Trig identities

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GCSE Maths Geometry and Measure Trigonometry Formula

Trig Identities

Here we will learn about trigonometric identities, including recognising and working with key trigonometric identities.

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

What are trigonometric identities?

Trigonometric identities show relationships between trigonometric functions. These are not examined in GCSE Mathematics but are covered in Level 2 Further Mathematics, Additional Mathematics and IGCSE Further Mathematics courses.

An identity is an equation that is always true.

For example, when we expand the bracket $2(x+1)$ we get $2x+2.$ These two functions of $x$ are identical and so we use the identity symbol (a triple equals symbol) to show that $2(x+1) \equiv 2x+2.$

An identity is an equation that is always true.

For example, when we expand the bracket $2(x+1)$ we get $2x+2.$ These two functions of $x$ are identical and so we use the identity symbol (a triple equals symbol) to show that $2(x+1) equiv 2x+2.$

We can do the same with three main functions of trigonometry: sine, cosine, and tangent.

$\sin(\theta)=\frac{O}{H}$
$\cos(\theta)=\frac{A}{H}$
$\tan(\theta)=\frac{O}{A}$

Where $O =$ Opposite, $A =$ Adjacent, and $H =$ Hypotenuse for the right-angled triangle with angle $\theta$ (theta).

There are two trigonometric identities that we will look at. These are,

$\tan(\theta)\equiv\frac{\sin(\theta)}{\cos(\theta)} \$ and $\ \sin^2(\theta)+\cos^2(\theta)\equiv1.$

Here $\sin^2(\theta)$ is another way of writing $(\sin(\theta))^2$ .

These two identities will help us prove further identities, using the trigonometric ratios when necessary.

Step-by-step guide: SOHCAHTOA

Step-by-step guide: Cosine rule

Step-by-step guide: Sin Cos Tan

See also: Area of a triangle trig

What are trigonometric identities?

Exact trig values worksheet (includes trig identities)

Get your free trigonometric identities worksheet of 20+ exact trig values questions and answers. Includes reasoning and applied questions.

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Exact trig values worksheet (includes trig identities)

Get your free trigonometric identities worksheet of 20+ exact trig values questions and answers. Includes reasoning and applied questions.

DOWNLOAD FREE

How to prove trigonometric identities

In order to prove trigonometric identities:

Substitute values using the trigonometric ratios or another trigonometric identity.
Simplify one side of the identity only.

In order to prove trigonometric identities:

Substitute values using the trigonometric ratios or another trigonometric identity.
Simplify one side of the identity only.

There are various identities in trigonometry that are essential for solving a wide range of trigonometric problems.

Explain how to prove trigonometric identities

Trigonometric identities examples

Example 1: the relationship between sin, cos and tan

Prove that $\ \frac{\sin(\theta)}{\cos(\theta)} \equiv \tan(\theta).$

Substitute values using the trigonometric ratios or another trigonometric identity.

We need to show that $\tan (\theta)$ is the same as $\sin (\theta) \div \cos (\theta).$ To do this we can use the trigonometric ratios.

We know that in a right angled triangle $\ \sin(\theta)=\frac{O}{H}, \ \cos(\theta)=\frac{A}{H}, \$ and $\ \tan(\theta)=\frac{O}{A}.$

As we need to divide $\sin(\theta)$ by $\cos(\theta)$ , this is the same as dividing $\frac{O}{H}$ by $\frac{A}{H}.$

So we have,

\sin(\theta) \div \cos(\theta) = \frac{O}{H} \div \frac{A}{H}.

2Simplify one side of the identity only.

Simplifying the right hand side, we have

\begin{aligned} &\frac{O}{H}\div\frac{A}{H} \\\\ = & \, \frac{O}{H}\times\frac{H}{A} \\\\ =& \, \frac{OH}{AH} \\\\ =& \, \frac{O}{A} \\\\ =& \, \tan(\theta) \end{aligned}

Therefore $\ \frac{\sin(\theta)}{\cos(\theta)} \equiv \tan(\theta).$

Example 2: using Pythagoras’ theorem to prove a trigonometric identity

Prove that $\ \sin^2(\theta)+\cos^2(\theta)\equiv1.$

Substitute values using the trigonometric ratios or another trigonometric identity.

We need to show that $\sin^2(\theta)+\cos^2(\theta)$ has a value of $1.$ To do this we can use the trigonometric ratios.

We know that $\sin(\theta)=\frac{O}{H}$ and $\cos(\theta)=\frac{A}{H}.$

As $\sin(\theta)=\frac{O}{H},$ multiplying both sides by $H$ to make $O$ the subject, we have $O=H\sin(\theta).$

As $\cos(\theta)=\frac{A}{H},$ multiplying both sides by $H$ to make $A$ the subject, we have $A=H\cos(\theta).$

We now have the length of the opposite side $O=H\sin(\theta),$ and the width of the adjacent side $A=H\cos(\theta).$

Labelling a triangle using these values, we have,

Pythagoras’ theorem states that $a^{2}+b^{2}=c^{2}$ where $a$ is the height, $b$ is the base, and $c$ is always the hypotenuse.

As the base is $H\cos(\theta),$ the height is $H\sin(\theta),$ and the hypotenuse is $H,$ we can substitute these into Pythagoras’ theorem to get

$(H\sin(\theta))^{2}+(H\cos(\theta))^{2}=H^{2}.$

Simplify one side of the identity only.

We can only simplify the left hand side and so by squaring both brackets, we have

$\begin{aligned} &(H\sin(\theta))^{2}+(H\cos(\theta))^{2}\\\\ =&H^{2}\sin^{2}(\theta)+H^{2}\cos^{2}(\theta)\\\\ =&H^{2}(\sin^{2}(\theta)+\cos^{2}(\theta)) \end{aligned}$

This means that as $H^{2}(\sin^{2}(\theta)+\cos^{2}(\theta))=H^{2},$ so $\ \sin^{2}(\theta)+\cos^{2}(\theta)=1$ as $H^{2}\times{1}=H^{2}.$

This proves the identity $\sin^{2}(\theta)+\cos^{2}(\theta)\equiv1.$

Example 3: proving identities using trig ratios

Prove that $\sin(\theta)\div\tan(\theta)\equiv\cos(\theta).$

Substitute values using the trigonometric ratios or another trigonometric identity.

We need to show that $\sin(\theta)\div\tan(\theta)$ is the same as $\cos(\theta).$

We know that $\ \sin(\theta)=\frac{O}{H}, \ \tan(\theta)=\frac{O}{A}, \$ and $\ \cos(\theta)=\frac{A}{H}.$

This means that $\sin(\theta)\div\tan(\theta)$ is equivalent to

$\frac{O}{H}\div\frac{O}{A}.$

Simplify one side of the identity only.

Converting this to one fraction, we have

$\begin{aligned} &\frac{O}{H}\div\frac{O}{A}\\\\ =&\frac{O}{H}\times\frac{A}{O}\\\\ =&\frac{OA}{HO}\\\\ =&\frac{A}{H}\\\\ =&\cos(\theta) \end{aligned}$

This proves that $\sin(\theta)\div\tan(\theta)\equiv\cos(\theta).$

Example 4: proving identities using exact trig values

Prove that $\cos(60)=1-\sin(30)$ using the right-angled triangle below.

Substitute values using the trigonometric ratios or another trigonometric identity.

We need to show that $\cos(60)$ gives us the same value as $1-\sin(30)$ using the triangle above.

Using the triangle, we can determine the exact trig value for $\cos(60)=\frac{A}{H}=\frac{1}{2}.$

We can also determine the exact trig value for $\sin(30)=\frac{O}{H}=\frac{1}{2}.$

Simplify one side of the identity only.

Calculating $1-\sin(30)$ we have

$1-\sin(30)=1-\frac{1}{2}=\frac{1}{2}.$

As $\cos(60)$ is the same as $1-\sin(30),$ we have proved that $\cos(60)=1-\sin(30).$

Example 5: proving identities featuring all three trig ratios

Prove that $\cos(\theta)\times\tan(\theta)\times\sin(\theta)\equiv\sin^{2}(\theta).$

Substitute values using the trigonometric ratios or another trigonometric identity.

We need to show that $\cos(\theta)\times\tan(\theta)\times\sin(\theta)$ is the same as $\sin^{2}(\theta).$

We can substitute $\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}$ into the identity to get

$\cos(\theta)\times\tan(\theta)\times\sin(\theta)=\cos(\theta)\times\frac{\sin(\theta)}{\cos(\theta)}\times\sin(\theta).$

Simplify one side of the identity only.

Simplifying this, we have

$\begin{aligned} & \cos(\theta)\times\frac{\sin(\theta)}{\cos(\theta)}\times\sin(\theta)\\\\ =& \; \frac{\cos(\theta)\times\sin(\theta)\times\sin(\theta)}{\cos(\theta)}\\\\ =& \; \frac{\cos(\theta)\times\sin^{2}(\theta)}{\cos(\theta)} \\\\ =& \; \sin^{2}(\theta) \end{aligned}$

Example 6: using the trigonometric identity sin²(θ) + cos²(θ) = 1

Prove that $6(3\sin^{2}(\theta))+2(9\cos^{2}(\theta))=18.$

Substitute values using the trigonometric ratios or another trigonometric identity.

We need to show that $6(3\sin^{2}(\theta))+2(9\cos^{2}(\theta))$ can be simplified to make $18.$

As there are no substitutions necessary, we can move on to step 2.

Simplify one side of the identity only.

Expanding both brackets gives us

$18\sin^{2}(\theta)+18\cos^{2}(\theta).$

Removing a common factor of $18,$ we get

$18(\sin^{2}(\theta)+\cos^{2}(\theta)).$

As $\sin^{2}(\theta)+\cos^{2}(\theta)=1, \ 18\times{1}=18.$

We have therefore proven that $6(3\sin^{2}(\theta))+2(9\cos^{2}(\theta))=18.$

Common misconceptions

Treating an identity as an equation that needs balancing

An identity cannot be balanced. It is a statement that two expressions or functions are mathematically identical. For example, $3x+6\equiv{3}(x+2)$ as when we factorise $3(x+6)$ we get the expression $3(x+2).$

Mixing up SOHCAHTOA

It is key to remember the three trig ratios that are written as the mnemonic SOHCAHTOA. This way you will not confuse the relationships between the sides and the angles.

Practice trigonometric identities questions and answers

\sin(\theta)\times\cos(\theta)=\tan(\theta)

\sin(\theta)+\cos(\theta)=\tan(\theta)

\sin(\theta)\div\cos(\theta)=\tan(\theta)

\sin(\theta)-\cos(\theta)=\tan(\theta)

As $\sin(\theta)=\frac{O}{H}$ and $\cos(\theta)=\frac{A}{H},$

\sin(\theta)\div\cos(\theta)=\frac{O}{H}\div\frac{A}{H}=\frac{O}{H}\times\frac{H}{A}=\frac{OH}{HA}=\frac{O}{A}=\tan(\theta).

\sin^{2}(\theta)-\cos^{2}(\theta)=1

\sin^{2}(\theta)+\cos^{2}(\theta)=1

\sin(\theta)-\cos(\theta)=1

\sin(\theta)+\cos(\theta)=1

As $\sin(\theta)=\frac{O}{H},$ multiplying both sides by $H$ to make $O$ the subject,
we have $O=H\sin(\theta).$

As $\cos(\theta)=\frac{A}{H},$ multiplying both sides by $H$ to make $A$ the subject,
we have $A=H\cos(\theta).$

As the base is $H\cos(\theta),$ the height is $H\sin(\theta),$ and the hypotenuse is $H,$ we can substitute these into Pythagoras’ theorem to get,

(H\sin(\theta))^{2}+(H\cos(\theta))^{2}=H^{2}

\begin{aligned} &(H\sin(\theta))^{2}+(H\cos(\theta))^{2}\\\\ =&H^{2}\sin^{2}(\theta)+H^{2}\cos^{2}(\theta)\\\\ =&H^{2}(\sin^{2}(\theta)+\cos^{2}(\theta)) \end{aligned}

This means that as $H^{2}(\sin^{2}(\theta)+\cos^{2}(\theta))=H^{2}, \ \sin^{2}(\theta)+\cos^{2}(\theta)=1$ as $H^{2}\times{1}=H^{2}.$

\cos(\theta)

\sin^{-1}(\theta)

\frac{1}{\cos(\theta)}

\cos^{2}(\theta)

As $\sin(\theta)=\frac{O}{H}, \ \tan(\theta)=\frac{O}{A},$ and $\cos(\theta)=\frac{A}{H},$

$\tan(\theta)\div\sin(\theta)$ is equivalent to

\begin{aligned} &\frac{O}{A}\div\frac{O}{H}\\\\ =\; & \frac{O}{A}\times\frac{H}{O}\\\\ =\; & \frac{OH}{AO}\\\\ =\; & \frac{H}{A}\\\\ =\; & \frac{1}{\cos(\theta)} \end{aligned}

\sqrt{2}

1

2\sqrt{2}

2

2\cos(45)\times\sin(45)=2\times\frac{1}{\sqrt{2}}\times\frac{1}{\sqrt{2}}=2\times\frac{1}{2}=1

1

\tan^{2}(\theta)

\frac{1}{\tan(\theta)}

\frac{\cos^{2}(\theta)}{\sin^{2}(\theta)}

\cos(\theta)\times\tan(\theta)\div\sin(\theta)=\cos(\theta)\times\frac{\sin(\theta)}{\cos(\theta)}\div\sin(\theta)=\frac{\cos(\theta)\sin(\theta)}{\cos(\theta)\sin(\theta)}=1

\sin(\theta)

\cos(\theta)

\tan(\theta)

\cos^{2}(\theta)

\frac{\sin(\theta)}{\sqrt{1-\sin^{2}(\theta)}}=\frac{\sin(\theta)}{\sqrt{\cos^{2}(\theta)}}=\frac{\sin(\theta)}{\cos(\theta)}=\tan(\theta)

Trigonometric identities questions

1. Prove that $\frac{\sqrt{1-\cos^{2}(\theta)}}{\cos(\theta)}\equiv \tan(\theta).$

(2 marks)

Show answer

\sqrt{1-\cos^{2}(\theta)} = \sin(\theta)

(1)

\frac{\sin(\theta)}{\cos(\theta)}= \tan(\theta)

(1)

2. (a) Expand and simplify $(\sin x + \cos x)^{2}.$

(b) Use your solution to part a) to prove that $(\sin x + \cos x)^{2} + (\sin x – \cos x)^{2} \equiv 2.$

(6 marks)

Show answer

(a)

(\sin x + \cos x) (\sin x + \cos x) =

(1)

\sin^{2} x + 2 \cos x \sin x + \cos^{2} x

(1)

(b)

(\sin x-\cos x)^{2} = \sin^{2} x-2\cos x \sin x + cos^{2} x

(1)

\sin^{2}x+2\cos x \sin x+ \cos^{2}x+ \sin^{2} x-2 \cos x \sin x+\cos^{2}x=2\sin^{2}x+2\cos^{2}x

(1)

2\sin^{2}x+2 \cos^{2}x =2(\sin^{2}x+\cos^{2}x)

(1)

2\sin^{2}x+2 \cos^{2}x =2(1) =2

(1)

3. The axes below show a circle where the point $P(H \cos x, \ H \sin x)$ is a point on the circumference of the circle.

Trigonometric identities GCSE question 3

Prove that the radius of the circle is $3.$

(4 marks)

Show answer

(3 \cos x)^{2}+(3 \sin x)^{2}=r^{2}

(1)

9 \cos^{2} x + 9 \sin^{2} x =r^{2}

(1)

9 (\cos^{2} x + \sin^{2} x) = 9

(1)

r=\sqrt{9}=3

(1)

Learning checklist

You have now learned how to:

Use Pythagoras’ theorem,

a^2+b^2=c^2

to prove the trigonometric ratios

\sin(\theta)=\frac{O}{H}, \ \cos(\theta)=\frac{A}{H}

and

\tan(\theta)=\frac{O}{A}

Use the trigonometric ratios to prove trigonometric identities

Extend and formalise your knowledge of ratio and proportion, including trigonometric ratios

The next lessons are

More trigonometric functions and identities

Trigonometric functions are mathematical tools that describe the relationships between the angles and side lengths of a right-angled triangle. These functions are fundamental in trigonometry and are used to solve various problems involving triangles and circles. Other trigonmetric functions that are studied in A level Mathematics include:

Sine (sin): This function represents the ratio of the length of the opposite side to the hypotenuse. For an angle θ in a right-angled triangle, sin(θ) = Opposite / Hypotenuse.
Cosine (cos): This function is the ratio of the length of the adjacent side to the hypotenuse. For an angle θ, cos(θ) = Adjacent / Hypotenuse.
Tangent (tan): This function is the ratio of the length of the opposite side to the adjacent side. For an angle θ, tan(θ) = Opposite / Adjacent.
Cotangent (cot): The cotangent function is the reciprocal of the tangent function. For an angle θ, cot(θ) = 1 / tan(θ) = Adjacent / Opposite.
Secant (sec): The secant function is the reciprocal of the cosine function. For an angle θ, sec(θ) = 1 / cos(θ) = Hypotenuse / Adjacent.
Cosecant (csc): The cosecant function is the reciprocal of the sine function. For an angle θ, csc(θ) = 1 / sin(θ) = Hypotenuse / Opposite.

These trigonometric functions are essential for understanding the properties of right triangles and are widely used in various fields, including physics, engineering, and computer science.

We also study trigonometric values beyond $90$ degrees (outside of the limits of a right-angled triangle). These include double-angle formulas (double-angle identities), and conversely half-angle formulas (half-angle identities). These allow us to form trigonometric equations.

At A-level, we use a different measurement for an angle called radians where $2\pi$ radians is a full turn $(360^{\circ}), \pi$ radians is a half turn $(180^{\circ}), \frac{\pi}{2}$ radians is a quarter turn $(90^{\circ}),$ etc.

We also work with the integrals and derivatives of different trigonometric functions to form other trig equations.

Step-by-step guide: Trigonometric functions

See also: Sine rule

See also: Trigonometry formula

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