Edexcel A-level Further Maths revision notes

Further Trigonometry (Further Pure 1)

Section 1

Year 1

3 specification points

Notes and three levels of exam-style practice for each registered specification point in this section.

Checked against Edexcel 9FM0 section 1

Checked against Edexcel 9FM0 section 1. Review basis: the qualification registry sourced from the Pearson Edexcel Level 3 Advanced GCE in Further Mathematics (9FM0) specification; registry verification recorded 17 July 2026.

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1.1

Year 1

The t-formulae.

Notes

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Explanation

Write $t=\tan\tfrac{\theta}{2}$ .
The double-angle formulae applied to $\tfrac{\theta}{2}$ give $\sin\theta=\dfrac{2t}{1+t^2}$ , $\cos\theta=\dfrac{1-t^2}{1+t^2}$ and $\tan\theta=\dfrac{2t}{1-t^2}$ .
The derivations start from $\sin\theta=2\sin\tfrac{\theta}{2}\cos\tfrac{\theta}{2}$ and $\cos\theta=\cos^2\tfrac{\theta}{2}-\sin^2\tfrac{\theta}{2}$ , then divide numerator and denominator by $\cos^2\tfrac{\theta}{2}$ using $\sec^2\tfrac{\theta}{2}=1+t^2$ .
Every trigonometric function of $\theta$ therefore becomes a rational function of the single variable $t$ , which is what makes the substitution so powerful.
The price is that $t$ is undefined when $\tfrac{\theta}{2}$ is an odd multiple of $\tfrac{\pi}{2}$ , that is when $\theta$ is an odd multiple of $\pi$ , so $\theta=\pi$ is never produced by the substitution and must be tested separately.

The unit circle: the chord from A(-1, 0) to P makes half the central angle with the axis, so its gradient is t.

Worked example

Given that $\tan\tfrac{\theta}{2}=\tfrac12$ , find the exact values of $\sin\theta$ , $\cos\theta$ and $\tan\theta$ .

1.Take $t=\tfrac12$ , so $1+t^2=\tfrac54$ and $1-t^2=\tfrac34$ .
2. $\sin\theta=\dfrac{2t}{1+t^2}=\dfrac{1}{5/4}=\dfrac45$ .
3. $\cos\theta=\dfrac{1-t^2}{1+t^2}=\dfrac{3/4}{5/4}=\dfrac35$ .
4. $\tan\theta=\dfrac{2t}{1-t^2}=\dfrac{1}{3/4}=\dfrac43$ , which agrees with $\sin\theta\div\cos\theta$ .

Answer: $\sin\theta=\tfrac45$ , $\cos\theta=\tfrac35$ and $\tan\theta=\tfrac43$ .

Common mistakes

Don't fall into the trap of writing $t=\tan\theta$ rather than $t=\tan\tfrac{\theta}{2}$ , so every substituted expression is wrong.
Don't fall into the trap of putting $1-t^2$ in the denominator of $\sin\theta$ ; only $\tan\theta$ has $1-t^2$ underneath.
Don't fall into the trap of forgetting that $\theta=\pi$ has no finite $t$ , so a solution there is silently lost.

Exam tip

Quote $t=\tan\tfrac{\theta}{2}$ in words before substituting, then state $1+t^2$ once and reuse it as the common denominator.

Tier 1 · Easy

ORIGINAL

Given that

t=\tan\tfrac{\theta}{2}=\tfrac12

, find the exact values of

\sin\theta

and

\cos\theta

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

Starting from

\sin\theta=2\sin\tfrac{\theta}{2}\cos\tfrac{\theta}{2}

, derive the t-formula

\sin\theta=\dfrac{2t}{1+t^2}

, where

t=\tan\tfrac{\theta}{2}

(3)

(Total for Question 1 is 3 marks)

Tier 3 · Hard

ORIGINAL

Derive the t-formula

\cos\theta=\dfrac{1-t^2}{1+t^2}

and hence state every value of

\theta

0\le\theta<2\pi

at which the three t-formulae cannot all be applied.

(5)

(Total for Question 1 is 5 marks)

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1.2

Year 1

Applications of t-formulae to trigonometric identities.

Notes

Evidence from your answers: none yet

Your confidence:

Explanation

To prove an identity with the t-substitution, replace every trigonometric function of $\theta$ by its rational expression in $t=\tan\tfrac{\theta}{2}$ , put each side over the common denominator $1+t^2$ , and simplify.
Two conversions do most of the work: $1+\cos\theta=\dfrac{2}{1+t^2}$ and $1-\cos\theta=\dfrac{2t^2}{1+t^2}$ .
Dividing either by $\sin\theta=\dfrac{2t}{1+t^2}$ immediately gives the two headline results $\operatorname{cosec}\theta+\cot\theta=\dfrac1t$ and $\operatorname{cosec}\theta-\cot\theta=t$ .
Work on one side only, state the identity you are proving at the top, and finish by writing that side in the required form; an identity is not proved by manipulating both sides until they meet unless every step is reversible.
Because $t$ does not exist at $\theta=\pi$ , always record the excluded values when the question asks for the domain of validity.

Worked example

Prove that $\operatorname{cosec}\theta-\cot\theta\equiv\tan\tfrac{\theta}{2}$ .

1. $\operatorname{cosec}\theta-\cot\theta=\dfrac{1-\cos\theta}{\sin\theta}$ .
2.In terms of $t$ the numerator is $\dfrac{2t^2}{1+t^2}$ and the denominator is $\dfrac{2t}{1+t^2}$ .
3.The factor $\dfrac{2}{1+t^2}$ cancels, leaving $\dfrac{t^2}{t}=t$ .

Answer: $\operatorname{cosec}\theta-\cot\theta\equiv t\equiv\tan\tfrac{\theta}{2}$ , valid whenever $\sin\theta\ne0$ .

Common mistakes

Don't fall into the trap of converting both sides at once and writing a chain of equals signs, which assumes the identity being proved.
Don't fall into the trap of cancelling $t$ from a numerator and denominator without noting that $t=0$ is excluded when the identity is divided through.
Don't fall into the trap of leaving the answer as $\dfrac{2t^2/(1+t^2)}{2t/(1+t^2)}$ instead of simplifying to $t$ .

Exam tip

Memorise $1+\cos\theta=\dfrac{2}{1+t^2}$ and $1-\cos\theta=\dfrac{2t^2}{1+t^2}$ ; almost every t-identity collapses once one of them appears.

Tier 1 · Easy

ORIGINAL

Using

t=\tan\tfrac{\theta}{2}

, prove that

\operatorname{cosec}\theta+\cot\theta\equiv\cot\tfrac{\theta}{2}

(3)

(Total for Question 1 is 3 marks)

Tier 2 · Standard

ORIGINAL

Prove that

\cot\tfrac{\theta}{2}-\tan\tfrac{\theta}{2}\equiv2\cot\theta

(4)

(Total for Question 1 is 4 marks)

Tier 3 · Hard

ORIGINAL

Prove that

\dfrac{1+\sin\theta-\cos\theta}{1+\sin\theta+\cos\theta}\equiv\tan\tfrac{\theta}{2}

(5)

(Total for Question 1 is 5 marks)

1.3

Year 1

Applications of t-formulae to solve trigonometric equations.

Notes

Evidence from your answers: none yet

Your confidence:

Explanation

An equation of the form $a\cos x+b\sin x=c$ becomes a quadratic in $t=\tan\tfrac{x}{2}$ .
Substituting and multiplying through by $1+t^2$ gives $a(1-t^2)+2bt=c(1+t^2)$ , that is $(a+c)t^2-2bt+(c-a)=0$ .
Its discriminant is $4\left(a^2+b^2-c^2\right)$ , so real solutions exist exactly when $c^2\le a^2+b^2$ — the same condition the $R\cos(x-\alpha)$ method gives.
Solve for $t$ , then recover $x=2\arctan t$ and add multiples of $2\pi$ to land in the required interval.
The one trap is that $t$ does not exist at $x=\pi$ , so $x=\pi$ can never appear as a root of the quadratic: always substitute $x=\pi$ into the original equation as a separate final check.

Worked example

Solve $\sin x-\cos x=1$ for $0\le x<2\pi$ .

1.Substituting gives $\dfrac{2t}{1+t^2}-\dfrac{1-t^2}{1+t^2}=1$ , so $2t-1+t^2=1+t^2$ .
2.Hence $2t=2$ and $t=1$ , giving $\tfrac{x}{2}=\tfrac{\pi}{4}$ and $x=\tfrac{\pi}{2}$ .
3.Test the excluded value $x=\pi$ in the original equation: $\sin\pi-\cos\pi=0-(-1)=1$ , so it is also a solution.

Answer: $x=\tfrac{\pi}{2}$ and $x=\pi$ .

Common mistakes

Don't fall into the trap of losing the root $x=\pi$ because the t-substitution can never produce it.
Don't fall into the trap of writing $x=\arctan t$ instead of $x=2\arctan t$ .
Don't fall into the trap of giving only the principal value of $2\arctan t$ and missing a second root inside the interval.

Exam tip

Finish every t-substitution solution with the line 'check $x=\pi$ separately'; the substitution can never produce that root, and dropping it costs the accuracy mark for an incomplete solution set.

Tier 1 · Easy

ORIGINAL

Use the substitution

t=\tan\tfrac{x}{2}

to solve

\sin x+\cos x=1

for

0\le x<2\pi

(4)

(Total for Question 1 is 4 marks)

Tier 2 · Standard

ORIGINAL

Solve

\sin x-\cos x=1

for

0\le x<2\pi

, explaining carefully why the t-substitution alone does not give the complete solution set.

(5)

(Total for Question 1 is 5 marks)

Tier 3 · Hard

ORIGINAL

Solve

2\cos x-\sin x=1

for

0\le x<2\pi

, giving non-exact answers to

3

significant figures.

(6)

(Total for Question 1 is 6 marks)

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