Mathematics 9709 · Pure Mathematics 1

Trigonometric identities

An identity is an equation that is true for every angle.
Just two of them matter for the exam. Almost every hard trig question wants you to use one of them. You turn a messy equation into a quadratic, which you already know how to solve.

You use two identities again and again in this chapter: $\sin^2\theta+\cos^2\theta\equiv1$ (Pythagoras) and $\tan\theta\equiv\tfrac{\sin\theta}{\cos\theta}$ (the meaning of tan). Everything else is just these two.

The two identities

Here they are, and where they come from:

$\sin^2\theta+\cos^2\theta\equiv1 \qquad\text{and}\qquad \tan\theta\equiv\dfrac{\sin\theta}{\cos\theta}$

The first one is Pythagoras on the unit circle. The point $(\cos\theta,\sin\theta)$ is distance 1 from the centre. So $\cos^2\theta+\sin^2\theta=1$ .
The second one is just the meaning of tan (height over across).
Watch the notation: $\sin^2\theta$ means $(\sin\theta)^2$ , never $\sin(\theta^2)$ .

sin²θ + cos²θ = 1 is Pythagoras on the unit circle: the point (cos θ, sin θ) sits at distance 1 from the centre.

The step that earns marks: rearrange the first one into $\sin^2\theta=1-\cos^2\theta$ (or the other way round). This puts the whole equation in one ratio. Then it is a quadratic.

Simplifying: swap one ratio for the other

The easiest use: if you have a mix of $\sin^2$ and $\cos^2$ , swap out the ratio you do not want.

Worked example

Express $4\cos^2 x - 3\sin^2 x$ in terms of $\cos x$

This is the textbook's Worked Example 5.18. There is only one $\sin^2$ : replace it using $\sin^2x=1-\cos^2x$ :

Substitute: $4\cos^2x-3(1-\cos^2x)$ .
Expand the bracket: $4\cos^2x-3+3\cos^2x$ .
Gather the $\cos^2x$ terms.

Answer

7\cos^2x-3

Proving an identity

A “prove that” or “show that” question gives you both sides. Your job is to change one side until it matches the other.

Start from the harder side. Turn every $\tan$ into $\tfrac{\sin}{\cos}$ . Put it all over one fraction. Then use $\sin^2+\cos^2=1$ to make it smaller.

Worked example

Prove $\dfrac{\sin^2 x - \cos x - 1}{1 + \cos x} \equiv -\cos x$

This is 9709/11 Jun 2024 Q5(a). The top has a single $\sin^2x$ . That is the clue to use Pythagoras:

Replace $\sin^2x$ with $1-\cos^2x$ , so the numerator becomes $(1-\cos^2x)-\cos x-1=-\cos^2x-\cos x$ .
Factorise the numerator: $-\cos x(\cos x+1)$ .
The denominator is $1+\cos x$ : the same bracket, so it cancels: $\dfrac{-\cos x(1+\cos x)}{1+\cos x}$ .

Answer

\equiv -\cos x

How the exam uses it: prove, then “hence solve”

A common two-part question proves an identity in (a). Then it says “hence solve” in (b). The identity you proved is the key to part (b):

Worked example

Hence solve $\dfrac{\sin^2 x - \cos x - 1}{2 + 2\cos x} = \tfrac14$ for $0^\circ \le x \le 360^\circ$

This is 9709/11 Jun 2024 Q5(b). The bottom is $2+2\cos x=2(1+\cos x)$ , so the left side is half of the part-(a) expression:

Use part (a): $\dfrac{-\cos x}{2}=\tfrac14$ .
Solve for the ratio: $\cos x=-\tfrac12$ .
Basic angle $60^\circ$ ; cosine is negative in quadrants 2 and 3, so $x=180-60$ and $x=180+60$ .

Answer

x=120^\circ,\ 240^\circ

This prove-then-solve shape comes up again and again. For example, see 9709/12 Mar 2022 Q7. (9709/12 Mar 2022 Q7)

Where the marks go

Common mistake

Notation:

\sin^2x=(\sin x)^2

. Writing

\sin x^2

means

\sin(x^2)

. That is a completely different thing.

Common mistake

In a proof, work each side down to the same thing. Do not move terms across the

\equiv

sign as if it were an equation to solve. The mark scheme wants “all necessary detail”. So show the factorising and the cancelling. (9709/11 Jun 2024 Q5)

Now you try

Express $2\sin^2 x-7\cos^2 x+4$ in terms of $\sin x$ . This is the same “put it in one ratio” step that opens a proof like 9709/11 Jun 2022 Q4. (9709/11 Jun 2022 Q4)

Your turn— tap to reveal the worked answer (swap cos² for 1 − sin²)

Use $\cos^2x=1-\sin^2x$ :

$2\sin^2x-7(1-\sin^2x)+4=2\sin^2x-7+7\sin^2x+4$

Answer

9\sin^2x-3

PreviousInverse trigonometric functions NextSolving trigonometric equations

Trigonometric identities

An identity is an equation that is true for every angle.
Just two of them matter for the exam. Almost every hard trig question wants you to use one of them. You turn a messy equation into a quadratic, which you already know how to solve.

The two identities

Here they are, and where they come from:

$\sin^2\theta+\cos^2\theta\equiv1 \qquad\text{and}\qquad \tan\theta\equiv\dfrac{\sin\theta}{\cos\theta}$

The first one is Pythagoras on the unit circle. The point $(\cos\theta,\sin\theta)$ is distance 1 from the centre. So $\cos^2\theta+\sin^2\theta=1$ .
The second one is just the meaning of tan (height over across).
Watch the notation: $\sin^2\theta$ means $(\sin\theta)^2$ , never $\sin(\theta^2)$ .

sin²θ + cos²θ = 1 is Pythagoras on the unit circle: the point (cos θ, sin θ) sits at distance 1 from the centre.

The step that earns marks: rearrange the first one into $\sin^2\theta=1-\cos^2\theta$ (or the other way round). This puts the whole equation in one ratio. Then it is a quadratic.

Simplifying: swap one ratio for the other

The easiest use: if you have a mix of $\sin^2$ and $\cos^2$ , swap out the ratio you do not want.

Worked example

Express $4\cos^2 x - 3\sin^2 x$ in terms of $\cos x$

This is the textbook's Worked Example 5.18. There is only one $\sin^2$ : replace it using $\sin^2x=1-\cos^2x$ :

Substitute: $4\cos^2x-3(1-\cos^2x)$ .
Expand the bracket: $4\cos^2x-3+3\cos^2x$ .
Gather the $\cos^2x$ terms.

Answer

7\cos^2x-3

Proving an identity

A “prove that” or “show that” question gives you both sides. Your job is to change one side until it matches the other.

Start from the harder side. Turn every $\tan$ into $\tfrac{\sin}{\cos}$ . Put it all over one fraction. Then use $\sin^2+\cos^2=1$ to make it smaller.

Worked example

Prove $\dfrac{\sin^2 x - \cos x - 1}{1 + \cos x} \equiv -\cos x$

This is 9709/11 Jun 2024 Q5(a). The top has a single $\sin^2x$ . That is the clue to use Pythagoras:

Replace $\sin^2x$ with $1-\cos^2x$ , so the numerator becomes $(1-\cos^2x)-\cos x-1=-\cos^2x-\cos x$ .
Factorise the numerator: $-\cos x(\cos x+1)$ .
The denominator is $1+\cos x$ : the same bracket, so it cancels: $\dfrac{-\cos x(1+\cos x)}{1+\cos x}$ .

Answer

\equiv -\cos x

How the exam uses it: prove, then “hence solve”

A common two-part question proves an identity in (a). Then it says “hence solve” in (b). The identity you proved is the key to part (b):

Worked example

Hence solve $\dfrac{\sin^2 x - \cos x - 1}{2 + 2\cos x} = \tfrac14$ for $0^\circ \le x \le 360^\circ$

This is 9709/11 Jun 2024 Q5(b). The bottom is $2+2\cos x=2(1+\cos x)$ , so the left side is half of the part-(a) expression:

Use part (a): $\dfrac{-\cos x}{2}=\tfrac14$ .
Solve for the ratio: $\cos x=-\tfrac12$ .
Basic angle $60^\circ$ ; cosine is negative in quadrants 2 and 3, so $x=180-60$ and $x=180+60$ .

Answer

x=120^\circ,\ 240^\circ

This prove-then-solve shape comes up again and again. For example, see 9709/12 Mar 2022 Q7. (9709/12 Mar 2022 Q7)

Where the marks go

Common mistake

Notation:

\sin^2x=(\sin x)^2

. Writing

\sin x^2

means

\sin(x^2)

. That is a completely different thing.

Common mistake

In a proof, work each side down to the same thing. Do not move terms across the

\equiv

sign as if it were an equation to solve. The mark scheme wants “all necessary detail”. So show the factorising and the cancelling. (9709/11 Jun 2024 Q5)

Now you try

Express $2\sin^2 x-7\cos^2 x+4$ in terms of $\sin x$ . This is the same “put it in one ratio” step that opens a proof like 9709/11 Jun 2022 Q4. (9709/11 Jun 2022 Q4)

Your turn— tap to reveal the worked answer (swap cos² for 1 − sin²)

Use $\cos^2x=1-\sin^2x$ :

$2\sin^2x-7(1-\sin^2x)+4=2\sin^2x-7+7\sin^2x+4$

Answer

9\sin^2x-3