Trigonometric Identities and Equations

What you'll learn

How to find all angles that satisfy a trigonometric equation in a given interval.
How to use the key identities $sin⁡2x+cos⁡2x≡1\sin^2 x+\cos^2 x\equiv 1$ and $tan⁡x≡sin⁡xcos⁡x\tan x\equiv \frac{\sin x}{\cos x}$ .
How to turn trig equations into quadratics in $sin⁡x\sin x$ or $cos⁡x\cos x$ .
How to avoid losing solutions when rearranging equations involving $tan⁡x\tan x$ .

1. The basics: angles, periods and quadrants

In AS Maths, trigonometric equations are usually in degrees. Before doing anything else, make sure your calculator is in degree mode.

Definition

Period and reference angle

The period of a trig function is the angle after which its values repeat. Sine and cosine have period 360°, while tangent has period 180°.
A reference angle is the acute angle made with the x-axis. It helps you find the matching angle in another quadrant.

The unit circle explains why one trig value often gives more than one answer. For example, $cos⁡x=0.4\cos x=0.4$ has two solutions between 0° and 360° because cosine is positive in Quadrants I and IV.

$A unit circle diagram showing where cosine is positive and why \cos x=0.4 gives two angles in one full turn.$

Unit circle showing CAST signs and reference angle

Key Idea

CAST rule

Use CAST to remember signs: All positive in Quadrant I, Sine positive in Quadrant II, Tangent positive in Quadrant III, Cosine positive in Quadrant IV.

Example

Solving a shifted cosine equation

Solve $5cos⁡(x−35)=25\cos(x-35)=2$ for $0∘≤x<360∘0^\circ \leq x < 360^\circ$ , giving your answers to two decimal places.

Divide both sides by 5:
$cos⁡(x−35)=0.4\cos(x-35)=0.4$
Let $u = x - 35$ . Since $0∘≤x<360∘0^\circ \leq x < 360^\circ$ , the new interval is:

The substitution u=x-35 shifts the whole solving interval 35° to the left.

$$
-35^\circ \leq u < 325^\circ
$$

3. Find the reference angle using inverse cosine:

$$
\alpha=\cos^{-1}(0.4)=66.4218\ldots^\circ
$$

4. Cosine is positive in Quadrants I and IV, so within the interval for $u$ :

$$
u=66.4218\ldots^\circ,\quad u=360^\circ-66.4218\ldots^\circ=293.5781\ldots^\circ
$$

5. Add 35° to return to $x$ :

$$
x=101.42^\circ,\quad x=328.58^\circ
$$

Tip

Calculator warning

If you use $sin^{-1}$ , $cos^{-1}$ or $tan^{-1}$ , your calculator gives only one angle. Your job is to use quadrants and the interval to find the rest.

2. Composite angles: change the interval carefully

The argument of a trig function is the expression inside it. In $sin⁡(3θ−20)\sin(3\theta-20)$ , the argument is $3θ−203\theta-20$ .

When the argument is not just the variable, solve using a substitution such as $u=3θ−20u=3\theta-20$ . The most important step is to transform the interval.

Example

Solving a sine equation with a composite angle

Solve $sin⁡(3θ−20)=0.7\sin(3\theta-20)=0.7$ for $0∘≤θ<180∘0^\circ \leq \theta < 180^\circ$ , giving your answers to two decimal places.

Let $u=3θ−20u=3\theta-20$ .
Convert the interval. When $θ=0∘\theta=0^\circ$ , $u=−20∘u=-20^\circ$ . When $θ\theta$ approaches 180°, $u$ approaches 520°:

$For u=3\theta-20, multiplying by 3 stretches the interval before shifting it by 20°.$

$$
-20^\circ \leq u < 520^\circ
$$

3. Find the reference angle:

$$
\alpha=\sin^{-1}(0.7)=44.4270\ldots^\circ
$$

4. Sine is positive in Quadrants I and II, so the possible values of $u$ are:

Over the extended interval, the positive sine solutions repeat after 360° and give four possible u values.

$$
u=44.4270\ldots^\circ,\quad 135.5729\ldots^\circ,\quad 404.4270\ldots^\circ,\quad 495.5729\ldots^\circ
$$

5. Use $u=3θ−20u=3\theta-20$ , so $θ=u+203\theta=\frac{u+20}{3}$ :

$$
\theta=21.48^\circ,\quad 51.86^\circ,\quad 141.48^\circ,\quad 171.86^\circ
$$

3. Core identities

Definition

Trigonometric identity

A trigonometric identity is an equation that is true for all allowed values of the variable. The symbol $≡\equiv$ means “identically equal to”.

The two identities you use most in this topic are:

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

and

tan⁡x≡sin⁡xcos⁡x,cos⁡x≠0\tan x\equiv \frac{\sin x}{\cos x},\quad \cos x\neq 0

You can rearrange the first one as:

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

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

These are especially useful when an equation contains both $sin^2 x$ and $cos⁡x\cos x$ , or both $cos^2 x$ and $sin⁡x\sin x$ .

Example

Turning a trig equation into a quadratic

Show that $3\sin^2 x=4\cos x+2$ can be written as $3\cos^2 x+4\cos x-1=0$ , then solve it for $0∘≤x<360∘0^\circ \leq x < 360^\circ$ .

Replace $sin^2 x$ with $1-\cos^2 x$ :
$3(1-\cos^2 x)=4\cos x+2$
Expand and rearrange into a quadratic:
$3−3cos⁡2x=4cos⁡x+23cos⁡2x+4cos⁡x−1=0\begin{aligned} 3-3\cos^2 x&=4\cos x+2\\ 3\cos^2 x+4\cos x-1&=0 \end{aligned}$
Factorise:
$(3cos⁡x−1)(cos⁡x+1)=0(3\cos x-1)(\cos x+1)=0$
Solve each part:
$cos⁡x=13orcos⁡x=−1\cos x=\frac{1}{3}\quad \text{or}\quad \cos x=-1$
For $cos⁡x=13\cos x=\frac{1}{3}$ , cosine is positive in Quadrants I and IV. Also, $cos⁡x=−1\cos x=-1$ gives $x=180∘x=180^\circ$ :

$The quadratic gives one pair of cosine-positive solutions and one exact solution at 180^\circ.$

$$
x=70.5^\circ,\quad 180^\circ,\quad 289.5^\circ
$$

Common Mistake

Accepting impossible trig values

If a quadratic gives $sin⁡x=1.4\sin x=1.4$ or $cos⁡x=−2\cos x=-2$ , reject it immediately. Sine and cosine values must lie between -1 and 1.

4. Equations involving tangent

Tangent often appears with sine and cosine because:

tan⁡x=sin⁡xcos⁡x\tan x=\frac{\sin x}{\cos x}

But be careful: $tan⁡x\tan x$ is undefined when $cos⁡x=0\cos x=0$ .

$Tangent is undefined where the unit-circle x-coordinate, \cos x, is zero.$

Example

Do not divide by a possible zero factor

Solve $3tan⁡x=5sin⁡x3\tan x=5\sin x$ for $0∘≤x<360∘0^\circ \leq x < 360^\circ$ , giving answers to one decimal place where appropriate.

Replace $tan⁡x\tan x$ with $sin⁡xcos⁡x\frac{\sin x}{\cos x}$ :
$3⋅sin⁡xcos⁡x=5sin⁡x3\cdot \frac{\sin x}{\cos x}=5\sin x$
Multiply by $cos⁡x\cos x$ . Values with $cos⁡x=0\cos x=0$ are not valid anyway because $tan⁡x\tan x$ would be undefined:
$3sin⁡x=5sin⁡xcos⁡x3\sin x=5\sin x\cos x$
Bring everything to one side and factorise:

$Factoring keeps the \sin x=0 branch, which would be lost by dividing by \sin x.$

$$
\sin x(3-5\cos x)=0
$$

4. Solve each factor:

$$
\sin x=0\quad \text{or}\quad \cos x=\frac{3}{5}
$$

5. Find all values in the interval:

$$
x=0^\circ,\quad 53.1^\circ,\quad 180^\circ,\quad 306.9^\circ
$$

Common Mistake

Cancelling away solutions

Do not divide both sides by $sin⁡x\sin x$ unless you have separately checked $sin⁡x=0\sin x=0$ . In this example, dividing by $sin⁡x\sin x$ would lose $x=0∘x=0^\circ$ and $x=180∘x=180^\circ$ .

5. Squared trig equations

If you see something like $tan^2 x=3$ , remember that this means:

tan⁡x=3ortan⁡x=−3\tan x=\sqrt{3}\quad \text{or}\quad \tan x=-\sqrt{3}

So squared equations often produce more solutions than expected.

Example

Solving a squared tangent equation

Solve $tan^2(2x)=3$ for $0∘≤x<180∘0^\circ \leq x < 180^\circ$ .

Take the square root carefully:
$tan⁡(2x)=±3\tan(2x)=\pm\sqrt{3}$
Let $u = 2 x$ . The interval becomes:
$0∘≤u<360∘0^\circ \leq u < 360^\circ$
Since $tan⁡u=3\tan u=\sqrt{3}$ or $tan⁡u=−3\tan u=-\sqrt{3}$ , the solutions for $u$ are:
$u=60∘,120∘,240∘,300∘u=60^\circ,\quad 120^\circ,\quad 240^\circ,\quad 300^\circ$
Divide by 2:
$x=30∘,60∘,120∘,150∘x=30^\circ,\quad 60^\circ,\quad 120^\circ,\quad 150^\circ$

6. Graphs and transformations

The graph of $y=sin⁡(x−30)y=\sin(x-30)$ is the graph of $y=sin⁡xy=\sin x$ shifted 30° to the right. This helps you check whether your number of solutions is sensible.

The transformed sine graph is the standard sine graph shifted 30° to the right.

Graph comparing y equals sin x and y equals sin x minus 30 degrees

Example

Using a transformed sine graph idea

Find all solutions of $sin⁡(x−30)=0.4\sin(x-30)=0.4$ for $0∘≤x<360∘0^\circ \leq x < 360^\circ$ .

Let $u = x - 30$ , so:
$−30∘≤u<330∘-30^\circ \leq u < 330^\circ$
Find the reference angle:
$α=sin⁡−1(0.4)=23.5781…∘\alpha=\sin^{-1}(0.4)=23.5781\ldots^\circ$
Sine is positive in Quadrants I and II, so:
$u=23.5781…∘,156.4218…∘u=23.5781\ldots^\circ,\quad 156.4218\ldots^\circ$
Add 30° to return to $x$ :
$x=53.6∘,186.4∘x=53.6^\circ,\quad 186.4^\circ$

7. Proving identities by simplifying one side

To prove an identity, usually start with the more complicated side and use known identities until it becomes the simpler side.

Example

Proving a fractional identity

Prove that $3sin⁡x−cos⁡2x+13+sin⁡x≡sin⁡x\frac{3\sin x-\cos^2 x+1}{3+\sin x}\equiv \sin x$ .

Start with the left-hand side:
$3sin⁡x−cos⁡2x+13+sin⁡x\frac{3\sin x-\cos^2 x+1}{3+\sin x}$
Replace $cos^2 x$ with $1-\sin^2 x$ :
$3sin⁡x−(1−sin⁡2x)+13+sin⁡x\frac{3\sin x-(1-\sin^2 x)+1}{3+\sin x}$
Simplify the numerator:
$3sin⁡x+sin⁡2x3+sin⁡x\frac{3\sin x+\sin^2 x}{3+\sin x}$
Factorise the numerator:
$sin⁡x(3+sin⁡x)3+sin⁡x\frac{\sin x(3+\sin x)}{3+\sin x}$
Cancel the common factor:
$sin⁡x\sin x$

Exam technique

In the exam

Write the transformed interval when you use a substitution like $u = 2 x + 15$ .
Use CAST or a sketch to find every solution, not just the calculator’s first answer.
Avoid dividing by $sin⁡x\sin x$ or $cos⁡x\cos x$ unless you have checked the zero case and any undefined tangent values.

Self review

Check yourself

Can you explain why $cos⁡x=0.4\cos x=0.4$ has two solutions between 0° and 360°?
When solving $sin⁡(3x−10)=0.5\sin(3x-10)=0.5$ , what interval should you use for the argument?
Why might dividing an equation by $sin⁡x\sin x$ cause you to lose a solution?

Trigonometric Identities and Equations

What you'll learn

How to find all angles that satisfy a trigonometric equation in a given interval.
How to use the key identities $sin⁡2x+cos⁡2x≡1\sin^2 x+\cos^2 x\equiv 1$ and $tan⁡x≡sin⁡xcos⁡x\tan x\equiv \frac{\sin x}{\cos x}$ .
How to turn trig equations into quadratics in $sin⁡x\sin x$ or $cos⁡x\cos x$ .
How to avoid losing solutions when rearranging equations involving $tan⁡x\tan x$ .

1. The basics: angles, periods and quadrants

In AS Maths, trigonometric equations are usually in degrees. Before doing anything else, make sure your calculator is in degree mode.

Definition

Period and reference angle

The period of a trig function is the angle after which its values repeat. Sine and cosine have period 360°, while tangent has period 180°.
A reference angle is the acute angle made with the x-axis. It helps you find the matching angle in another quadrant.

$A unit circle diagram showing where cosine is positive and why \cos x=0.4 gives two angles in one full turn.$

Unit circle showing CAST signs and reference angle

Key Idea

CAST rule

Use CAST to remember signs: All positive in Quadrant I, Sine positive in Quadrant II, Tangent positive in Quadrant III, Cosine positive in Quadrant IV.

Example

Solving a shifted cosine equation

Solve $5cos⁡(x−35)=25\cos(x-35)=2$ for $0∘≤x<360∘0^\circ \leq x < 360^\circ$ , giving your answers to two decimal places.

Divide both sides by 5:
$cos⁡(x−35)=0.4\cos(x-35)=0.4$
Let $u = x - 35$ . Since $0∘≤x<360∘0^\circ \leq x < 360^\circ$ , the new interval is:

The substitution u=x-35 shifts the whole solving interval 35° to the left.

$$
-35^\circ \leq u < 325^\circ
$$

3. Find the reference angle using inverse cosine:

$$
\alpha=\cos^{-1}(0.4)=66.4218\ldots^\circ
$$

4. Cosine is positive in Quadrants I and IV, so within the interval for $u$ :

$$
u=66.4218\ldots^\circ,\quad u=360^\circ-66.4218\ldots^\circ=293.5781\ldots^\circ
$$

5. Add 35° to return to $x$ :

$$
x=101.42^\circ,\quad x=328.58^\circ
$$

Tip

Calculator warning

If you use $sin^{-1}$ , $cos^{-1}$ or $tan^{-1}$ , your calculator gives only one angle. Your job is to use quadrants and the interval to find the rest.

2. Composite angles: change the interval carefully

The argument of a trig function is the expression inside it. In $sin⁡(3θ−20)\sin(3\theta-20)$ , the argument is $3θ−203\theta-20$ .

When the argument is not just the variable, solve using a substitution such as $u=3θ−20u=3\theta-20$ . The most important step is to transform the interval.

Example

Solving a sine equation with a composite angle

Solve $sin⁡(3θ−20)=0.7\sin(3\theta-20)=0.7$ for $0∘≤θ<180∘0^\circ \leq \theta < 180^\circ$ , giving your answers to two decimal places.

Let $u=3θ−20u=3\theta-20$ .
Convert the interval. When $θ=0∘\theta=0^\circ$ , $u=−20∘u=-20^\circ$ . When $θ\theta$ approaches 180°, $u$ approaches 520°:

$For u=3\theta-20, multiplying by 3 stretches the interval before shifting it by 20°.$

$$
-20^\circ \leq u < 520^\circ
$$

3. Find the reference angle:

$$
\alpha=\sin^{-1}(0.7)=44.4270\ldots^\circ
$$

4. Sine is positive in Quadrants I and II, so the possible values of $u$ are:

Over the extended interval, the positive sine solutions repeat after 360° and give four possible u values.

$$
u=44.4270\ldots^\circ,\quad 135.5729\ldots^\circ,\quad 404.4270\ldots^\circ,\quad 495.5729\ldots^\circ
$$

5. Use $u=3θ−20u=3\theta-20$ , so $θ=u+203\theta=\frac{u+20}{3}$ :

$$
\theta=21.48^\circ,\quad 51.86^\circ,\quad 141.48^\circ,\quad 171.86^\circ
$$

3. Core identities

Definition

Trigonometric identity

A trigonometric identity is an equation that is true for all allowed values of the variable. The symbol $≡\equiv$ means “identically equal to”.

The two identities you use most in this topic are:

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

and

tan⁡x≡sin⁡xcos⁡x,cos⁡x≠0\tan x\equiv \frac{\sin x}{\cos x},\quad \cos x\neq 0

You can rearrange the first one as:

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

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

These are especially useful when an equation contains both $sin^2 x$ and $cos⁡x\cos x$ , or both $cos^2 x$ and $sin⁡x\sin x$ .

Example

Turning a trig equation into a quadratic

Show that $3\sin^2 x=4\cos x+2$ can be written as $3\cos^2 x+4\cos x-1=0$ , then solve it for $0∘≤x<360∘0^\circ \leq x < 360^\circ$ .

Replace $sin^2 x$ with $1-\cos^2 x$ :
$3(1-\cos^2 x)=4\cos x+2$
Expand and rearrange into a quadratic:
$3−3cos⁡2x=4cos⁡x+23cos⁡2x+4cos⁡x−1=0\begin{aligned} 3-3\cos^2 x&=4\cos x+2\\ 3\cos^2 x+4\cos x-1&=0 \end{aligned}$
Factorise:
$(3cos⁡x−1)(cos⁡x+1)=0(3\cos x-1)(\cos x+1)=0$
Solve each part:
$cos⁡x=13orcos⁡x=−1\cos x=\frac{1}{3}\quad \text{or}\quad \cos x=-1$
For $cos⁡x=13\cos x=\frac{1}{3}$ , cosine is positive in Quadrants I and IV. Also, $cos⁡x=−1\cos x=-1$ gives $x=180∘x=180^\circ$ :

$The quadratic gives one pair of cosine-positive solutions and one exact solution at 180^\circ.$

$$
x=70.5^\circ,\quad 180^\circ,\quad 289.5^\circ
$$

Common Mistake

Accepting impossible trig values

If a quadratic gives $sin⁡x=1.4\sin x=1.4$ or $cos⁡x=−2\cos x=-2$ , reject it immediately. Sine and cosine values must lie between -1 and 1.

4. Equations involving tangent

Tangent often appears with sine and cosine because:

tan⁡x=sin⁡xcos⁡x\tan x=\frac{\sin x}{\cos x}

But be careful: $tan⁡x\tan x$ is undefined when $cos⁡x=0\cos x=0$ .

$Tangent is undefined where the unit-circle x-coordinate, \cos x, is zero.$

Example

Do not divide by a possible zero factor

Solve $3tan⁡x=5sin⁡x3\tan x=5\sin x$ for $0∘≤x<360∘0^\circ \leq x < 360^\circ$ , giving answers to one decimal place where appropriate.

Replace $tan⁡x\tan x$ with $sin⁡xcos⁡x\frac{\sin x}{\cos x}$ :
$3⋅sin⁡xcos⁡x=5sin⁡x3\cdot \frac{\sin x}{\cos x}=5\sin x$
Multiply by $cos⁡x\cos x$ . Values with $cos⁡x=0\cos x=0$ are not valid anyway because $tan⁡x\tan x$ would be undefined:
$3sin⁡x=5sin⁡xcos⁡x3\sin x=5\sin x\cos x$
Bring everything to one side and factorise:

$Factoring keeps the \sin x=0 branch, which would be lost by dividing by \sin x.$

$$
\sin x(3-5\cos x)=0
$$

4. Solve each factor:

$$
\sin x=0\quad \text{or}\quad \cos x=\frac{3}{5}
$$

5. Find all values in the interval:

$$
x=0^\circ,\quad 53.1^\circ,\quad 180^\circ,\quad 306.9^\circ
$$

Common Mistake

Cancelling away solutions

5. Squared trig equations

If you see something like $tan^2 x=3$ , remember that this means:

tan⁡x=3ortan⁡x=−3\tan x=\sqrt{3}\quad \text{or}\quad \tan x=-\sqrt{3}

So squared equations often produce more solutions than expected.

Example

Solving a squared tangent equation

Solve $tan^2(2x)=3$ for $0∘≤x<180∘0^\circ \leq x < 180^\circ$ .

Take the square root carefully:
$tan⁡(2x)=±3\tan(2x)=\pm\sqrt{3}$
Let $u = 2 x$ . The interval becomes:
$0∘≤u<360∘0^\circ \leq u < 360^\circ$
Since $tan⁡u=3\tan u=\sqrt{3}$ or $tan⁡u=−3\tan u=-\sqrt{3}$ , the solutions for $u$ are:
$u=60∘,120∘,240∘,300∘u=60^\circ,\quad 120^\circ,\quad 240^\circ,\quad 300^\circ$
Divide by 2:
$x=30∘,60∘,120∘,150∘x=30^\circ,\quad 60^\circ,\quad 120^\circ,\quad 150^\circ$

6. Graphs and transformations

The graph of $y=sin⁡(x−30)y=\sin(x-30)$ is the graph of $y=sin⁡xy=\sin x$ shifted 30° to the right. This helps you check whether your number of solutions is sensible.

The transformed sine graph is the standard sine graph shifted 30° to the right.

Graph comparing y equals sin x and y equals sin x minus 30 degrees

Example

Using a transformed sine graph idea

Find all solutions of $sin⁡(x−30)=0.4\sin(x-30)=0.4$ for $0∘≤x<360∘0^\circ \leq x < 360^\circ$ .

Let $u = x - 30$ , so:
$−30∘≤u<330∘-30^\circ \leq u < 330^\circ$
Find the reference angle:
$α=sin⁡−1(0.4)=23.5781…∘\alpha=\sin^{-1}(0.4)=23.5781\ldots^\circ$
Sine is positive in Quadrants I and II, so:
$u=23.5781…∘,156.4218…∘u=23.5781\ldots^\circ,\quad 156.4218\ldots^\circ$
Add 30° to return to $x$ :
$x=53.6∘,186.4∘x=53.6^\circ,\quad 186.4^\circ$

7. Proving identities by simplifying one side

To prove an identity, usually start with the more complicated side and use known identities until it becomes the simpler side.

Example

Proving a fractional identity

Prove that $3sin⁡x−cos⁡2x+13+sin⁡x≡sin⁡x\frac{3\sin x-\cos^2 x+1}{3+\sin x}\equiv \sin x$ .

Start with the left-hand side:
$3sin⁡x−cos⁡2x+13+sin⁡x\frac{3\sin x-\cos^2 x+1}{3+\sin x}$
Replace $cos^2 x$ with $1-\sin^2 x$ :
$3sin⁡x−(1−sin⁡2x)+13+sin⁡x\frac{3\sin x-(1-\sin^2 x)+1}{3+\sin x}$
Simplify the numerator:
$3sin⁡x+sin⁡2x3+sin⁡x\frac{3\sin x+\sin^2 x}{3+\sin x}$
Factorise the numerator:
$sin⁡x(3+sin⁡x)3+sin⁡x\frac{\sin x(3+\sin x)}{3+\sin x}$
Cancel the common factor:
$sin⁡x\sin x$

Exam technique

In the exam

Write the transformed interval when you use a substitution like $u = 2 x + 15$ .
Use CAST or a sketch to find every solution, not just the calculator’s first answer.
Avoid dividing by $sin⁡x\sin x$ or $cos⁡x\cos x$ unless you have checked the zero case and any undefined tangent values.

Self review

Check yourself

Can you explain why $cos⁡x=0.4\cos x=0.4$ has two solutions between 0° and 360°?
When solving $sin⁡(3x−10)=0.5\sin(3x-10)=0.5$ , what interval should you use for the argument?
Why might dividing an equation by $sin⁡x\sin x$ cause you to lose a solution?

Trigonometric Identities and Equations

What you'll learn

1. The basics: angles, periods and quadrants

2. Composite angles: change the interval carefully

3. Core identities

4. Equations involving tangent

5. Squared trig equations

6. Graphs and transformations

7. Proving identities by simplifying one side

Recap questions

You've reached the end

Trigonometric Identities and Equations

What you'll learn

1. The basics: angles, periods and quadrants

2. Composite angles: change the interval carefully

3. Core identities

4. Equations involving tangent

5. Squared trig equations

6. Graphs and transformations

7. Proving identities by simplifying one side

Recap questions

You've reached the end

Trigonometric Identities and Equations

What you'll learn

1. The basics: angles, periods and quadrants

2. Composite angles: change the interval carefully

3. Core identities

4. Equations involving tangent

5. Squared trig equations

6. Graphs and transformations

7. Proving identities by simplifying one side

Recap questions

You've reached the end

Trigonometric Identities and Equations Revision Guide

Trigonometric Identities and Equations

What you'll learn

1. The basics: angles, periods and quadrants

2. Composite angles: change the interval carefully

3. Core identities

4. Equations involving tangent

5. Squared trig equations

6. Graphs and transformations

7. Proving identities by simplifying one side

Recap questions

You've reached the end

Trigonometric Identities and Equations Revision Guide