1.5.10 Trigonometric identities - Revision Guide

What you'll learn

What makes a trigonometric identity different from an equation.
How $tan⁡θ\tan\theta$ , $sec⁡θ\sec\theta$ , $cosec⁡θ\operatorname{cosec}\theta$ and $cot⁡θ\cot\theta$ are connected to sine and cosine.
How to derive and use the three Pythagorean identities.
How to simplify expressions and prove identities systematically.

Prerequisites: the trigonometric ratios

You should already know the three basic trigonometric ratios:

sin⁡θ=oppositehypotenuse,cos⁡θ=adjacenthypotenuse,tan⁡θ=oppositeadjacent.\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}, \qquad \cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}, \qquad \tan\theta=\frac{\text{opposite}}{\text{adjacent}}.

At A-level, you also use three reciprocal trigonometric functions. A reciprocal is obtained by dividing 1 by a quantity.

Definition

Reciprocal trigonometric functions

sec⁡θ=1cos⁡θ,cosec⁡θ=1sin⁡θ,cot⁡θ=1tan⁡θ.\sec\theta=\frac{1}{\cos\theta}, \qquad \operatorname{cosec}\theta=\frac{1}{\sin\theta}, \qquad \cot\theta=\frac{1}{\tan\theta}.

For example, $sec⁡2θ\sec^2\theta$ means $(sec⁡θ)2(\sec\theta)^2$ . It does not mean $sec⁡(θ2)\sec(\theta^2)$ .

Common Mistake

Reciprocal is not inverse

The reciprocal $sec⁡θ=1cos⁡θ\sec\theta=\frac{1}{\cos\theta}$ is different from the inverse function $cos⁡−1θ\cos^{-1}\theta$ , which is used to find an angle.

What is an identity?

A trigonometric identity is a statement that is true for every value of the angle for which both sides are defined.

The symbol $≡\equiv$ means “is identically equal to”. For example,

sin⁡2θ+cos⁡2θ≡1.\sin^2\theta+\cos^2\theta\equiv 1.

This is different from an equation such as $sin⁡θ=12\sin\theta=\frac12$ , which is true only for particular values of $θ\theta$ .

Key Idea

Identities versus equations

An equation is solved to find particular values. An identity is rearranged, applied or proved because it is true throughout its domain.

The tangent identity

The first identity connects tangent to sine and cosine:

tan⁡θ≡sin⁡θcos⁡θ.\tan\theta\equiv\frac{\sin\theta}{\cos\theta}.

It follows directly from the right-angled triangle definitions:

sin⁡θcos⁡θ=oppositehypotenuseadjacenthypotenuse=oppositeadjacent=tan⁡θ.\frac{\sin\theta}{\cos\theta} = \frac{\frac{\text{opposite}}{\text{hypotenuse}}} {\frac{\text{adjacent}}{\text{hypotenuse}}} = \frac{\text{opposite}}{\text{adjacent}} = \tan\theta.

Common Mistake

When tangent is undefined

The fraction $sin⁡θcos⁡θ\frac{\sin\theta}{\cos\theta}$ is undefined when $cos⁡θ=0\cos\theta=0$ . Therefore, tangent is also undefined at these angles.

Example

Simplifying an expression containing tangent

Simplify

sin⁡θtan⁡θ.\frac{\sin\theta}{\tan\theta}.

Replace tangent using $tan⁡θ=sin⁡θcos⁡θ\tan\theta=\frac{\sin\theta}{\cos\theta}$ :
$sin⁡θtan⁡θ=sin⁡θsin⁡θcos⁡θ.\frac{\sin\theta}{\tan\theta} = \frac{\sin\theta}{\frac{\sin\theta}{\cos\theta}}.$
Dividing by a fraction means multiplying by its reciprocal:
$sin⁡θ⋅cos⁡θsin⁡θ.\sin\theta\cdot\frac{\cos\theta}{\sin\theta}.$
Cancel the common factor $sin⁡θ\sin\theta$ to obtain
$sin⁡θtan⁡θ=cos⁡θ.\frac{\sin\theta}{\tan\theta}=\cos\theta.$

The first Pythagorean identity

On the unit circle, the point at angle $θ\theta$ has coordinates $(cos⁡θ,sin⁡θ)(\cos\theta,\sin\theta)$ . Because the circle has radius 1, Pythagoras’ theorem gives

(cos⁡θ)2+(sin⁡θ)2=12.(\cos\theta)^2+(\sin\theta)^2=1^2.

Therefore,

sin⁡2θ+cos⁡2θ≡1.\sin^2\theta+\cos^2\theta\equiv 1.

Unit circle showing the coordinates cos theta and sin theta

This identity can be rearranged in either direction:

sin⁡2θ≡1−cos⁡2θ,cos⁡2θ≡1−sin⁡2θ.\sin^2\theta\equiv 1-\cos^2\theta, \qquad \cos^2\theta\equiv 1-\sin^2\theta.

Example

Rewriting in terms of cosine

Simplify

3sin⁡2θ+3cos⁡2θ−2cos⁡2θ.3\sin^2\theta+3\cos^2\theta-2\cos^2\theta.

Group the first two terms because they contain the complete Pythagorean identity:
$3(sin⁡2θ+cos⁡2θ)−2cos⁡2θ.3(\sin^2\theta+\cos^2\theta)-2\cos^2\theta.$
Substitute $sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1$ :
$3(1)−2cos⁡2θ.3(1)-2\cos^2\theta.$
The simplified expression is
$3−2cos⁡2θ.3-2\cos^2\theta.$

Deriving the secant identity

Begin with

sin⁡2θ+cos⁡2θ≡1.\sin^2\theta+\cos^2\theta\equiv 1.

Divide every term by $cos⁡2θ\cos^2\theta$ :

sin⁡2θcos⁡2θ+cos⁡2θcos⁡2θ≡1cos⁡2θ.\frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\cos^2\theta} \equiv \frac{1}{\cos^2\theta}.

Using $tan⁡θ=sin⁡θcos⁡θ\tan\theta=\frac{\sin\theta}{\cos\theta}$ and $sec⁡θ=1cos⁡θ\sec\theta=\frac{1}{\cos\theta}$ gives

tan⁡2θ+1≡sec⁡2θ.\tan^2\theta+1\equiv\sec^2\theta.

It is usually written as

sec⁡2θ≡1+tan⁡2θ.\sec^2\theta\equiv 1+\tan^2\theta.

Key Idea

Choosing the secant identity

Use $sec⁡2θ≡1+tan⁡2θ\sec^2\theta\equiv 1+\tan^2\theta$ when an expression involves only secant and tangent, or when you want to change one of these functions into the other.

Example

Simplifying with secant and tangent

Simplify

sec⁡2θ−1tan⁡θ.\frac{\sec^2\theta-1}{\tan\theta}.

Rearrange $sec⁡2θ=1+tan⁡2θ\sec^2\theta=1+\tan^2\theta$ to obtain
$sec⁡2θ−1=tan⁡2θ.\sec^2\theta-1=\tan^2\theta.$
Substitute this into the numerator:
$sec⁡2θ−1tan⁡θ=tan⁡2θtan⁡θ.\frac{\sec^2\theta-1}{\tan\theta} = \frac{\tan^2\theta}{\tan\theta}.$
Cancel one factor of $tan⁡θ\tan\theta$ :
$sec⁡2θ−1tan⁡θ=tan⁡θ.\frac{\sec^2\theta-1}{\tan\theta}=\tan\theta.$

Deriving the cosecant identity

Again begin with

sin⁡2θ+cos⁡2θ≡1.\sin^2\theta+\cos^2\theta\equiv 1.

This time, divide every term by $sin⁡2θ\sin^2\theta$ :

sin⁡2θsin⁡2θ+cos⁡2θsin⁡2θ≡1sin⁡2θ.\frac{\sin^2\theta}{\sin^2\theta} + \frac{\cos^2\theta}{\sin^2\theta} \equiv \frac{1}{\sin^2\theta}.

Since $cot⁡θ=cos⁡θsin⁡θ\cot\theta=\frac{\cos\theta}{\sin\theta}$ and $cosec⁡θ=1sin⁡θ\operatorname{cosec}\theta=\frac{1}{\sin\theta}$ , this becomes

1+cot⁡2θ≡cosec⁡2θ.1+\cot^2\theta\equiv\operatorname{cosec}^2\theta.

Hence,

cosec⁡2θ≡1+cot⁡2θ.\operatorname{cosec}^2\theta\equiv 1+\cot^2\theta.

Tip

Remembering the pairings

Secant pairs with tangent, while cosecant pairs with cotangent:

sec⁡2θ≡1+tan⁡2θ,cosec⁡2θ≡1+cot⁡2θ.\sec^2\theta\equiv 1+\tan^2\theta, \qquad \operatorname{cosec}^2\theta\equiv 1+\cot^2\theta.

Example

Finding a cosecant value

Given that $cot⁡θ=2\cot\theta=2$ and $cosec⁡θ>0\operatorname{cosec}\theta>0$ , find $cosec⁡θ\operatorname{cosec}\theta$ .

Apply the appropriate identity:
$cosec⁡2θ=1+cot⁡2θ.\operatorname{cosec}^2\theta=1+\cot^2\theta.$
Substitute $cot⁡θ=2\cot\theta=2$ :
$cosec⁡2θ=1+22=5.\operatorname{cosec}^2\theta=1+2^2=5.$
Taking square roots gives $cosec⁡θ=±5\operatorname{cosec}\theta=\pm\sqrt5$ . The given sign condition selects
$cosec⁡θ=5.\operatorname{cosec}\theta=\sqrt5.$

Common Mistake

Forgetting both square roots

From $cosec⁡2θ=5\operatorname{cosec}^2\theta=5$ , you initially obtain $cosec⁡θ=±5\operatorname{cosec}\theta=\pm\sqrt5$ . Use information about the angle or the required sign to choose the correct value.

Proving trigonometric identities

To prove an identity, usually start with the more complicated side and transform it until it matches the other side. Avoid changing both sides at once, as this can hide gaps in your reasoning.

Converting tangent, cotangent, secant and cosecant into sine and cosine often makes the route clearer.

Example

Proving an identity using sine and cosine

Prove that

1−cos⁡2θsin⁡θ≡sin⁡θ.\frac{1-\cos^2\theta}{\sin\theta}\equiv\sin\theta.

Use the rearranged Pythagorean identity $1−cos⁡2θ=sin⁡2θ1-\cos^2\theta=\sin^2\theta$ :
$1−cos⁡2θsin⁡θ=sin⁡2θsin⁡θ.\frac{1-\cos^2\theta}{\sin\theta} = \frac{\sin^2\theta}{\sin\theta}.$
Write $sin⁡2θ\sin^2\theta$ as a product:
$sin⁡θsin⁡θsin⁡θ.\frac{\sin\theta\sin\theta}{\sin\theta}.$
Cancel the common factor, wherever the original expression is defined:
$1−cos⁡2θsin⁡θ=sin⁡θ,\frac{1-\cos^2\theta}{\sin\theta} = \sin\theta,$
which is the required right-hand side.

Common Mistake

Treating sums as factors

You cannot cancel terms across addition. For example, nothing cancels directly in $1+sin⁡θsin⁡θ\frac{1+\sin\theta}{\sin\theta}$ . Cancellation is allowed only between common factors.

Exam technique

In the exam

Identify which functions appear: sine and cosine suggest $sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1$ , while secant and tangent suggest $sec⁡2θ=1+tan⁡2θ\sec^2\theta=1+\tan^2\theta$ .
In a proof, start with the more complicated side and show each substitution or factorisation clearly.
Convert everything to sine and cosine if the route is unclear, and check that you have not divided by a quantity that could be zero.

Self review

Check yourself

Can you derive $sec⁡2θ≡1+tan⁡2θ\sec^2\theta\equiv1+\tan^2\theta$ from the first Pythagorean identity?
How would you simplify $cosec⁡2θ−cot⁡2θ\operatorname{cosec}^2\theta-\cot^2\theta$ ?
Why is $tan⁡θ≡sin⁡θcos⁡θ\tan\theta\equiv\frac{\sin\theta}{\cos\theta}$ not defined when $cos⁡θ=0\cos\theta=0$ ?

What you'll learn

What makes a trigonometric identity different from an equation.
How $tan⁡θ\tan\theta$ , $sec⁡θ\sec\theta$ , $cosec⁡θ\operatorname{cosec}\theta$ and $cot⁡θ\cot\theta$ are connected to sine and cosine.
How to derive and use the three Pythagorean identities.
How to simplify expressions and prove identities systematically.

Prerequisites: the trigonometric ratios

You should already know the three basic trigonometric ratios:

sin⁡θ=oppositehypotenuse,cos⁡θ=adjacenthypotenuse,tan⁡θ=oppositeadjacent.\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}, \qquad \cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}, \qquad \tan\theta=\frac{\text{opposite}}{\text{adjacent}}.

At A-level, you also use three reciprocal trigonometric functions. A reciprocal is obtained by dividing 1 by a quantity.

Definition

Reciprocal trigonometric functions

sec⁡θ=1cos⁡θ,cosec⁡θ=1sin⁡θ,cot⁡θ=1tan⁡θ.\sec\theta=\frac{1}{\cos\theta}, \qquad \operatorname{cosec}\theta=\frac{1}{\sin\theta}, \qquad \cot\theta=\frac{1}{\tan\theta}.

For example, $sec⁡2θ\sec^2\theta$ means $(sec⁡θ)2(\sec\theta)^2$ . It does not mean $sec⁡(θ2)\sec(\theta^2)$ .

Common Mistake

Reciprocal is not inverse

The reciprocal $sec⁡θ=1cos⁡θ\sec\theta=\frac{1}{\cos\theta}$ is different from the inverse function $cos⁡−1θ\cos^{-1}\theta$ , which is used to find an angle.

What is an identity?

A trigonometric identity is a statement that is true for every value of the angle for which both sides are defined.

The symbol $≡\equiv$ means “is identically equal to”. For example,

sin⁡2θ+cos⁡2θ≡1.\sin^2\theta+\cos^2\theta\equiv 1.

This is different from an equation such as $sin⁡θ=12\sin\theta=\frac12$ , which is true only for particular values of $θ\theta$ .

Key Idea

Identities versus equations

An equation is solved to find particular values. An identity is rearranged, applied or proved because it is true throughout its domain.

The tangent identity

The first identity connects tangent to sine and cosine:

tan⁡θ≡sin⁡θcos⁡θ.\tan\theta\equiv\frac{\sin\theta}{\cos\theta}.

It follows directly from the right-angled triangle definitions:

sin⁡θcos⁡θ=oppositehypotenuseadjacenthypotenuse=oppositeadjacent=tan⁡θ.\frac{\sin\theta}{\cos\theta} = \frac{\frac{\text{opposite}}{\text{hypotenuse}}} {\frac{\text{adjacent}}{\text{hypotenuse}}} = \frac{\text{opposite}}{\text{adjacent}} = \tan\theta.

Common Mistake

When tangent is undefined

The fraction $sin⁡θcos⁡θ\frac{\sin\theta}{\cos\theta}$ is undefined when $cos⁡θ=0\cos\theta=0$ . Therefore, tangent is also undefined at these angles.

Example

Simplifying an expression containing tangent

Simplify

sin⁡θtan⁡θ.\frac{\sin\theta}{\tan\theta}.

Replace tangent using $tan⁡θ=sin⁡θcos⁡θ\tan\theta=\frac{\sin\theta}{\cos\theta}$ :
$sin⁡θtan⁡θ=sin⁡θsin⁡θcos⁡θ.\frac{\sin\theta}{\tan\theta} = \frac{\sin\theta}{\frac{\sin\theta}{\cos\theta}}.$
Dividing by a fraction means multiplying by its reciprocal:
$sin⁡θ⋅cos⁡θsin⁡θ.\sin\theta\cdot\frac{\cos\theta}{\sin\theta}.$
Cancel the common factor $sin⁡θ\sin\theta$ to obtain
$sin⁡θtan⁡θ=cos⁡θ.\frac{\sin\theta}{\tan\theta}=\cos\theta.$

The first Pythagorean identity

On the unit circle, the point at angle $θ\theta$ has coordinates $(cos⁡θ,sin⁡θ)(\cos\theta,\sin\theta)$ . Because the circle has radius 1, Pythagoras’ theorem gives

(cos⁡θ)2+(sin⁡θ)2=12.(\cos\theta)^2+(\sin\theta)^2=1^2.

Therefore,

sin⁡2θ+cos⁡2θ≡1.\sin^2\theta+\cos^2\theta\equiv 1.

Unit circle showing the coordinates cos theta and sin theta

This identity can be rearranged in either direction:

sin⁡2θ≡1−cos⁡2θ,cos⁡2θ≡1−sin⁡2θ.\sin^2\theta\equiv 1-\cos^2\theta, \qquad \cos^2\theta\equiv 1-\sin^2\theta.

Example

Rewriting in terms of cosine

Simplify

3sin⁡2θ+3cos⁡2θ−2cos⁡2θ.3\sin^2\theta+3\cos^2\theta-2\cos^2\theta.

Group the first two terms because they contain the complete Pythagorean identity:
$3(sin⁡2θ+cos⁡2θ)−2cos⁡2θ.3(\sin^2\theta+\cos^2\theta)-2\cos^2\theta.$
Substitute $sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1$ :
$3(1)−2cos⁡2θ.3(1)-2\cos^2\theta.$
The simplified expression is
$3−2cos⁡2θ.3-2\cos^2\theta.$

Deriving the secant identity

Begin with

sin⁡2θ+cos⁡2θ≡1.\sin^2\theta+\cos^2\theta\equiv 1.

Divide every term by $cos⁡2θ\cos^2\theta$ :

sin⁡2θcos⁡2θ+cos⁡2θcos⁡2θ≡1cos⁡2θ.\frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\cos^2\theta} \equiv \frac{1}{\cos^2\theta}.

Using $tan⁡θ=sin⁡θcos⁡θ\tan\theta=\frac{\sin\theta}{\cos\theta}$ and $sec⁡θ=1cos⁡θ\sec\theta=\frac{1}{\cos\theta}$ gives

tan⁡2θ+1≡sec⁡2θ.\tan^2\theta+1\equiv\sec^2\theta.

It is usually written as

sec⁡2θ≡1+tan⁡2θ.\sec^2\theta\equiv 1+\tan^2\theta.

Key Idea

Choosing the secant identity

Use $sec⁡2θ≡1+tan⁡2θ\sec^2\theta\equiv 1+\tan^2\theta$ when an expression involves only secant and tangent, or when you want to change one of these functions into the other.

Example

Simplifying with secant and tangent

Simplify

sec⁡2θ−1tan⁡θ.\frac{\sec^2\theta-1}{\tan\theta}.

Rearrange $sec⁡2θ=1+tan⁡2θ\sec^2\theta=1+\tan^2\theta$ to obtain
$sec⁡2θ−1=tan⁡2θ.\sec^2\theta-1=\tan^2\theta.$
Substitute this into the numerator:
$sec⁡2θ−1tan⁡θ=tan⁡2θtan⁡θ.\frac{\sec^2\theta-1}{\tan\theta} = \frac{\tan^2\theta}{\tan\theta}.$
Cancel one factor of $tan⁡θ\tan\theta$ :
$sec⁡2θ−1tan⁡θ=tan⁡θ.\frac{\sec^2\theta-1}{\tan\theta}=\tan\theta.$

Deriving the cosecant identity

Again begin with

sin⁡2θ+cos⁡2θ≡1.\sin^2\theta+\cos^2\theta\equiv 1.

This time, divide every term by $sin⁡2θ\sin^2\theta$ :

sin⁡2θsin⁡2θ+cos⁡2θsin⁡2θ≡1sin⁡2θ.\frac{\sin^2\theta}{\sin^2\theta} + \frac{\cos^2\theta}{\sin^2\theta} \equiv \frac{1}{\sin^2\theta}.

Since $cot⁡θ=cos⁡θsin⁡θ\cot\theta=\frac{\cos\theta}{\sin\theta}$ and $cosec⁡θ=1sin⁡θ\operatorname{cosec}\theta=\frac{1}{\sin\theta}$ , this becomes

1+cot⁡2θ≡cosec⁡2θ.1+\cot^2\theta\equiv\operatorname{cosec}^2\theta.

Hence,

cosec⁡2θ≡1+cot⁡2θ.\operatorname{cosec}^2\theta\equiv 1+\cot^2\theta.

Tip

Remembering the pairings

Secant pairs with tangent, while cosecant pairs with cotangent:

sec⁡2θ≡1+tan⁡2θ,cosec⁡2θ≡1+cot⁡2θ.\sec^2\theta\equiv 1+\tan^2\theta, \qquad \operatorname{cosec}^2\theta\equiv 1+\cot^2\theta.

Example

Finding a cosecant value

Given that $cot⁡θ=2\cot\theta=2$ and $cosec⁡θ>0\operatorname{cosec}\theta>0$ , find $cosec⁡θ\operatorname{cosec}\theta$ .

Apply the appropriate identity:
$cosec⁡2θ=1+cot⁡2θ.\operatorname{cosec}^2\theta=1+\cot^2\theta.$
Substitute $cot⁡θ=2\cot\theta=2$ :
$cosec⁡2θ=1+22=5.\operatorname{cosec}^2\theta=1+2^2=5.$
Taking square roots gives $cosec⁡θ=±5\operatorname{cosec}\theta=\pm\sqrt5$ . The given sign condition selects
$cosec⁡θ=5.\operatorname{cosec}\theta=\sqrt5.$

Common Mistake

Forgetting both square roots

Proving trigonometric identities

To prove an identity, usually start with the more complicated side and transform it until it matches the other side. Avoid changing both sides at once, as this can hide gaps in your reasoning.

Converting tangent, cotangent, secant and cosecant into sine and cosine often makes the route clearer.

Example

Proving an identity using sine and cosine

Prove that

1−cos⁡2θsin⁡θ≡sin⁡θ.\frac{1-\cos^2\theta}{\sin\theta}\equiv\sin\theta.

Use the rearranged Pythagorean identity $1−cos⁡2θ=sin⁡2θ1-\cos^2\theta=\sin^2\theta$ :
$1−cos⁡2θsin⁡θ=sin⁡2θsin⁡θ.\frac{1-\cos^2\theta}{\sin\theta} = \frac{\sin^2\theta}{\sin\theta}.$
Write $sin⁡2θ\sin^2\theta$ as a product:
$sin⁡θsin⁡θsin⁡θ.\frac{\sin\theta\sin\theta}{\sin\theta}.$
Cancel the common factor, wherever the original expression is defined:
$1−cos⁡2θsin⁡θ=sin⁡θ,\frac{1-\cos^2\theta}{\sin\theta} = \sin\theta,$
which is the required right-hand side.

Common Mistake

Treating sums as factors

You cannot cancel terms across addition. For example, nothing cancels directly in $1+sin⁡θsin⁡θ\frac{1+\sin\theta}{\sin\theta}$ . Cancellation is allowed only between common factors.

Exam technique

In the exam

Identify which functions appear: sine and cosine suggest $sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1$ , while secant and tangent suggest $sec⁡2θ=1+tan⁡2θ\sec^2\theta=1+\tan^2\theta$ .
In a proof, start with the more complicated side and show each substitution or factorisation clearly.
Convert everything to sine and cosine if the route is unclear, and check that you have not divided by a quantity that could be zero.

Self review

Check yourself

Can you derive $sec⁡2θ≡1+tan⁡2θ\sec^2\theta\equiv1+\tan^2\theta$ from the first Pythagorean identity?
How would you simplify $cosec⁡2θ−cot⁡2θ\operatorname{cosec}^2\theta-\cot^2\theta$ ?
Why is $tan⁡θ≡sin⁡θcos⁡θ\tan\theta\equiv\frac{\sin\theta}{\cos\theta}$ not defined when $cos⁡θ=0\cos\theta=0$ ?

1.8.5 Trigonometric identities

What you'll learn

Prerequisites: the trigonometric ratios

What is an identity?

The tangent identity

The first Pythagorean identity

Deriving the secant identity

Deriving the cosecant identity

Proving trigonometric identities

Review 1.5.10 Trigonometric identities by teaching Genie

1.8.5 Trigonometric identities

What you'll learn

Prerequisites: the trigonometric ratios

What is an identity?

The tangent identity

The first Pythagorean identity

Deriving the secant identity

Deriving the cosecant identity

Proving trigonometric identities

Review 1.5.10 Trigonometric identities by teaching Genie

1.8.5 Trigonometric identities

What you'll learn

Prerequisites: the trigonometric ratios

What is an identity?

The tangent identity

The first Pythagorean identity

Deriving the secant identity

Deriving the cosecant identity

Proving trigonometric identities

Review 1.5.10 Trigonometric identities by teaching Genie

1.5.10 Trigonometric identities Revision Guide

1.8.5 Trigonometric identities

What you'll learn

Prerequisites: the trigonometric ratios

What is an identity?

The tangent identity

The first Pythagorean identity

Deriving the secant identity

Deriving the cosecant identity

Proving trigonometric identities

Review 1.5.10 Trigonometric identities by teaching Genie

1.5.10 Trigonometric identities Revision Guide