Trigonometric Ratios

What you'll learn

How sine, cosine and tangent connect angles to side lengths.
When to use the sine rule, cosine rule and area formula.
How to handle exact values and the “two possible angles” situation.
How to sketch simple sine and cosine graphs from 0° to 360°.

1. Starting point: right-angled triangle ratios

A ratio compares two quantities by division. In trigonometry, the main ratios compare side lengths in a right-angled triangle.

For an angle $θ\theta$ :

A right-angled triangle labelled with hypotenuse, opposite and adjacent sides relative to angle θ.

The hypotenuse is the longest side, opposite the right angle.
The opposite side is opposite $θ\theta$ .
The adjacent side touches $θ\theta$ , but is not the hypotenuse.

Definition

SOH CAH TOA

In a right-angled triangle,

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}}

Tip

Calculator mode

For these notes, angles are in degrees. Make sure your calculator is in degree mode before using $sin⁡\sin$ , $cos⁡\cos$ , $tan⁡\tan$ or their inverse functions.

Example

Using a basic trigonometric ratio

A right-angled triangle has hypotenuse 13 cm and an angle of 28°. Find the side opposite the 28° angle.

The given right-angled triangle for finding an unknown opposite side using sine.

Choose the ratio involving opposite and hypotenuse: $sin⁡θ\sin \theta$ .
Substitute the values:
$sin⁡28∘=x13\sin 28^\circ = \frac{x}{13}$
Rearrange to find the side length:
$13\sin 28^\circ \approx 6.10$

2. Labelling non-right-angled triangles

At AS Level, you often work with triangles that are not right-angled. We use a standard labelling convention:

Standard triangle notation showing each side opposite its matching capital-letter angle.

Angle $A$ is opposite side $a$ .
Angle $B$ is opposite side $b$ .
Angle $C$ is opposite side $c$ .

This matters because the sine rule and cosine rule are built around opposite angle-side pairs.

Diagram showing standard triangle notation with sides opposite angles

Key Idea

Label before calculating

Before choosing a formula, mark the angle you know and the side opposite it. Most mistakes in this topic come from using the wrong opposite pair.

3. Area of a triangle using sine

You already know:

triangle=12×base×height\text{Area of a triangle} = \frac{1}{2} \times \text{base} \times \text{height}

For a non-right-angled triangle, the height may not be given directly. If you know two sides and the included angle between them, use:

The sine area formula comes from using the perpendicular height formed from one side and the included angle.

Area=12absin⁡C\text{Area} = \frac{1}{2}ab\sin C

Here, $a$ and $b$ are the two known sides, and $C$ is the angle between them.

Definition

Included angle

The included angle is the angle between two given sides. For example, if you know sides $A B$ and $A C$ , the included angle is angle $B A C$ .

Example

Finding an area from two sides and an included angle

In triangle $A B C$ , $14\text{ cm}$ , $9\text{ cm}$ and angle $35^\circ$ . Find the area.

The known sides meet at $A$ , so the included angle is 35°.
Substitute into the area formula:
$Area=12×14×9×sin⁡35∘\text{Area} = \frac{1}{2} \times 14 \times 9 \times \sin 35^\circ$
Simplify and calculate:
$cm2\text{Area} = 63\sin 35^\circ \approx 36.1\text{ cm}^2$

Common Mistake

Using the wrong angle in the area formula

The angle in $12absin⁡C\frac{1}{2}ab\sin C$ must be between the two sides you are multiplying. If the angle is not included, do not use this formula directly.

4. The cosine rule

Use the cosine rule when you know:

two sides and the included angle, and want the third side; or
all three sides, and want an angle.

The main form is:

a^2 = b^2 + c^2 - 2bc\cos A

This finds side $a$ , which is opposite angle $A$ .

To find an angle, rearrange it:

cos⁡A=b2+c2−a22bc\cos A = \frac{b^2 + c^2 - a^2}{2bc}

Example

Finding a side using the cosine rule

In triangle $P QR$ , $10\text{ cm}$ , $7\text{ cm}$ and angle $65^\circ$ . Find $QR$ to 3 significant figures.

The cosine rule setup with two known sides and the included angle, and the opposite side unknown.

The side opposite angle $P$ is $QR$ , so use the cosine rule with $QR$ as the unknown side.
Substitute carefully:
$QR2=102+72−2×10×7×cos⁡65∘QR^2 = 10^2 + 7^2 - 2 \times 10 \times 7 \times \cos 65^\circ$
Calculate $QR^2$ first:
$QR2≈89.83QR^2 \approx 89.83$
Square root to find the length:
$\approx 9.48\text{ cm}$

Example

Finding an angle and then an exact area

In triangle $X Y Z$ , $5\text{ cm}$ , $8\text{ cm}$ and $7\text{ cm}$ . Find $cos⁡Y\cos Y$ , then find the exact area.

Angle $Y$ is between sides $X Y$ and $Y Z$ . The side opposite angle $Y$ is $X Z$ .
Use the rearranged cosine rule:
$cos⁡Y=52+82−722×5×8\cos Y = \frac{5^2 + 8^2 - 7^2}{2 \times 5 \times 8}$
Simplify:
$cos⁡Y=25+64−4980=12\cos Y = \frac{25 + 64 - 49}{80} = \frac{1}{2}$
Since $cos⁡Y=12\cos Y = \frac{1}{2}$ , we have $60^\circ$ , so use the area formula:
$cm2\text{Area} = \frac{1}{2} \times 5 \times 8 \times \sin 60^\circ = 10\sqrt{3}\text{ cm}^2$

5. Algebra with the cosine rule

Sometimes the side lengths contain an unknown, such as $x$ . The method is the same: substitute, expand, simplify and solve.

Example

Solving for an unknown side expression

In triangle $A B C$ , $(x+5)\text{ cm}$ , $x\text{ cm}$ , $(x+3)\text{ cm}$ and angle $60^\circ$ . Find $x$ .

The side opposite angle $A$ is $B C$ , so $B C = x + 3$ goes on the left of the cosine rule.
Substitute into $a^2 = b^2 + c^2 - 2bc\cos A$ :
$(x+3)2=(x+5)2+x2−2(x+5)xcos⁡60∘(x+3)^2 = (x+5)^2 + x^2 - 2(x+5)x\cos 60^\circ$
Use $cos⁡60∘=12\cos 60^\circ = \frac{1}{2}$ and simplify:
$x+3)^2 = (x+5)^2 + x^2 - (x+5)x$
Expand both sides:
$x^2 + 6x + 9 = x^2 + 5x + 25$
Solve:
$x = 16$

6. The sine rule and the ambiguous case

Use the sine rule when you know an opposite angle-side pair and another side or angle.

asin⁡A=bsin⁡B=csin⁡C\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}

Equivalently,

sin⁡Aa=sin⁡Bb=sin⁡Cc\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}

The second version is often neater when finding an angle.

Example

Finding two possible angles

In triangle $A B C$ , $12\text{ cm}$ , $8\text{ cm}$ and angle $35^\circ$ . Find the two possible values of angle $A B C$ to one decimal place.

The ambiguous sine-rule case can form two different triangles from the same given side, side and angle.

Match opposite pairs: $B C$ is opposite angle $A$ , and $A B$ is opposite angle $C$ .
Use the sine rule to find angle $C$ :
$sin⁡C12=sin⁡35∘8\frac{\sin C}{12} = \frac{\sin 35^\circ}{8}$
Rearrange:
$sin⁡C=12sin⁡35∘8≈0.8604\sin C = \frac{12\sin 35^\circ}{8} \approx 0.8604$
Find the two possible values of $C$ :
$\approx 59.4^\circ \quad \text{or} \quad C \approx 120.6^\circ$
Use angles in a triangle to find angle $B$ :
$180^\circ - 35^\circ - C$
Therefore the two possible values are:
$\approx 85.6^\circ \quad \text{or} \quad B \approx 24.4^\circ$

Common Mistake

Forgetting the second sine angle

If $sin⁡θ=k\sin \theta = k$ , your calculator gives one angle. In a triangle, the other possible angle is $180∘−θ180^\circ - \theta$ , as long as the angle sum still works.

7. Exact values from $sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1$

The identity

sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1

is useful when you are given one trigonometric ratio and need another.

If the angle is acute, then both sine and cosine are positive.

Example

Finding an exact cosine value

Angle $θ\theta$ is acute and $sin⁡θ=23\sin \theta = \frac{2}{3}$ . Find the exact value of $cos⁡θ\cos \theta$ .

An acute-angle right triangle representation of sin θ = 2/3, leaving the adjacent side to be found exactly.

Start with the identity:
$sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1$
Substitute $sin⁡θ=23\sin \theta = \frac{2}{3}$ :
$(23)2+cos⁡2θ=1\left(\frac{2}{3}\right)^2 + \cos^2 \theta = 1$
Rearrange:
$cos⁡2θ=1−49=59\cos^2 \theta = 1 - \frac{4}{9} = \frac{5}{9}$
Since $θ\theta$ is acute, choose the positive square root:
$cos⁡θ=53\cos \theta = \frac{\sqrt{5}}{3}$

8. Sketching sine and cosine graphs

For $\le x \le 360^\circ$ , remember the key values:

$\sin x$ starts at 0, reaches 1 at 90°, returns to 0 at 180°, reaches -1 at 270°, then returns to 0 at 360°.
$\cos x$ starts at 1, reaches 0 at 90°, reaches -1 at 180°, returns to 0 at 270°, then reaches 1 at 360°.

A vertical shift moves the graph up or down. For example, $\sin x + 1$ is the sine graph shifted up by 1.

Graph showing y equals sin x and y equals sin x plus 1 from 0 to 360 degrees

Example

Sketching a transformed cosine graph

Sketch $\cos(x+90^\circ)$ for $\le x \le 360^\circ$ .

The transformed cosine graph passes through the five key points and matches the shape of −sin x.

Work out the value of the expression inside cosine at key values of $x$ :
$x=0∘:y=cos⁡90∘=0x=90∘:y=cos⁡180∘=−1x=180∘:y=cos⁡270∘=0x=270∘:y=cos⁡360∘=1x=360∘:y=cos⁡450∘=0\begin{aligned} x = 0^\circ &: \quad y = \cos 90^\circ = 0 \\ x = 90^\circ &: \quad y = \cos 180^\circ = -1 \\ x = 180^\circ &: \quad y = \cos 270^\circ = 0 \\ x = 270^\circ &: \quad y = \cos 360^\circ = 1 \\ x = 360^\circ &: \quad y = \cos 450^\circ = 0 \end{aligned}$
Plot the points $(0, 0)$ , $(90, - 1)$ , $(180, 0)$ , $(270, 1)$ and $(360, 0)$ .
Join the points with a smooth wave-shaped curve.
Notice that $\cos(x+90^\circ)$ has the same shape as $-\sin x$ .

Exam technique

In the exam

Label opposite angle-side pairs before choosing a formula.
Use the cosine rule for two sides with the included angle, or for all three sides.
Use the sine rule when you have an opposite pair, and always check for a second possible angle.
For graph sketches, plot the five key x-values: 0°, 90°, 180°, 270° and 360°.

Self review

Check yourself

Which formula would you use if you knew two sides and the angle between them?
Why can $sin⁡θ=0.6\sin \theta = 0.6$ lead to two possible angles in a triangle?
What are the key points of $\sin x + 1$ between 0° and 360°?

What you'll learn

How sine, cosine and tangent connect angles to side lengths.
When to use the sine rule, cosine rule and area formula.
How to handle exact values and the “two possible angles” situation.
How to sketch simple sine and cosine graphs from 0° to 360°.

1. Starting point: right-angled triangle ratios

A ratio compares two quantities by division. In trigonometry, the main ratios compare side lengths in a right-angled triangle.

For an angle $θ\theta$ :

A right-angled triangle labelled with hypotenuse, opposite and adjacent sides relative to angle θ.

The hypotenuse is the longest side, opposite the right angle.
The opposite side is opposite $θ\theta$ .
The adjacent side touches $θ\theta$ , but is not the hypotenuse.

Definition

SOH CAH TOA

In a right-angled triangle,

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}}

Tip

Calculator mode

For these notes, angles are in degrees. Make sure your calculator is in degree mode before using $sin⁡\sin$ , $cos⁡\cos$ , $tan⁡\tan$ or their inverse functions.

Example

Using a basic trigonometric ratio

A right-angled triangle has hypotenuse 13 cm and an angle of 28°. Find the side opposite the 28° angle.

The given right-angled triangle for finding an unknown opposite side using sine.

Choose the ratio involving opposite and hypotenuse: $sin⁡θ\sin \theta$ .
Substitute the values:
$sin⁡28∘=x13\sin 28^\circ = \frac{x}{13}$
Rearrange to find the side length:
$13\sin 28^\circ \approx 6.10$

2. Labelling non-right-angled triangles

At AS Level, you often work with triangles that are not right-angled. We use a standard labelling convention:

Standard triangle notation showing each side opposite its matching capital-letter angle.

Angle $A$ is opposite side $a$ .
Angle $B$ is opposite side $b$ .
Angle $C$ is opposite side $c$ .

This matters because the sine rule and cosine rule are built around opposite angle-side pairs.

Diagram showing standard triangle notation with sides opposite angles

Key Idea

Label before calculating

Before choosing a formula, mark the angle you know and the side opposite it. Most mistakes in this topic come from using the wrong opposite pair.

3. Area of a triangle using sine

You already know:

triangle=12×base×height\text{Area of a triangle} = \frac{1}{2} \times \text{base} \times \text{height}

For a non-right-angled triangle, the height may not be given directly. If you know two sides and the included angle between them, use:

The sine area formula comes from using the perpendicular height formed from one side and the included angle.

Area=12absin⁡C\text{Area} = \frac{1}{2}ab\sin C

Here, $a$ and $b$ are the two known sides, and $C$ is the angle between them.

Definition

Included angle

The included angle is the angle between two given sides. For example, if you know sides $A B$ and $A C$ , the included angle is angle $B A C$ .

Example

Finding an area from two sides and an included angle

In triangle $A B C$ , $14\text{ cm}$ , $9\text{ cm}$ and angle $35^\circ$ . Find the area.

The known sides meet at $A$ , so the included angle is 35°.
Substitute into the area formula:
$Area=12×14×9×sin⁡35∘\text{Area} = \frac{1}{2} \times 14 \times 9 \times \sin 35^\circ$
Simplify and calculate:
$cm2\text{Area} = 63\sin 35^\circ \approx 36.1\text{ cm}^2$

Common Mistake

Using the wrong angle in the area formula

The angle in $12absin⁡C\frac{1}{2}ab\sin C$ must be between the two sides you are multiplying. If the angle is not included, do not use this formula directly.

4. The cosine rule

Use the cosine rule when you know:

two sides and the included angle, and want the third side; or
all three sides, and want an angle.

The main form is:

a^2 = b^2 + c^2 - 2bc\cos A

This finds side $a$ , which is opposite angle $A$ .

To find an angle, rearrange it:

cos⁡A=b2+c2−a22bc\cos A = \frac{b^2 + c^2 - a^2}{2bc}

Example

Finding a side using the cosine rule

In triangle $P QR$ , $10\text{ cm}$ , $7\text{ cm}$ and angle $65^\circ$ . Find $QR$ to 3 significant figures.

The cosine rule setup with two known sides and the included angle, and the opposite side unknown.

The side opposite angle $P$ is $QR$ , so use the cosine rule with $QR$ as the unknown side.
Substitute carefully:
$QR2=102+72−2×10×7×cos⁡65∘QR^2 = 10^2 + 7^2 - 2 \times 10 \times 7 \times \cos 65^\circ$
Calculate $QR^2$ first:
$QR2≈89.83QR^2 \approx 89.83$
Square root to find the length:
$\approx 9.48\text{ cm}$

Example

Finding an angle and then an exact area

In triangle $X Y Z$ , $5\text{ cm}$ , $8\text{ cm}$ and $7\text{ cm}$ . Find $cos⁡Y\cos Y$ , then find the exact area.

Angle $Y$ is between sides $X Y$ and $Y Z$ . The side opposite angle $Y$ is $X Z$ .
Use the rearranged cosine rule:
$cos⁡Y=52+82−722×5×8\cos Y = \frac{5^2 + 8^2 - 7^2}{2 \times 5 \times 8}$
Simplify:
$cos⁡Y=25+64−4980=12\cos Y = \frac{25 + 64 - 49}{80} = \frac{1}{2}$
Since $cos⁡Y=12\cos Y = \frac{1}{2}$ , we have $60^\circ$ , so use the area formula:
$cm2\text{Area} = \frac{1}{2} \times 5 \times 8 \times \sin 60^\circ = 10\sqrt{3}\text{ cm}^2$

5. Algebra with the cosine rule

Sometimes the side lengths contain an unknown, such as $x$ . The method is the same: substitute, expand, simplify and solve.

Example

Solving for an unknown side expression

In triangle $A B C$ , $(x+5)\text{ cm}$ , $x\text{ cm}$ , $(x+3)\text{ cm}$ and angle $60^\circ$ . Find $x$ .

The side opposite angle $A$ is $B C$ , so $B C = x + 3$ goes on the left of the cosine rule.
Substitute into $a^2 = b^2 + c^2 - 2bc\cos A$ :
$(x+3)2=(x+5)2+x2−2(x+5)xcos⁡60∘(x+3)^2 = (x+5)^2 + x^2 - 2(x+5)x\cos 60^\circ$
Use $cos⁡60∘=12\cos 60^\circ = \frac{1}{2}$ and simplify:
$x+3)^2 = (x+5)^2 + x^2 - (x+5)x$
Expand both sides:
$x^2 + 6x + 9 = x^2 + 5x + 25$
Solve:
$x = 16$

6. The sine rule and the ambiguous case

Use the sine rule when you know an opposite angle-side pair and another side or angle.

asin⁡A=bsin⁡B=csin⁡C\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}

Equivalently,

sin⁡Aa=sin⁡Bb=sin⁡Cc\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}

The second version is often neater when finding an angle.

Example

Finding two possible angles

In triangle $A B C$ , $12\text{ cm}$ , $8\text{ cm}$ and angle $35^\circ$ . Find the two possible values of angle $A B C$ to one decimal place.

The ambiguous sine-rule case can form two different triangles from the same given side, side and angle.

Match opposite pairs: $B C$ is opposite angle $A$ , and $A B$ is opposite angle $C$ .
Use the sine rule to find angle $C$ :
$sin⁡C12=sin⁡35∘8\frac{\sin C}{12} = \frac{\sin 35^\circ}{8}$
Rearrange:
$sin⁡C=12sin⁡35∘8≈0.8604\sin C = \frac{12\sin 35^\circ}{8} \approx 0.8604$
Find the two possible values of $C$ :
$\approx 59.4^\circ \quad \text{or} \quad C \approx 120.6^\circ$
Use angles in a triangle to find angle $B$ :
$180^\circ - 35^\circ - C$
Therefore the two possible values are:
$\approx 85.6^\circ \quad \text{or} \quad B \approx 24.4^\circ$

Common Mistake

Forgetting the second sine angle

If $sin⁡θ=k\sin \theta = k$ , your calculator gives one angle. In a triangle, the other possible angle is $180∘−θ180^\circ - \theta$ , as long as the angle sum still works.

7. Exact values from $sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1$

The identity

sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1

is useful when you are given one trigonometric ratio and need another.

If the angle is acute, then both sine and cosine are positive.

Example

Finding an exact cosine value

Angle $θ\theta$ is acute and $sin⁡θ=23\sin \theta = \frac{2}{3}$ . Find the exact value of $cos⁡θ\cos \theta$ .

An acute-angle right triangle representation of sin θ = 2/3, leaving the adjacent side to be found exactly.

Start with the identity:
$sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1$
Substitute $sin⁡θ=23\sin \theta = \frac{2}{3}$ :
$(23)2+cos⁡2θ=1\left(\frac{2}{3}\right)^2 + \cos^2 \theta = 1$
Rearrange:
$cos⁡2θ=1−49=59\cos^2 \theta = 1 - \frac{4}{9} = \frac{5}{9}$
Since $θ\theta$ is acute, choose the positive square root:
$cos⁡θ=53\cos \theta = \frac{\sqrt{5}}{3}$

8. Sketching sine and cosine graphs

For $\le x \le 360^\circ$ , remember the key values:

$\sin x$ starts at 0, reaches 1 at 90°, returns to 0 at 180°, reaches -1 at 270°, then returns to 0 at 360°.
$\cos x$ starts at 1, reaches 0 at 90°, reaches -1 at 180°, returns to 0 at 270°, then reaches 1 at 360°.

A vertical shift moves the graph up or down. For example, $\sin x + 1$ is the sine graph shifted up by 1.

Graph showing y equals sin x and y equals sin x plus 1 from 0 to 360 degrees

Example

Sketching a transformed cosine graph

Sketch $\cos(x+90^\circ)$ for $\le x \le 360^\circ$ .

The transformed cosine graph passes through the five key points and matches the shape of −sin x.

Work out the value of the expression inside cosine at key values of $x$ :
$x=0∘:y=cos⁡90∘=0x=90∘:y=cos⁡180∘=−1x=180∘:y=cos⁡270∘=0x=270∘:y=cos⁡360∘=1x=360∘:y=cos⁡450∘=0\begin{aligned} x = 0^\circ &: \quad y = \cos 90^\circ = 0 \\ x = 90^\circ &: \quad y = \cos 180^\circ = -1 \\ x = 180^\circ &: \quad y = \cos 270^\circ = 0 \\ x = 270^\circ &: \quad y = \cos 360^\circ = 1 \\ x = 360^\circ &: \quad y = \cos 450^\circ = 0 \end{aligned}$
Plot the points $(0, 0)$ , $(90, - 1)$ , $(180, 0)$ , $(270, 1)$ and $(360, 0)$ .
Join the points with a smooth wave-shaped curve.
Notice that $\cos(x+90^\circ)$ has the same shape as $-\sin x$ .

Exam technique

In the exam

Label opposite angle-side pairs before choosing a formula.
Use the cosine rule for two sides with the included angle, or for all three sides.
Use the sine rule when you have an opposite pair, and always check for a second possible angle.
For graph sketches, plot the five key x-values: 0°, 90°, 180°, 270° and 360°.

Self review

Check yourself

Which formula would you use if you knew two sides and the angle between them?
Why can $sin⁡θ=0.6\sin \theta = 0.6$ lead to two possible angles in a triangle?
What are the key points of $\sin x + 1$ between 0° and 360°?

What you'll learn

1. Starting point: right-angled triangle ratios

2. Labelling non-right-angled triangles

3. Area of a triangle using sine

4. The cosine rule

5. Algebra with the cosine rule

6. The sine rule and the ambiguous case

7. Exact values from $sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1$

8. Sketching sine and cosine graphs

Recap questions

Review Trigonometric Ratios by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Trigonometric Ratios

What you'll learn

1. Starting point: right-angled triangle ratios

2. Labelling non-right-angled triangles

3. Area of a triangle using sine

4. The cosine rule

5. Algebra with the cosine rule

6. The sine rule and the ambiguous case

7. Exact values from $sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1$

8. Sketching sine and cosine graphs

Recap questions

Review Trigonometric Ratios by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Trigonometric Ratios

What you'll learn

1. Starting point: right-angled triangle ratios

2. Labelling non-right-angled triangles

3. Area of a triangle using sine

4. The cosine rule

5. Algebra with the cosine rule

6. The sine rule and the ambiguous case

7. Exact values from sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1sin2θ+cos2θ=1

8. Sketching sine and cosine graphs

Recap questions

Review Trigonometric Ratios by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Trigonometric Ratios Revision Guide

Trigonometric Ratios

What you'll learn

1. Starting point: right-angled triangle ratios

2. Labelling non-right-angled triangles

3. Area of a triangle using sine

4. The cosine rule

5. Algebra with the cosine rule

6. The sine rule and the ambiguous case

7. Exact values from sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1sin2θ+cos2θ=1

8. Sketching sine and cosine graphs

Recap questions

Review Trigonometric Ratios by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Trigonometric Ratios Revision Guide

7. Exact values from $sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1$

7. Exact values from $sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1$