1−sin⁡x.\text{LHS} = \frac{\cos x}{1+\sin x} \times \frac{1-\sin x}{1-\sin x}.$
Multiply out the denominator and apply the Pythagorean identity:
$LHS=cos⁡x(1−sin⁡x)1−sin⁡2x=cos⁡x(1−sin⁡x)cos⁡2x.\begin{aligned} \text{LHS} &= \frac{\cos x(1-\sin x)} {1-\sin^2x}\\ &= \frac{\cos x(1-\sin x)} {\cos^2x}. \end{aligned}$
Cancel a factor of $cos⁡x\cos x$ :
$LHS=1−sin⁡xcos⁡x=RHS.\text{LHS} = \frac{1-\sin x}{\cos x} =\text{RHS}.$

Factorising trigonometric expressions

Trigonometric expressions can be factorised using the same algebraic techniques as polynomials.

For example,

sin^2x-\sin x = \sin x(\sin x-1).

You may also need to recognise a difference of two squares:

A^2-B^2=(A-B)(A+B).

Example

Factorising before simplifying

Prove that

sin⁡2x−cos⁡2xsin⁡x−cos⁡x≡sin⁡x+cos⁡x.\frac{\sin^2x-\cos^2x}{\sin x-\cos x}\equiv\sin x+\cos x.

Recognise the numerator as a difference of two squares:
$sin^2x-\cos^2x = (\sin x-\cos x)(\sin x+\cos x).$
Substitute this factorisation into the left-hand side:
$LHS=(sin⁡x−cos⁡x)(sin⁡x+cos⁡x)sin⁡x−cos⁡x.\text{LHS} = \frac{(\sin x-\cos x)(\sin x+\cos x)} {\sin x-\cos x}.$
Cancel the common factor, provided $sin⁡x−cos⁡x≠0\sin x-\cos x\neq 0$ :
$LHS=sin⁡x+cos⁡x=RHS.\text{LHS}=\sin x+\cos x=\text{RHS}.$

Building a clear proof

A good proof is a connected chain of equal expressions. Each line should follow from the previous one by a recognised identity or a valid algebraic operation.

You do not need to name every elementary algebraic step, but the important transformations must be visible. In particular, show when you:

substitute a standard identity;
take a common denominator;
factorise an expression;
multiply by a conjugate;
cancel a common factor.

Common Mistake

Starting from both sides

Avoid simplifying the LHS and RHS separately and meeting somewhere in the middle unless the argument remains completely clear. Starting from one side and reaching the other gives a more direct and convincing proof.

Key Idea

A reliable decision process

First decide which side is more complicated. Then look for fractions to combine, functions to rewrite in sine and cosine, expressions matching standard identities, or algebraic structures that can be factorised.

Exam technique

In the exam

Write LHS = or RHS = and begin with the more complicated side.
Change one feature at a time, keeping enough working to justify every significant step.
Look for the Pythagorean identities, common denominators, conjugates and opportunities to factorise.
Never verify an identity using only selected numerical values; that is a check, not a proof.
Keep exact trigonometric expressions and finish explicitly with the other side of the identity.

Self review

Check yourself

Why does checking an identity at three values of $x$ not constitute a proof?
How would you begin proving an identity containing both $tan⁡x\tan x$ and $sec⁡x\sec x$ ?
Which conjugate would you use if a denominator contained $1−cos⁡x1-\cos x$ ?

What you'll learn

How to distinguish an identity from an equation.
How to select and apply the main trigonometric identities.
How to construct a proof by simplifying one side of a statement.
How to use fractions, factorisation and conjugates in harder proofs.

What is a trigonometric identity?

A trigonometric function is a function such as sine, cosine or tangent. These are written as $sin⁡x\sin x$ , $cos⁡x\cos x$ and $tan⁡x\tan x$ .

An equation is true only for particular values of its variable. For example, $sin⁡x=0\sin x=0$ is true when $x$ is an integer multiple of $π\pi$ , but it is not true for every value of $x$ .

An identity is an equality that is true for every value of the variable for which both sides are defined. The identity symbol is $≡\equiv$ .

For example,

sin⁡2x+cos⁡2x≡1.\sin^2 x+\cos^2 x\equiv 1.

Here, $sin^2x$ means $sin x)^2$ .

Definition

Trigonometric identity

A trigonometric identity is a statement involving trigonometric functions that is true for every value in the common domain of its two sides.

Key Idea

What a proof must show

A few successful substitutions do not prove an identity. You must use accepted identities and valid algebra to show that one side is equal to the other for all permitted values of the variable.

The identities you need

The fundamental Pythagorean identity is

sin⁡2x+cos⁡2x≡1.\sin^2x+\cos^2x\equiv 1.

Rearranging it gives two useful forms:

1−sin⁡2x≡cos⁡2x1-\sin^2x\equiv\cos^2x

and

1−cos⁡2x≡sin⁡2x.1-\cos^2x\equiv\sin^2x.

The quotient identity is

tan⁡x≡sin⁡xcos⁡x.\tan x\equiv\frac{\sin x}{\cos x}.

The reciprocal functions are defined by

sec⁡x≡1cos⁡x,cosec⁡x≡1sin⁡x,cot⁡x≡cos⁡xsin⁡x.\sec x\equiv\frac{1}{\cos x}, \qquad \cosec x\equiv\frac{1}{\sin x}, \qquad \cot x\equiv\frac{\cos x}{\sin x}.

Dividing the Pythagorean identity by $cos^2x$ gives

1+tan⁡2x≡sec⁡2x.1+\tan^2x\equiv\sec^2x.

Dividing it by $sin^2x$ gives

1+cot⁡2x≡cosec⁡2x.1+\cot^2x\equiv\cosec^2x.

Tip

Choosing an identity

Look at the expression you want to remove. Expressions such as $1-\sin^2x$ , $1+\tan^2x$ and $sin⁡xcos⁡x\frac{\sin x}{\cos x}$ are strong signals for a standard identity.

The basic proof method

When asked to prove an identity, begin with the more complicated side and transform it into the simpler side.

Write either LHS for the left-hand side or RHS for the right-hand side. Work on that side only until you reach the other side.

Do not begin by writing the entire identity and then altering both sides simultaneously. That can hide circular reasoning, where you accidentally assume the result you are meant to prove.

Example

Using the Pythagorean identity

Prove that

1−sin⁡2xcos⁡x≡cos⁡x.\frac{1-\sin^2x}{\cos x}\equiv\cos x.

Start with the more complicated left-hand side:
$LHS=1−sin⁡2xcos⁡x.\text{LHS}=\frac{1-\sin^2x}{\cos x}.$
Use $1−sin⁡2x≡cos⁡2x1-\sin^2x\equiv\cos^2x$ :
$LHS=cos⁡2xcos⁡x.\text{LHS}=\frac{\cos^2x}{\cos x}.$
Cancel the common factor of $cos⁡x\cos x$ :
$LHS=cos⁡x=RHS.\text{LHS}=\cos x=\text{RHS}.$
Therefore, the identity is proved wherever the original expression is defined.

Common Mistake

Cancelling terms across subtraction

You can cancel common factors, not separate terms. For example, you cannot cancel $sin^2x$ from $1-\sin^2x$ . Factorise or apply an identity first.

Converting to sine and cosine

If an expression contains several different trigonometric functions, rewriting everything in terms of $sin⁡x\sin x$ and $cos⁡x\cos x$ often reveals the underlying algebra.

This is especially useful when $tan⁡x\tan x$ , $sec⁡x\sec x$ , $cosec⁡x\cosec x$ or $cot⁡x\cot x$ appears.

Example

Converting tangent and secant

Prove that

tan⁡xsec⁡x≡sin⁡x.\frac{\tan x}{\sec x}\equiv\sin x.

Start with the left-hand side and replace each function using its definition:
$LHS=sin⁡xcos⁡x1cos⁡x.\text{LHS} =\frac{\frac{\sin x}{\cos x}}{\frac{1}{\cos x}}.$
Dividing by a fraction is equivalent to multiplying by its reciprocal:
$LHS=sin⁡xcos⁡x×cos⁡x.\text{LHS} =\frac{\sin x}{\cos x}\times\cos x.$
Cancel the common factor of $cos⁡x\cos x$ :
$LHS=sin⁡x=RHS.\text{LHS}=\sin x=\text{RHS}.$

Common Mistake

Restrictions on the variable

Combining fractions

When the chosen side contains added or subtracted fractions, a common denominator is often the key step.

For fractions with denominators $a$ and $b$ ,

pa+qb=pb+qaab.\frac{p}{a}+\frac{q}{b} = \frac{pb+qa}{ab}.

The same algebra applies when the denominators contain trigonometric functions.

Example

Combining trigonometric fractions

Prove that

11−sin⁡x+11+sin⁡x≡2sec⁡2x.\frac{1}{1-\sin x}+\frac{1}{1+\sin x}\equiv 2\sec^2x.

Combine the fractions using the common denominator $(1−sin⁡x)(1+sin⁡x)(1-\sin x)(1+\sin x)$ :
$LHS=1+sin⁡x+1−sin⁡x(1−sin⁡x)(1+sin⁡x)=21−sin⁡2x.\begin{aligned} \text{LHS} &= \frac{1+\sin x+1-\sin x} {(1-\sin x)(1+\sin x)}\\ &= \frac{2}{1-\sin^2x}. \end{aligned}$
Apply the difference of two squares and then the Pythagorean identity:
$1−sin⁡2x≡cos⁡2x.1-\sin^2x\equiv\cos^2x.$
Therefore,
$LHS=2cos⁡2x.\text{LHS}=\frac{2}{\cos^2x}.$
Since $sec⁡x≡1cos⁡x\sec x\equiv\frac{1}{\cos x}$ ,
$2cos⁡2x=2sec⁡2x=RHS.\frac{2}{\cos^2x}=2\sec^2x=\text{RHS}.$

Tip

Useful denominator pattern

Multiplying by a conjugate

Expressions such as $1+sin⁡x1+\sin x$ and $1−sin⁡x1-\sin x$ form a conjugate pair: they have the same terms but the sign between them is reversed.

Multiplying conjugates removes the middle terms:

1+\sin x)(1-\sin x)=1-\sin^2x=\cos^2x.

This technique is useful when a denominator contains $1+sin⁡x1+\sin x$ , $1−sin⁡x1-\sin x$ , $1+cos⁡x1+\cos x$ or $1−cos⁡x1-\cos x$ .

Example

Using a conjugate

Prove that

cos⁡x1+sin⁡x≡1−sin⁡xcos⁡x.\frac{\cos x}{1+\sin x}\equiv\frac{1-\sin x}{\cos x}.

Start with the left-hand side and multiply the numerator and denominator by the conjugate $1−sin⁡x1-\sin x$ :
$LHS=cos⁡x1+sin⁡x×1−sin⁡x1−sin⁡x.\text{LHS} = \frac{\cos x}{1+\sin x} \times \frac{1-\sin x}{1-\sin x}.$
Multiply out the denominator and apply the Pythagorean identity:
$LHS=cos⁡x(1−sin⁡x)1−sin⁡2x=cos⁡x(1−sin⁡x)cos⁡2x.\begin{aligned} \text{LHS} &= \frac{\cos x(1-\sin x)} {1-\sin^2x}\\ &= \frac{\cos x(1-\sin x)} {\cos^2x}. \end{aligned}$
Cancel a factor of $cos⁡x\cos x$ :
$LHS=1−sin⁡xcos⁡x=RHS.\text{LHS} = \frac{1-\sin x}{\cos x} =\text{RHS}.$

Factorising trigonometric expressions

Trigonometric expressions can be factorised using the same algebraic techniques as polynomials.

For example,

sin^2x-\sin x = \sin x(\sin x-1).

You may also need to recognise a difference of two squares:

A^2-B^2=(A-B)(A+B).

Example

Factorising before simplifying

Prove that

sin⁡2x−cos⁡2xsin⁡x−cos⁡x≡sin⁡x+cos⁡x.\frac{\sin^2x-\cos^2x}{\sin x-\cos x}\equiv\sin x+\cos x.

Recognise the numerator as a difference of two squares:
$sin^2x-\cos^2x = (\sin x-\cos x)(\sin x+\cos x).$
Substitute this factorisation into the left-hand side:
$LHS=(sin⁡x−cos⁡x)(sin⁡x+cos⁡x)sin⁡x−cos⁡x.\text{LHS} = \frac{(\sin x-\cos x)(\sin x+\cos x)} {\sin x-\cos x}.$
Cancel the common factor, provided $sin⁡x−cos⁡x≠0\sin x-\cos x\neq 0$ :
$LHS=sin⁡x+cos⁡x=RHS.\text{LHS}=\sin x+\cos x=\text{RHS}.$

Building a clear proof

A good proof is a connected chain of equal expressions. Each line should follow from the previous one by a recognised identity or a valid algebraic operation.

You do not need to name every elementary algebraic step, but the important transformations must be visible. In particular, show when you:

substitute a standard identity;
take a common denominator;
factorise an expression;
multiply by a conjugate;
cancel a common factor.

Common Mistake

Starting from both sides

Key Idea

A reliable decision process

Exam technique

In the exam

Write LHS = or RHS = and begin with the more complicated side.
Change one feature at a time, keeping enough working to justify every significant step.
Look for the Pythagorean identities, common denominators, conjugates and opportunities to factorise.
Never verify an identity using only selected numerical values; that is a check, not a proof.
Keep exact trigonometric expressions and finish explicitly with the other side of the identity.

Self review

Check yourself

Why does checking an identity at three values of $x$ not constitute a proof?
How would you begin proving an identity containing both $tan⁡x\tan x$ and $sec⁡x\sec x$ ?
Which conjugate would you use if a denominator contained $1−cos⁡x1-\cos x$ ?

What you'll learn

What is a trigonometric identity?

The identities you need

The basic proof method

Converting to sine and cosine

Combining fractions

Multiplying by a conjugate

Factorising trigonometric expressions

Building a clear proof

Review 1.8.8 Proofs with trigonometric identities by teaching Genie

1.8.8 Proofs with trigonometric identities

What you'll learn

What is a trigonometric identity?

The identities you need

The basic proof method

Converting to sine and cosine

Combining fractions

Multiplying by a conjugate

Factorising trigonometric expressions

Building a clear proof

Review 1.8.8 Proofs with trigonometric identities by teaching Genie

1.8.8 Proofs with trigonometric identities

What you'll learn

What is a trigonometric identity?

The identities you need

The basic proof method

Converting to sine and cosine

Combining fractions

Multiplying by a conjugate

Factorising trigonometric expressions

Building a clear proof

Review 1.8.8 Proofs with trigonometric identities by teaching Genie

1.8.8 Proofs with trigonometric identities Revision Guide

1.8.8 Proofs with trigonometric identities

What you'll learn

What is a trigonometric identity?

The identities you need

The basic proof method

Converting to sine and cosine

Combining fractions

Multiplying by a conjugate

Factorising trigonometric expressions

Building a clear proof

Review 1.8.8 Proofs with trigonometric identities by teaching Genie

1.8.8 Proofs with trigonometric identities Revision Guide