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Here we will learn about trigonometry formulas, including the three trigonometric functions; the sine function, \text{Sin}, the cosine function, \text{Cos}, and the tangent function, \text{Tan}.

We will also look at the inverse trigonometric functions, the sine rule, cosine rule and the trigonometric formula for the area of a triangle.

There are also trigonometry formulas* *worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck.

A** trigonometry formula** is a formula that is used to represent relationships between the **parts of a triangle** including the **side lengths**,** angles** and the **area**.

To solve problems using trigonometric formulas, we need to be able to use the information provided to deduce which trigonometric formula we need to use.

Right-Angled Triangles\sin{\theta}=\frac{O}{H} \cos{\theta}=\frac{A}{H} \tan{\theta}=\frac{O}{A} | Non-Right Angled TrianglesSine Rule\frac{a}{\sin{A}} = \frac{b}{\sin{B}} = \frac{c}{\sin{C}} Cosine Rulea^2=b^2+c^2-2bc \cos(A) |

Area of a Triangle\text{Area }=\frac{1}{2}ab \sin{(C)} |

Basic trigonometry focuses on the **right-angled triangle**.

The three trigonometric functions we use are **sine** \text{(sin)}, **cosine** \text{(cos)} and **tangent** \text{(tan)}. These three functions relate an angle to two sides of a right-angled triangle, giving us the three **trigonometric ratios**.

Trigonometric Function | Sine
f(x)=\sin(x) | Cosine
f(x)=\cos(x) | Tangent
f(x)=\tan(x) |

Trigonometric Ratio | \sin{\theta}=\frac{O}{H} | \cos{\theta}=\frac{A}{H} | \tan{\theta}=\frac{O}{A} |

The letter \text{x} represents a value, hence, \text{f(x)} is a function of the value \text{x}.

The symbol \theta represents the **angle**.

The letter \text{O} represents the **opposite side** to the angle \theta .

The letter \text{A} represents the **adjacent side **to the angle \theta .

The letter \text{H} represents the **hypotenuse** of the right-angled triangle (which is always the side of the triangle facing the right-angle). Also known as the longest side of the triangle.

**Example 1**

Here,

- the
**angle**\theta=30^{\circ} - the side
**opposite**the angle O=4 \ cm, - the unknown length of the
**hypotenuse**H=x \ cm.

We know \theta and O, and we want to find H.

Using SOHCAHTOA, SOH is the **trigonometric ratio**, so we need to use

**Example 2**

Here,

- the
**angle**\theta=28^{\circ}, - the
**hypotenuse**H=5 \ cm, - the unknown length of the
**adjacent**side A=x \ cm.

We know \theta and H, and we want to find A.

Using SOHCAHTOA, CAH is the **trig ratio**, so we need to use

**Example 3**

Here,

- the
**angle**\theta=63^{\circ}, - the
**adjacent**side A=7 \ cm, - the unknown length of the
**opposite**side O=x \ cm.

We know \theta and A, and we want to find O.

Using SOHCAHTOA, TOA is the **trig ratio**, so we need to use

The solutions are,

Example 1: H=8 \ cm

Example 2: A=4.14 \ cm \ (2dp)

Example 3: O=13.74 \ cm \ (2dp)

**Step-by-step guide:** SOHCAHTOA

**Step-by-step guide:** Trigonometric Functions

When we know two of the three sides of a right angled triangle and we want to find a missing angle, we need to use inverse trigonometric functions.

The three inverse trigonometric functions are

Inverse trigonometric function | Inverse Sine
f^{-1}(x)=\sin^{-1}(x) | Inverse Cosine
f^{-1}(x)=\cos^{-1}(x) | Inverse Tangent
f^{-1}(x)=\tan^{-1}(x) |
---|---|---|---|

Inverse trig of a ratio | \theta=\sin^{-1}(\frac{O}{H}) | \theta=\cos^{-1}\left(\frac{A}{H}\right) | \theta=\tan^{-1}\left(\frac{O}{A}\right) |

**Example 1**

Here,

- the side
**opposite**the angle O=5 \ cm, - the length of the
**hypotenuse**H=10 \ cm, - the unknown
**angle**is \theta.

We know O and H, and we want to find \theta .

Using SOHCAHTOA, SOH is the **trig ratio**, so as we need to find the value of \theta, we must use the inverse trig formula \theta=\sin^{-1}\left(\frac{O}{H}\right).

**Example 2**

Here,

- the side
**adjacent**to the angle A=4 \ cm, - the length of the
**hypotenuse**H=5 \ cm, - the unknown
**angle**is \theta.

We know A and H, and we want to find \theta .

Using SOHCAHTOA, CAH is the **trig ratio**, so as we need to find the value of \theta , we must use the inverse trig formula \theta=\cos^{-1}\left(\frac{A}{H}\right).

**Example 3**

Here,

- the side
**opposite**the angle O=12 \ cm, - the side
**adjacent**to the angle A=8 \ cm, - the unknown
**angle**is \theta .

We know O and A, and we want to find \theta .

Using SOHCAHTOA, TOA is the **trig ratio**, so as we need to find the value of \theta, we must use the inverse trig formula \theta=\tan^{-1}\left(\frac{O}{A}\right).

The solutions are,

Example 1: \theta=30^{\circ}

Example 2: \theta=36.87^{\circ}\text{ (2dp)}

Example 3: \theta=56.31^{\circ} \text{ (2dp)}

**Step-by-step guide:** Trigonometric functions

We can use trigonometry in all types of triangles including scalene, equilateral and isosceles.

__Sine rule__

The sine rule (sometimes called the **law of sines**) is the relationship between the **sine of an angle**, and the side **opposite **that angle.

More specifically the ratio of the sine of an angle and the opposing side to that angle is the same for all three angles/sides. As a formula, this is written as

\frac{a}{\sin{A}}=\frac{b}{\sin{B}}=\frac{c}{\sin{C}}.Where a, \ b, and c are the three sides of a triangle, and A, \ B, and C are the angles that face their opposing side (see the diagram below).

Notice that the upper case letters denote the angles, and the lower case letters denote the side lengths where a is opposite A, \ b is opposite B, and c is opposite C.

The equation above shows how all three side lengths, divided by the sine of the angle facing that side, are the same value. We only need to work with a pair of letters, giving us the working **formula for the law of sines**.

We can use the sine rule to find

- a missing side using this arrangement of the formula.

By rearranging the formula to make the unknown side a the subject, we have

a=\frac{b\sin{A}}{\sin{B}}.- a missing angle using this arrangement of the formula.

By rearranging the formula to make the unknown angle \text{A} the subject, we have

A=\sin^{-1}\left(\frac{a\sin{B}}{b}\right).**Step-by-step guide**: Sine Rule

**Step-by-step guide:** Changing the subject

For example,

Below is a non-right angled triangle. We can use the law of sines to find the length \text{AB}.

\begin{aligned} & \frac{a}{\sin A}=\frac{b}{\sin B} \\\\ & \frac{A B}{\sin (28)}=\frac{10.2}{\sin (134)} \\\\ & A B=\sin (28) \times \frac{10.2}{\sin (134)} \\\\ & A B=6.656951191 \mathrm{~cm} \end{aligned}The **cosine rule **(sometimes called the **law of cosines** ) is the relationship between the **cosine of an angle**, and all **three sides** of the triangle.

As a formula, this is written as

a^{2}=b^{2}+c^{2}-2bc\cos(A).Where a, \ b, and c are the three sides of a triangle, and A is the angle between the two sides of b and c (see the diagram below).

Notice that the upper case A is an angle, and the lower case letters are the side lengths where a is opposite A, and b and c are the two sides that surround the angle A.

We need to use the cosine rule when we don’t know one of the four values of a, \ b, \ c, or A.

We can use the cosine rule to find

- a missing side by using this arrangement of the formula,

- a missing angle by using this arrangement of the formula,

**Step-by-step guide**: Cosine Rule

**Step-by-step guide:** Changing the subject

For example,

Below is a non-right angled triangle. We can use the law of cosine to find the length \text{BC}.

\begin{aligned} & a^2=b^2+c^2-2 b c \cos (A) \\\\ & B C^2=4.8^2+6.4^2-2 \times 4.8 \times 6.4 \times \cos (123) \\\\ & B C^2=97.46262231 \\\\ & B C=\sqrt{97.46262231} \\\\ & B C=9.872315955 \mathrm{~cm} \end{aligned}We can calculate the area of any triangle if we need to know **two sides** and the **included** angle between those two sides. The formula is

Where C is the angle between the two sides of a and b (see the diagram below).

We can also find a missing angle knowing the area of the triangle and by rearranging the formula to find the angle \text{C}.

C=\sin^{-1}(\frac{2Area}{ab})We can find a missing side length knowing the area of the triangle and rearranging the formula to find a missing side length a.

a=\frac{2Area}{b\sin(C)}**Step-by-step guide:** Area of a triangle trig

For example,

We can find the area of the triangle below using the trigonometric formula for the area of any triangle.

\begin{aligned} & Area=\frac{1}{2} a b \sin (C) \\\\ & Area=\frac{1}{2} \times 5 \times 7 \times \sin (40) \\\\ & Area=11.24878317 \mathrm{~m}^2 \end{aligned}In order to determine which trigonometric formula you need to use, the flow chart below can help you decide.

Remembering that, to find the area of any triangle, we can use the sine rule for area.

In order to solve problems using trigonometry formulas:

**Determine which trigonometry formula is needed to answer the question.****Label each side and angle appropriately.****Substitute known values into the formula.****Solve for the unknown value.**

Get your free trigonometry formulas worksheet of 20+ SOHCAHTOA questions and answers. Includes reasoning and applied questions.

DOWNLOAD FREEGet your free trigonometry formulas worksheet of 20+ SOHCAHTOA questions and answers. Includes reasoning and applied questions.

DOWNLOAD FREECalculate the length of the side labelled x. Write your answer correctly to 2 decimal places.

**Determine which trigonometry formula is needed to answer the question.**

Using the flowchart,

This means that we need to use either of the three trig ratios of sine, cosine or tangent. We can work out which one in the next step.

2**Label each side and angle appropriately.**

From the angle of 20^{\circ},

- the side opposite the angle O=10 \ cm,
- the hypotenuse H=x \ cm,
- the adjacent side has no value and so is not used.

As we know the opposite side and we want to find the hypotenuse, the trigonometric ratio that we have to use is \sin(\theta)=\frac{O}{H}.

Here, we used the abbreviation \text{SOHCAHTOA}, where \text{SOH} is the sine ratio \text{S}, of the opposite \text{O}, and the hypotenuse \text{H}.

3**Substitute known values into the formula.**

As O=10, \ H=x, and \theta=20, substituting these values into the formula

\sin(\theta)=\frac{O}{H}, we have

\sin(20)=\frac{10}{x}.4**Solve for the unknown value.**

Multiplying both sides by x gives us

x\times\sin(20)=10.Dividing both sides by \sin(20) gives us

x=\frac{10}{\sin(20)}=29.238044…This means that the length x=29.24 \ cm \ (2dp).

Determine the size of the missing angle, marked \theta in the triangle below. Write your answer correctly to 2 decimal places.

**Determine which trigonometry formula is needed to answer the question.**

Using the flowchart,

This means that we need to use either of the three trig ratios of sine, cosine or tangent. We can work out which one in the next step.

**Label each side and angle appropriately.**

From the angle of \theta,

- the side adjacent to the angle A=2 \ m,
- the hypotenuse H=2.8 \ m,
- the opposite side has no value and so is not used.

As we know the adjacent side and the hypotenuse, the trigonometric ratio that we have to use is \cos(\theta)=\frac{A}{H}.

However, as we need to find the missing angle, we need to use the inverse trigonometric formula \theta=\cos^{-1}\left(\frac{A}{H}\right).

Here, we used the abbreviation \text{SOHCAHTOA}, where
\text{CAH} is the cosine ratio \text{C}, of the adjacent \text{A}, and the hypotenuse \text{H}.

**Substitute known values into the formula.**

As A=2, \ H=2.8, and \theta is unknown, substituting these values into the formula

\theta=\cos^{-1}\left(\frac{A}{H}\right), we have

\theta=\cos^{-1}\left(\frac{2}{2.8}\right).

**Solve for the unknown value.**

\theta=\cos^{-1}\left(\frac{2}{2.8}\right)=44.4153086…

This means that the angle \theta=44.42^{\circ} \ (2dp).

Determine the missing length x of the right-angled triangle below. Write your answer to 2 decimal places.

**Determine which trigonometry formula is needed to answer the question.**

Using the flowchart,

This means that we need to use either of the three trig ratios of sine, cosine or tangent. We can work out which one in the next step.

**Label each side and angle appropriately.**

From the angle of 15^{\circ},

- the side adjacent to the angle A=7 \ cm,
- the opposite side to the angle O=x \ cm,
- the hypotenuse has no value and so is not used.

As we know the opposite side and the adjacent side, the trigonometric ratio that we have to use is \tan(\theta)=\frac{O}{A}.

**Tip**: Here, we used the abbreviation \text{SOHCAHTOA}, where TOA is the tangent ratio T, of the opposite O, and the adjacent A.

**Substitute known values into the formula.**

As A=7, \ O=x, and \theta=15, substituting these values into the formula

\tan(\theta)=\frac{O}{A}, we have

\tan(15)=\frac{x}{7}.

**Solve for the unknown value.**

Multiplying both sides by 7 gives us

7\times\tan(15)=x.

This means that x=7\tan(15)=1.875644347...

The length x=1.88 \ cm \ (2dp).

Determine the length of the side labelled x for the triangle below. Write your answer correctly to 2 decimal places.

**Determine which trigonometry formula is needed to answer the question.**

Using the flowchart,

For this question, we can only use the cosine rule (the law of cosines),

a^{2}=b^{2}+c^{2}-2bc\cos(A).

**Label each side and angle appropriately.**

From the angle at A of 27^{\circ},

- the side b next to the angle A is b=15 \ cm,
- the side c next to the angle A is c=10 \ cm,
- the side a opposite the angle A is a=x \ cm,
- the angles at B and C have no values and so they are not used.

**Substitute known values into the formula.**

As A=27, \ a=x, \ b=15, and c=10, substituting these into the cosine rule

a^{2}=b^{2}+c^{2}-2bc\cos(A), we have

x^{2}=15^{2}+10^{2}-2\times{15}\times{10}\times\cos(27).

**Solve for the unknown value.**

\begin{aligned}
x^{2}&=225+100-300\times\cos(27)\\\\
&=325-300\cos(27)\\\\
&=57.69804274…\\\\
x&=\sqrt{57.69804274…}\\\\
&=7.595922771…\\\\
&=7.60\text{ (2dp)}
\end{aligned}

The length x=7.60 \ cm \ (2dp).

Determine the size of the missing angle \theta. Write your answer to 2 decimal places.

**Determine which trigonometry formula is needed to answer the question.**

Using the flowchart,

For this question, we can only use the sine rule (the law of sines),

\frac{a}{\sin{A}}=\frac{b}{\sin{B}}=\frac{c}{\sin{C}}.

**Label each side and angle appropriately.**

From the angle at A of \theta,

- the side a opposite the angle A is a=6.5 \ mm,
- the side b next to the angle A is b=7 \ mm,
- the angle at B=52^{\circ},
- the angle at C and the length of side c have no values and so they are not used.

**Substitute known values into the formula.**

As a=6.5, \ A=\theta, \ b=7, and B=52^{\circ}, substituting these into the sine rule

\frac{a}{\sin{A}}=\frac{b}{\sin{B}}, we have

\frac{6.5}{\sin{\theta}}=\frac{7}{\sin{52}}.

Remember that we do not need to use all three parts of the formula for the sine rule so we only have one equals sign.

**Solve for the unknown value.**

Multiplying both sides by \sin(\theta), we have

6.5=\frac{7\sin{\theta}}{\sin{52}}.

Multiplying both sides by \sin(52), we have

6.5\sin{52}=7\sin{\theta}.

Dividing both sides by 7, we have

\frac{6.5\sin{52}}{7}=\sin{\theta}.

Calculating the inverse sine of each side, we have

\sin^{-1}\left(\frac{6.5\sin{52}}{7}\right)=\theta.

This gives us the value \theta=47.03114122…=47.03^{\circ}\text{ (2dp)}

Calculate the area of the triangle below. Write your answer correctly to 2 decimal places.

**Determine which trigonometry formula is needed to answer the question.**

As we are calculating the area of the triangle, we use the sine rule for area

Area=\frac{1}{2}ab\sin(C).

**Label each side and angle appropriately.**

From the angle at C of 39^{\circ},

- the side a next to the angle C is a=5 \ cm,
- the side b next to the angle C is b=7 \ cm,
- the angles at B and C and the length of side c have no values and so they are not used.

**Substitute known values into the formula.**

As a=5, \ b=7, and c=39, substituting these into the formula for the area of a triangle using the sine rule

Area=\frac{1}{2}ab\sin{C}, we have

A=\frac{1}{2}\times{5}\times{7}\times\sin{39}.

**Solve for the unknown value.**

\begin{aligned}
A&=\frac{1}{2}\times{35}\times\sin{39}\\\\
&=17.5\sin{39}\\\\
&=11.01310684…\\\\
&=11.01 (2dp)
\end{aligned}

As we are finding the area, the units are in cm^2.

This gives us the solution A=11.01 \ cm^2 \ (2dp).

**Labelling a triangle incorrectly**

For example, this triangle has been incorrectly labelled with the side c next to the angle rather than opposite it.

**Using the incorrect trigonometric function**

If the triangle is incorrectly labelled, it can lead to the use of the incorrect standard or inverse trigonometric functions and formulae.

**Rounding the decimal too early**

This can lose accuracy marks. Always use as many decimal places as possible throughout the calculation, then round the final solution.

**Using the sine rule instead of the cosine rule**

In order to use the sine rule, we need to have pairs of opposite angles and sides.

**Not using the included angle**

For the cosine rule and the area of a triangle using A=\frac{1}{2}ab\sin(C), the angle is included between the two sides. Using any other angle will result in an incorrect solution.

**Using**\bf{A=b\times{h}\div{2}}

If the vertical height of a triangle is not available then we cannot calculate the area by halving the base times the height.

**Using the inverse trig function instead, inducing a mathematical error**

If the inverse trig function is used instead of the standard trig function, the calculator may return a maths error as the solution does not exist.

**Trigonometry can only be used in right-angled triangles**

The use of trigonometry does not have to be limited to a right-angled triangle where all angles are acute angles. We can also use the law of cosines, the law of sines, and the area of a triangle using the sine rule on triangles that are not right-angled.

1. Determine the size of the missing side marked x in the triangle below. Write your answer to 2 decimal places.

x=5.03 \ cm

x=6.22 \ cm

x=10.29 \ cm

x=12.71 \ cm

After labelling the triangle with O, \ A and H,

- A=8 \ cm
- H=x \ cm
- \theta=51^{\circ}

~~SOH~~ CAH ~~TOA~~

\begin{aligned} x&=\frac{8}{\cos{51}}\\\\ &=12.71212583…\\\\ &=12.71 \ cm \ (2dp) \end{aligned}

2. Determine the size of the missing side marked \theta in the triangle below. Write your answer to 2 decimal places.

\theta=26.57^{\circ}

\theta=30.00^{\circ}

\theta=60.00^{\circ}

\theta=63.43^{\circ}

After labelling the triangle with O, \ A and H,

- O=8 \ cm
- A=16 \ cm

~~SOH~~ ~~CAH~~ TOA

\begin{aligned} \tan({\theta})&=\frac{8}{16}\\\\ \theta&=\tan^{-1}\left(\frac{8}{16}\right)\\\\ &=26.56505118…\\\\ &=26.57^{\circ} \ (2dp) \end{aligned}

3. Determine the size of the missing side marked \theta in the triangle below. Write your answer to a suitable degree of accuracy.

\theta=0.01^{\circ}

\theta=59.04^{\circ}

\theta=0.03^{\circ}

\theta=30.96^{\circ}

After labelling the triangle with O, \ A and H,

- O=5 \ cm
- A=3 \ cm

~~SOH~~ ~~CAH~~ TOA

\begin{aligned} \tan({\theta})&=\frac{5}{3}\\\\ \theta&=\tan^{-1}\left(\frac{5}{3}\right)\\\\ &=59.03624347… \\\\ &=59.04^{\circ} \ (2dp) \end{aligned}

4. Determine the size of the missing length marked x in the triangle below. Write your answer to 2 decimal places.

x=7.00 \ cm

x=4.95 \ cm

x=5.36 \ cm

x=28.70 \ cm

Labelling the sides and angles, we have

As we know the two side lengths of b and c and the included angle at A, we need to use the cosine rule (the law of cosines) to determine the length of side x.

\begin{aligned} a^{2}&=b^{2}+c^{2}-2bc\cos(A)\\\\ x^{2}&=7^{2}+7^{2}-2\times{7}\times{7}\times\cos(45)\\\\ x^2&=98-98\cos(45)\\\\ x^2&=28.70353544…\\\\ x&=\sqrt{28.70353544…}\\\\ x&=5.357568053…\\\\ x&=5.36 \ cm \ (2dp) \end{aligned}

5. Determine the size of the missing angle marked \theta in the triangle below. Write your answer to 2 decimal places.

\theta=40.72^{\circ}

\theta=0.65^{\circ}

\theta=11.76^{\circ}

\theta=29.14^{\circ}

Labelling the sides and angles, we have

As we know the angle at C the length of side c, the length of side b and we want to determine the angle at B, we need to use the sine rule (the law of sines) to determine the length of angle \theta.

Being careful with the labelling and the correct version of the sine rule, we have

\begin{aligned} \frac{\sin{B}}{b}&=\frac{\sin{C}}{c}\\\\ \frac{\sin{\theta}}{21}&=\frac{\sin{34}}{18}\\\\ \sin{\theta}&=\frac{\sin{34}}{18}\times{21}\\\\ \theta&=\sin^{-1}\left(\frac{21\sin{34}}{18}\right)\\\\ \theta&=40.72217103…\\\\ \theta&=40.72^{\circ} \ (2dp) \end{aligned}

6. Determine the area of the triangle below. Write your answer to 2 decimal places.

6.06\text{ cm}^{2}

12.96\text{ cm}^{2}

68.44\text{ cm}^{2}

8.50\text{ cm}^{2}

As we know the included angle between two side lengths, we can find the area of a triangle using sine where

a=3.2 \ cm, \ b=8.1 \ cm and C=41^{\circ}.

Being careful with the labelling and the correct version of the sine rule, we have

\begin{aligned} A&=\frac{1}{2}ab\sin{C}\\\\ &=\frac{1}{2}\times{3.2}\times{8.1}\times\sin{41}\\\\ &=12.96\sin{41}\\\\ &=8.502525016… \\\\ A&=8.50 \ cm^2 \ (2dp) \end{aligned}

1. Using the triangle below, complete the table. Give your answers in exact form.

**(3 marks)**

Show answer

6 correct

**(1)**

5 correct

**(1)**

4 correct

**(1)**

2. (a) Rearrange the equation \theta=\sin^{-1}(\frac{O}{H}) to make H the subject.

(b) Rearrange the equation A=\frac{1}{2}ab\sin{C} to make b the subject.

(c) Using your answer to part b) or otherwise, determine the missing length of the triangle ABC below. Write your answer to 2 decimal places.

**(7 marks)**

Show answer

(a)

\sin(\theta)=\frac{O}{H}

**(1)**

**(1)**

**(1)**

(b)

2A=ab\sin{C}

**(1)**

**(1)**

(c)

b=\frac{2\times214}{8.9\sin72}

**(1)**

**(1)**

3. (a) Show that the formula \frac{a}{\sin{A}}=\frac{b}{\sin{B}} rearranges to make B=\sin^{-1}(\frac{b\sin{A}}{a}).

(b) Determine the size of the missing angle \theta in the triangle below. Write your answer to 2 decimal places.

**(5 marks)**

Show answer

(a)

a\sin{B}=b\sin{A}

**(1)**

**(1)**

**(1)**

(b)

\theta =\sin^{-1}(\frac{b\sin{A}}{a})=\sin^{-1}(\frac{15.6\sin38}{12.4})

**(1)**

**(1)**

4. (a) Show that the formula a^2=b^2+c^2-2bc\cos{A} rearranges to make A=\cos^{-1}(\frac{b^2+c^2-a^2}{2bc}).

(b) Determine the size of the missing angle \theta in the triangle below. Write your answer to 2 decimal places.

**(5 marks)**

Show answer

(a)

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

**(1)**

**(1)**

**(1)**

(b)

\theta =\cos^{-1}(\frac{b^2+c^2-a^2}{2bc})=\cos^{-1}(\frac{13^2+15^2-12^2}{2\times13\times15})

**(1)**

**(1)**

You have now learned how to:

- Extend and formalise knowledge of ratio and proportion, including trigonometric ratios, in working with measures and geometry, and in working with proportional relations algebraically
- Know and apply the sine rule, \frac{a}{\sin{A}}=\frac{b}{\sin{B}}=\frac{c}{\sin{C}}, and cosine rule, a^2=b^2+c^2-2bc\cos{A}, to find unknown lengths and angles
- Know and apply Area = \frac{1}{2}ab\sin{C} to calculate the area, sides or angles of any triangle

Trigonometry is a large branch of mathematics. At GCSE, you barely scratch the surface of how widely trigonometry is used within mathematics and so this is developed further at A level. Other trigonometric functions you will look at at A level include

- Secant, cosecant and cotangent reciprocal identities
- Secant or sec \theta \ (\sec\theta=\frac{1}{\sin{\theta}})
- Cosecant or cosec \theta \ (\csc\theta=\frac{1}{\cos{\theta}})
- Cotangent or cot \theta \ (\cot{\theta}=\frac{1}{\tan{\theta}})

- Double angle formulas (the angle sum and difference formulas/sum and difference identities)
- \sin(A\pm{B})=\sin{A}\cos{B}\pm\cos{A}\sin{B}
- \cos(A\pm{B})=\cos{A}\cos{B}\mp\sin{A}\sin{B}
- \tan(A\pm{B})=\frac{\tan{A}\pm\tan{B}}{1\mp\tan{A}\tan{B}}

- Half angle identities
- \sin\left(\frac{A}{2}\right)=\pm\sqrt{\frac{1-\cos{A}}{2}}
- \cos\left(\frac{A}{2}\right)=\pm\sqrt{\frac{1+\cos{A}}{2}}
- \tan\left(\frac{A}{2}\right)=\pm\sqrt{\frac{1-\cos{A}}{1+\cos{A}}}

- Product identities
- \cos{A}\cos{B}=\frac{1}{2}[\cos(A+B)+\cos(A-B)]
- \sin{A}\sin{B}=\frac{1}{2}[\cos(A-B)-\cos(A+B)]
- \sin{A}\cos{B}=\frac{1}{2}[\sin(A+B)+\sin(A-B)]
- \cos{A}\sin{B}=\frac{1}{2}[\sin(A+B)-\sin(A-B)]

- Pythagorean identities
- \sin^{2}\theta+\cos^{2}\theta=1
- \sin^{2}\theta=1-\cos^{2}\theta
- \sin\theta=\sqrt{1-\cos^{2}\theta}

We also consider the graphs of the functions of \sin \ \theta, \ \cos \ \theta, and \tan \ \theta in greater detail in relation to when the functions are positive and negative for angles and the periodicity beyond the range of 0^{\circ} \leq \theta \leq 180^{\circ} by using the four quadrants of an xy set of axes as well as using a different unit of measure for angles, radians. This also includes the periodicity identities for trigonometric functions in radians.

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