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Pythagoras’ Theorem Rearranging equations SOHCAHTOAThis topic is relevant for:

Here we will learn what the** **trigonometric functions and inverse trigonometric functions are and how we can use them to calculate missing side lengths and angles in right-angled triangles.

Look out for the trigonometric functions practice problems, worksheets and exam questions at the end.

**Trigonometric functions** are functions that relate an angle in a right angled triangle to the ratio of two of its sides.

The angle can be called anything but it is often referred to as ϴ (theta).

- \sin (\theta)

When we input a given angle into \sin (\theta) it gives us the ratio between the opposite length and the hypotenuse.

\[\sin ( \theta ) = \frac{Opposite}{Hypotenuse}\]

It tells us how many times bigger the opposite is than the hypotenuse.

- \cos(\theta)

When we input a given angle into \cos(\theta) it gives us the ratio between the adjacent length and the hypotenuse.

\[\cos ( \theta ) = \frac{Adjacent}{Hypotenuse}\]

It tells us how many times bigger the adjacent is than the hypotenuse.

- \tan(\theta)

When we input a given angle into \tan(\theta) it gives us the ratio between the adjacent length and the opposite length.

\[\tan ( \theta ) = \frac{Opposite}{Adjacent}\]

It tells us how many times bigger the opposite is than the adjacent.

Trigonometric functions tell us how many times bigger one side is than another in triangles containing a right angle. Therefore, if we are given the length of one side and the value of one angle, we can set up equations using trigonometric functions to help us find the lengths of the other sides.

E.g.

- \sin(30)=\frac{x}{10}

We need to work out the value of

\[\text{We know }\sin(\theta)=\frac{Opposite}{Hypotenuse} \text{ therefore } \sin(30)=\frac{x}{10}\]

Now we can rearrange this equation to make

\[\begin{aligned}
\sin(30)&=\frac{x}{10}\\
10 \times \sin(30)&=x
\end{aligned}\]

Finally we can type

\[\begin{aligned}
x&=10\times\sin(30)\\
x&=5cm
\end{aligned}\]

- \cos(30)=\frac{x}{10}

\[\text{We know }\cos(\theta)=\frac{Adjacent}{Hypotenuse} \text{ therefore } \cos(30)=\frac{x}{10}\]

Now we can rearrange this equation to make

\[\begin{aligned}
\cos(30)&=\frac{x}{10}\\
10 \times \cos(30)&=x
\end{aligned}\]

Finally we can type

\[\begin{aligned}
x&=10\times\cos(30)\\
x&=8.660254038\\
x&=8.66cm (2dp)
\end{aligned}\]

- \tan(30)=\frac{5}{x}

\[\text{We know }\tan(\theta)=\frac{Opposite}{Adjacent} \text{ therefore } \tan(30)=\frac{5}{x}\]

Now we can rearrange this equation to make

\[\begin{aligned}
\tan(30)&=\frac{5}{x}\\
x \times \tan(30)&=5\\
x&=\frac{5}{\tan(30)}
\end{aligned}\]

Finally we can type \frac{5}{tan(30)} into a calculator to find the value for x .

\[\begin{aligned}
x&=\frac{5}{\tan(30)}\\
x&=8.660254038\\
x&=8.66\mathrm{cm} ~(2\mathrm{dp})
\end{aligned}\]

In order to use trigonometric functions:

**Set up an equation until involving**sin ,cos, ortan .**Rearrange the equation, making the value you are trying to find the subject.****Work out the answer using a calculator and round it as needed.**

Get your free trigonometric functions worksheet of 20+ trigonometry questions and answers. Includes reasoning and applied questions.

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

DOWNLOAD FREEWork out the value of x , given sin(72)=\frac{25}{x} . Give your answer to 1dp.

**Set up an equation until involving**.Sin ,Cos , orTan

We already have an equation involving

\[sin(72)=\frac{25}{x}\]

2**Rearrange the equation, making the value you are trying to find the subject.**

\[\begin{aligned}
\sin(72)&=\frac{25}{x}\\
x \times \sin(72)&=25\\
x&=\frac{25}{\sin(72)}
\end{aligned}\]

3**Work out the answer using a calculator and round it as needed.**

Type \frac{25}{\sin(72)} into a calculator to work out x .

\[\begin{aligned}
x&=\frac{25}{\sin(72)}\\
x&=26.28655561\\
x&=26.2~(1 \mathrm{dp})
\end{aligned}\]

Find the length labelled

**Set up an equation until involving Sin, Cos or Tan.**

Labelling the triangle, we can see that we know the hypotenuse and we want to know the adjacent. The trigonometric function which compares these two sides is

\[\text{We know }\cos(\theta)=\frac{Adjacent}{Hypotenuse} \text{ therefore } \cos(34)=\frac{x}{32}\]

**Rearrange the equation, making the value you are trying to find the subject.**

\[\begin{aligned}
\cos(34)&=\frac{x}{32}\\
32 \times \cos(34) &= x
\end{aligned}\]

**Work out the answer using a calculator and round it as needed.**

\[\begin{aligned}
x&=32 \times \cos(34)\\
x&=26.52920232\\
x&=26.5\mathrm{mm} \text{ (3sf)}
\end{aligned}\]

Find the length labelled

**Set up an equation until involving Sin, Cos or Tan.**

Labelling the triangle, we can see that we know the opposite and we want to know the adjacent. The trigonometric function which compares these two sides is

\[\text{We know }\tan(\theta)=\frac{Opposite}{Adjacent} \text{ therefore } \tan(29)=\frac{8}{x}\]

**Rearrange the equation, making the value you are trying to find the subject.**

\[\begin{aligned}
\tan(29)&=\frac{8}{x}\\
x \times \tan(29) &= 8\\
x&=\frac{8}{\tan(29)}
\end{aligned}\]

**Work out the answer using a calculator and round it as needed.**

\[\begin{aligned}
x&=\frac{8}{\tan(29)}\\
x&=14.43238204\\
x&=14.4 \text{ km (3sf)}
\end{aligned}\]

In the same way that addition and subtraction are inverse operations, inverse trigonometric functions do the opposite of regular trigonometric functions.

We can use the** inverse sine function**, the **inverse cosine function** and the **inverse tangent function** to work out the missing angle ϴ.

The** inverse trig functions** are:

\[\sin ^{-1} \qquad \cos ^{-1} \qquad \tan ^{-1}\]

\[\sin (\theta)=\frac{\text { Opposite }}{\text { Hypotenuse }} \qquad
\theta=\sin^{-1} (\frac{\text { Opposite }}{\text { Hypotenuse }} )\]

\[\cos (\theta)=\frac{\text { Adjacent }}{\text { Hypotenuse }} \qquad
\theta=\cos^{-1} (\frac{\text { Adjacent }}{\text { Hypotenuse }} )\]

\[\tan (\theta)=\frac{\text { Opposite }}{\text { Adjacent }} \qquad
\theta=\tan^{-1} (\frac{\text { Opposite }}{\text { Adjacent }} )\]

In order to work out missing angles in right-angled triangles we need use the inverse trigonometric functions:

- \sin (\theta)=\frac{3}{5}, \qquad \theta=\sin ^{-1} (\frac{3}{5})

\[\text{We know }\sin(\theta)=\frac{Opposite}{Hypotenuse} \text{ therefore } \sin(\theta)=\frac{3}{5}\]

We need to work out θ, but it is currently inside the function

To ‘undo’ ^{-1}.

We call this ‘’inverse

We can find this on the calculator by pressing* SHIFT* and then

- \cos (\theta)=\frac{4}{5}, \qquad \theta=\cos ^{-1} (\frac{4}{5}) .

\[\text{We know }\cos(\theta)=\frac{Adjacent}{Hypotenuse} \text{ therefore } \cos(\theta)=\frac{4}{5}\]

We need to work out θ, but it is currently inside the function

To ‘undo’ ^{-1}

We call this ‘’inverse

We can find this on the calculator by pressing* SHIFT* and then

- \tan (\theta)=\frac{3}{4}, \qquad \theta=\tan ^{-1} (\frac{3}{4}) .

\[\text{We know }\tan(\theta)=\frac{Opposite}{Adjacent} \text{ therefore } \tan(\theta)=\frac{3}{4}\]

We need to work out θ, but it is currently inside the function

To ‘undo’ ^{-1}

We call this ‘’inverse

We can find this on the calculator by pressing* SHIFT* and then

**Set up an equation involving**Sin ,Cos orTan and rearrange it until you are left with the trig function as the subject.**Apply the inverse trigonometric function.****Calculate the answer, using the SHIFT button on the calculator, and round it as needed.**

Work out the value of to

\[\sin (\theta)=0.9\]

**Set up an equation involving Sin, Cos or Tan and rearrange it until you are left with the trig function as the subject.**

\[\sin (\theta)=0.9\]

is already in this form.

**Apply the inverse trigonometric function.**

\[\begin{aligned}
\sin (\theta) &=0.9 \\
\theta &=\sin ^{-1}(0.9)
\end{aligned}\]

** Calculate the answer, using the SHIFT button on the calculator, and round it as needed.**

\[\theta=64.15806724\\
\theta=64.2^{\circ}~(1 \mathrm{dp})\]

Work out the value of to

\[2 \cos (\theta)=0.8\]

**Apply the inverse trigonometric function.**

\[\begin{aligned}
\cos (\theta) &=0.4 \\
\theta &=\cos ^{-1}(0.4)
\end{aligned}\]

** Calculate the answer, using the SHIFT button on the calculator, and round it as needed.**

\[\theta=66.42182152\\
\theta=66.42^{\circ} \text { (2dp) }\]

Calculate the unknown angle

After labeling the sides of the triangle we can see that we have information about the opposite side and the adjacent side.

We know that

\[\tan (x)=\frac{\text { Opposite }}{\text { Adjacent }}\]

So,

\[\tan (x)=\frac{8}{14.2}\]

**Apply the inverse trigonometric function.**

\[\begin{aligned}
\tan (x) &=\frac{8}{14.2} \\
x &=\tan ^{-1}\left(\frac{8}{14.2}\right)
\end{aligned}\]

** Calculate the answer, using the SHIFT button on the calculator, and round it as needed.**

\[\begin{array}{l}
x=29.39605285 \\
x=30^{\circ}\text{ (1sf)}
\end{array}\]

We can check our answer is sensible here. All angles (other than the right angle) in a right angled triangle should be acute angles. 30° is acute therefore the answer is sensible.

**Right Triangle**

Right-angled triangles are sometimes called right triangles.

**Using the wrong function**

We use

**Other Trig Identities**

There are lots of other trigonometric identities as well as

\[\begin{aligned}
\frac{1}{\sin x} &=\text{cosecant} x \text{ or cosec} x\\
\frac{1}{\cos x} &=\text{secant} x \text{ or sec} x\\
\frac{1}{\tan x}&=\text{cotangent} \text{ or cot} x
\end{aligned}\]

- \sin^{-1}(\theta) and \frac{1}{sin(\theta)} .

\sin^{-1}(\theta) does not mean \frac{1}{sin(\theta)} .

Raising a variable or number to the power of

E.g.

\[x^{-1}=\frac{1}{x}, \quad 5^{-1}=\frac{1}{5} \text { etc.}\]

Raising a function to the power of

E.g. ^{-1}(x)

^{-1}(x)

Referring to

^{-1}(x)

^{-1}(x)

^{-1}(x)

avoids this confusion. We don’t use these terms in GCSE Mathematics, but they are used in A level Maths and beyond!

**Degrees and radians**

At GCSE angles are always measured in degrees. Make sure that your calculation is in degrees before using it for trigonometry. We can tell if the calculator is in degrees if there is a ‘D’ at the top of the screen.

Trigonometric functions is part of our series of lessons to support revision on trigonometry. You may find it helpful to start with the main trigonometry lesson for a summary of what to expect, or use the step by step guides below for further detail on individual topics. Other lessons in this series include:

1. Work out the value of x when tan(42)=\frac{x}{5} , giving your answer to 1dp .

5.6

210

4.5

0.9

\begin{aligned}
\tan(42)&=\frac{x}{5}\\
5 \times \tan(42)&=x\\
x&=4.502020221\\
x&=4.5 \text{ (1dp)}
\end{aligned}

2. Work out the length of the side labelled a on this triangle, giving your answer to 2sf .

18 cm

33 cm

8.9 cm

15 cm

Labelling the triangle, we know H and we want to find O. The trigonometric function involving O and H is sin.

\text{We know } \sin(\theta)=\frac{O}{H}\text{ therefore } \sin(61)=\frac{16}{a}Rearranging this we get

\begin{aligned} a \times \sin(61)&=16\\ a&=\frac{16}{\sin(61)} \end{aligned}So a=18.29366509 or a=18cm (2sf)

3. Work out the length AB. Give your answer to 2dp .

3.82 cm

1.17 cm

1.22 cm

13.08 cm

We know A and we want to find O . The trigonometric function involving O and A is tan.

We know \tan(\theta)=\frac{O}{A} therefore \tan(73)=\frac{AB}{4}

Rearranging this we get 4 \times \tan(73)=AB

So AB=13.08341047 or AB=13.08cm (2dp)

4. Work out the value of x when cos x = 0.2 . Give your answer to 1dp .

x=1.0^{\circ} (1\mathrm{~d.p.})

x=11.5^{\circ} (1\mathrm{~d.p.})

x=1.4^{\circ} (1\mathrm{~d.p.})

x=78.5^{\circ} (1\mathrm{~d.p.})

\begin{aligned}
\cos(x)&=0.2\\
x&= \cos^{-1}(0.2)\\
x&= 78.46304097^{\circ}\\
x&=78.5^{\circ} ~(1\mathrm{~d.p.})
\end{aligned}

5. Solve the equation \sin(\theta)+5=5.7 . Give your answer to 1d.p .

\theta=-4.9^{\circ} (1\mathrm{~d.p.})

\theta=44.4^{\circ} (1\mathrm{~d.p.})

\theta=61.3^{\circ} (1\mathrm{~d.p.})

\theta=45.6^{\circ} (1\mathrm{~d.p.})

\begin{aligned}
\sin(\theta)+5 &=5.7\\
\sin(\theta)&=0.7\\
\theta&= \sin^{-1}(0.7)\\
\theta&= 44.427004^{\circ}\\
\theta&=44.4^{\circ} ~(1\mathrm{~d.p.})
\end{aligned}

6. Work out the value of \theta . Give your answer to 1d.p .

\theta=36.9^{\circ} (1\mathrm{~d.p.})

\theta=38.7^{\circ} (1\mathrm{~d.p.})

\theta=53.1^{\circ} (1\mathrm{~d.p.})

\theta=0.9^{\circ} (1\mathrm{~d.p.})

First we need to set up the equation using SOHCAHTOA.

We know the opposite and the hypotenuse so we need to use sin.

\begin{aligned} \sin(\theta)&=\frac{O}{H}\\ \sin(\theta)&= \frac{12}{15}\\ \theta&= \sin^{-1}(\frac{12}{15})\\ \theta&=53.13010235\\ \theta&=53.1^{\circ} (1\mathrm{~d.p.}) \end{aligned}1. The diagram shows triangle PQR .

PR=32cm and angle PRQ=32^{\circ} . Calculate the length PQ . Give your answer to 3sf .

**(3 marks)**

Show answer

\tan(32)=\frac{PQ}{23}

**(1)**

**(1)**

**(1)**

2. We can see from the diagram that \tan(x)=\frac{1}{3} . Find the size of angle a to two decimal places.

**(1 mark)**

Show answer

\begin{aligned}
a&=\tan^{-1}(\frac{1}{3})\\
a&=18.43^{\circ}
\end{aligned}

**(1)**

3. The diagram shows a right angled triangle. Find the value of x . Give your answer to 2dp .

**(2 marks)**

Show answer

\sin(x)=\frac{10}{12.8}

**(1)**

**(1)**

You have now learned how to:

- Apply trigonometric ratios to find angles in right-angled triangles in 2 dimensions

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