FP1-1 Further Trigonometry (Further Pure 1) — revision question pack

3 specification points · notes, questions, answers and worked methods

Checked against Edexcel 9FM0 section FP1-1. 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.

How this checking works

FP1-1.1 · The t-formulae.

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

1.
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)
2.
Given that $\tan\tfrac{\theta}{2}=3$ , find the exact values of $\sin\theta$ and $\cos\theta$ , and state which quadrant contains $\theta$ .
(3)
(Total for Question 2 is 3 marks)

Tier 2 · Standard

1.
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)
2.
Given that $\sin\theta=\tfrac{24}{25}$ and $0<\theta<\tfrac{\pi}{2}$ , find the exact value of $t=\tan\tfrac{\theta}{2}$ .
(4)
(Total for Question 2 is 4 marks)
3.
Given that $\cos\theta=-\tfrac{7}{25}$ and $0<\theta<\pi$ , find the exact value of $\tan\tfrac{\theta}{2}$ .
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
Derive the t-formula $\cos\theta=\dfrac{1-t^2}{1+t^2}$ and hence state every value of $\theta$ in $0\le\theta<2\pi$ at which the three t-formulae cannot all be applied.
(5)
(Total for Question 1 is 5 marks)
2.
Using $t=\tan\tfrac{\theta}{2}$ , express $\dfrac{3\sin\theta+4\cos\theta}{5+4\cos\theta}$ as a single algebraic fraction in $t$ , giving your answer in a fully factorised form.
(5)
(Total for Question 2 is 5 marks)
3.
Express $\sin\theta+\cos\theta$ as a single fraction in $t=\tan\tfrac{\theta}{2}$ , and hence find every value of $t$ for which $\sin\theta+\cos\theta=1$ .
(5)
(Total for Question 3 is 5 marks)
4.
The point $P$ lies on the unit circle $x^2+y^2=1$ at angle $\theta$ measured anticlockwise from the positive $x$ -axis, and $A$ is the point $(-1,0)$ . Show that the gradient of the chord $AP$ is $\tan\tfrac{\theta}{2}$ , and explain what this says about the t-substitution.
(6)
(Total for Question 4 is 6 marks)
5.
Given that $\tan\theta=\tfrac34$ and $0<\theta<\tfrac{\pi}{2}$ , find the exact value of $\tan\tfrac{\theta}{2}$ .
(5)
(Total for Question 5 is 5 marks)

FP1-1.2 · Applications of t-formulae to trigonometric identities.

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

1.
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)
2.
Show that $\dfrac{1-\cos\theta}{\sin\theta}\equiv\tan\tfrac{\theta}{2}$ .
(3)
(Total for Question 2 is 3 marks)

Tier 2 · Standard

1.
Prove that $\cot\tfrac{\theta}{2}-\tan\tfrac{\theta}{2}\equiv2\cot\theta$ .
(4)
(Total for Question 1 is 4 marks)
2.
Prove that $\dfrac{1-\cos\theta}{1+\cos\theta}\equiv\tan^2\tfrac{\theta}{2}$ .
(3)
(Total for Question 2 is 3 marks)
3.
Prove that $\tan\theta-\tan\tfrac{\theta}{2}\equiv\tan\tfrac{\theta}{2}\sec\theta$ .
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
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)
2.
Show that $\sec\theta+\tan\theta\equiv\dfrac{1+t}{1-t}$ , where $t=\tan\tfrac{\theta}{2}$ , and deduce that $\sec\theta+\tan\theta\equiv\tan\left(\tfrac{\pi}{4}+\tfrac{\theta}{2}\right)$ .
(5)
(Total for Question 2 is 5 marks)
3.
Prove that $\tan\tfrac{\theta}{2}+\cot\tfrac{\theta}{2}\equiv2\operatorname{cosec}\theta$ , and state the values of $\theta$ in $0<\theta<2\pi$ for which the identity is not defined.
(5)
(Total for Question 3 is 5 marks)
4.
Show that $\dfrac{1+\cos\theta}{1-\cos\theta}+\dfrac{1-\cos\theta}{1+\cos\theta}\equiv4\operatorname{cosec}^2\theta-2$ .
(6)
(Total for Question 4 is 6 marks)
5.
Show that $\operatorname{cosec}\theta\equiv\tfrac12\left(t+\dfrac1t\right)$ , where $t=\tan\tfrac{\theta}{2}$ , and hence prove that $\operatorname{cosec}\theta\ge1$ for all $\theta$ with $0<\theta<\pi$ , stating when equality holds.
(6)
(Total for Question 5 is 6 marks)

FP1-1.3 · Applications of t-formulae to solve trigonometric equations.

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

1.
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)
2.
Use the substitution $t=\tan\tfrac{x}{2}$ to solve $5\cos x+12\sin x=5$ for $0\le x<2\pi$ , giving non-exact answers to $3$ significant figures.
(4)
(Total for Question 2 is 4 marks)

Tier 2 · Standard

1.
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)
2.
Use the substitution $t=\tan\tfrac{x}{2}$ to solve $\cos x+\sin x=0$ for $0\le x<2\pi$ , giving your answers as exact multiples of $\pi$ .
(5)
(Total for Question 2 is 5 marks)
3.
Use the substitution $t=\tan\tfrac{x}{2}$ to solve $3\sin x+4\cos x=4$ for $0\le x<2\pi$ , giving non-exact answers to $3$ significant figures.
(5)
(Total for Question 3 is 5 marks)

Tier 3 · Hard

1.
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)
2.
Solve $\operatorname{cosec}x+\cot x=2$ for $0<x<2\pi$ , giving your answer to $3$ significant figures.
(5)
(Total for Question 2 is 5 marks)
3.
Show that, under the substitution $t=\tan\tfrac{x}{2}$ , the equation $a\cos x+b\sin x=c$ becomes $(a+c)t^2-2bt+(c-a)=0$ , and deduce that the equation has no real solution when $c^2>a^2+b^2$ .
(6)
(Total for Question 3 is 6 marks)
4.
Solve $7\cos x+6\sin x=2$ for $0\le x<2\pi$ , giving your answers to $3$ significant figures.
(6)
(Total for Question 4 is 6 marks)
5.
Solve $3\cos x+4\sin x=5$ for $0\le x<2\pi$ , and use the discriminant of the resulting quadratic in $t$ to explain why the equation has exactly one solution.
(6)
(Total for Question 5 is 6 marks)

Answer key

Answers begin on a new printed page so the question pack can be completed without the solutions alongside it.

FP1-1.1 · The t-formulae.

Tier 1 · Easy

Mark scheme for FP1-1.1 Tier 1 · Easy
Question	Scheme	Marks
1	$\sin\theta=\dfrac45$ $\cos\theta=\dfrac35$	2
(2 marks)		2
Notes
With $t=\tfrac12$ , $1+t^2=\tfrac54$ , so $\sin\theta=\dfrac{2(1/2)}{5/4}=\dfrac45$ and $\cos\theta=\dfrac{1-1/4}{5/4}=\dfrac35$ . Independent check: $\theta=2\arctan\tfrac12=0.9273$ radians gives $\sin\theta=0.8$ and $\cos\theta=0.6$ , and $\left(\tfrac45\right)^2+\left(\tfrac35\right)^2=1$ .
2	$\sin\theta=\dfrac35$ $\cos\theta=-\dfrac45$ $\theta$ lies in the second quadrant	3
(3 marks)		3
Notes
With $t=3$ , $1+t^2=10$ , so $\sin\theta=\dfrac{6}{10}=\dfrac35$ and $\cos\theta=\dfrac{1-9}{10}=-\dfrac45$ . Positive sine with negative cosine places $\theta$ in the second quadrant. Independent check: $\theta=2\arctan 3=2.4981$ radians, and $2.4981$ lies between $\tfrac{\pi}{2}$ and $\pi$ with $\sin\theta=0.6$ , $\cos\theta=-0.8$ .

Tier 2 · Standard

Mark scheme for FP1-1.1 Tier 2 · Standard
Question	Scheme	Marks
1	$\sin\theta=2\sin\tfrac{\theta}{2}\cos\tfrac{\theta}{2}=\dfrac{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{\cos^2\frac{\theta}{2}+\sin^2\frac{\theta}{2}}$ Dividing numerator and denominator by $\cos^2\tfrac{\theta}{2}$ gives $\dfrac{2t}{1+t^2}$	3
(3 marks)		3
Notes
Write $\sin\theta=2\sin\tfrac{\theta}{2}\cos\tfrac{\theta}{2}$ and divide by the Pythagorean identity $\cos^2\tfrac{\theta}{2}+\sin^2\tfrac{\theta}{2}=1$ . Dividing numerator and denominator by $\cos^2\tfrac{\theta}{2}$ turns the numerator into $2t$ and the denominator into $1+t^2$ . Independent check: at $\theta=\tfrac{\pi}{3}$ , $t=\tan\tfrac{\pi}{6}=\tfrac{1}{\sqrt3}$ and $\dfrac{2/\sqrt3}{1+1/3}=\dfrac{\sqrt3}{2}=\sin\tfrac{\pi}{3}$ .
2	$\dfrac{2t}{1+t^2}=\dfrac{24}{25}$ leads to $12t^2-25t+12=0$ $t=\dfrac43$ or $t=\dfrac34$ $0<\theta<\tfrac{\pi}{2}$ gives $0<\tfrac{\theta}{2}<\tfrac{\pi}{4}$ , so $t<1$ $t=\dfrac34$	4
(4 marks)		4
Notes
Substituting gives $25(2t)=24(1+t^2)$ , so $24t^2-50t+24=0$ and hence $12t^2-25t+12=0$ . Factorising, $(3t-4)(4t-3)=0$ , so $t=\tfrac43$ or $t=\tfrac34$ . Since $0<\theta<\tfrac{\pi}{2}$ we need $0<\tfrac{\theta}{2}<\tfrac{\pi}{4}$ , so $t<1$ and only $t=\tfrac34$ qualifies. Independent check: $\arcsin\tfrac{24}{25}=1.2870$ radians, and $\tan(0.6435)=0.75$ .
3	$\dfrac{1-t^2}{1+t^2}=-\dfrac{7}{25}$ gives $18t^2=32$ $t^2=\dfrac{16}{9}$ , so $t=\pm\dfrac43$ $0<\tfrac{\theta}{2}<\tfrac{\pi}{2}$ forces $t>0$ $t=\dfrac43$	4
(4 marks)		4
Notes
Cross-multiplying, $25(1-t^2)=-7(1+t^2)$ , so $25-25t^2=-7-7t^2$ and $18t^2=32$ , giving $t^2=\tfrac{16}{9}$ . Because $0<\theta<\pi$ we have $0<\tfrac{\theta}{2}<\tfrac{\pi}{2}$ , where the tangent is positive, so $t=\tfrac43$ . Independent check: $\arccos(-0.28)=1.8546$ radians and $\tan(0.9273)=1.3333=\tfrac43$ .

Tier 3 · Hard

Mark scheme for FP1-1.1 Tier 3 · Hard
Question	Scheme	Marks
1	$\cos\theta=\cos^2\tfrac{\theta}{2}-\sin^2\tfrac{\theta}{2}=\dfrac{\cos^2\frac{\theta}{2}-\sin^2\frac{\theta}{2}}{\cos^2\frac{\theta}{2}+\sin^2\frac{\theta}{2}}$ Dividing through by $\cos^2\tfrac{\theta}{2}$ gives $\dfrac{1-t^2}{1+t^2}$ $t$ is undefined at $\theta=\pi$ The formula for $\tan\theta$ also fails when $1-t^2=0$ , that is at $\theta=\tfrac{\pi}{2}$ and $\theta=\tfrac{3\pi}{2}$	5
(5 marks)		5
Notes
Use $\cos\theta=\cos^2\tfrac{\theta}{2}-\sin^2\tfrac{\theta}{2}$ , divide by $\cos^2\tfrac{\theta}{2}+\sin^2\tfrac{\theta}{2}=1$ , then divide numerator and denominator by $\cos^2\tfrac{\theta}{2}$ to get $\dfrac{1-t^2}{1+t^2}$ . The substitution needs $\tan\tfrac{\theta}{2}$ to exist, which fails at $\theta=\pi$ ; the tangent formula additionally needs $t\ne\pm1$ , which fails at $\theta=\tfrac{\pi}{2}$ and $\theta=\tfrac{3\pi}{2}$ , exactly where $\tan\theta$ itself is undefined. Independent check: at $\theta=\tfrac{2\pi}{3}$ , $t=\tan\tfrac{\pi}{3}=\sqrt3$ and $\dfrac{1-3}{1+3}=-\tfrac12=\cos\tfrac{2\pi}{3}$ .
2	Numerator $=\dfrac{6t+4(1-t^2)}{1+t^2}=\dfrac{4+6t-4t^2}{1+t^2}$ Denominator $=\dfrac{5(1+t^2)+4(1-t^2)}{1+t^2}=\dfrac{9+t^2}{1+t^2}$ The quotient is $\dfrac{4+6t-4t^2}{9+t^2}$ $=\dfrac{-2(2t+1)(t-2)}{t^2+9}$	5
(5 marks)		5
Notes
Substitute both t-formulae over the common denominator $1+t^2$ : the numerator becomes $6t+4-4t^2$ and the denominator becomes $5+5t^2+4-4t^2=t^2+9$ . The factor $1+t^2$ cancels, leaving $\dfrac{-4t^2+6t+4}{t^2+9}$ , and $-4t^2+6t+4=-2(2t^2-3t-2)=-2(2t+1)(t-2)$ . Independent check: at $\theta=\tfrac{\pi}{2}$ the original expression is $\dfrac{3}{5}$ , and $t=1$ gives $\dfrac{4+6-4}{10}=\dfrac{6}{10}=\dfrac35$ .
3	$\sin\theta+\cos\theta=\dfrac{2t+1-t^2}{1+t^2}$ Setting this equal to $1$ gives $2t+1-t^2=1+t^2$ $2t^2-2t=0$ , so $t(t-1)=0$ $t=0$ or $t=1$	5
(5 marks)		5
Notes
Over the common denominator, $\sin\theta+\cos\theta=\dfrac{2t+(1-t^2)}{1+t^2}$ . Equating to $1$ and clearing the denominator gives $2t+1-t^2=1+t^2$ , hence $2t^2=2t$ and $t=0$ or $t=1$ . Independent check: $t=0$ corresponds to $\theta=0$ where $\sin\theta+\cos\theta=0+1=1$ , and $t=1$ corresponds to $\theta=\tfrac{\pi}{2}$ where the sum is $1+0=1$ .
4	$P$ has coordinates $(\cos\theta,\ \sin\theta)$ Gradient of $AP=\dfrac{\sin\theta-0}{\cos\theta+1}=\dfrac{2t/(1+t^2)}{2/(1+t^2)}$ $=t=\tan\tfrac{\theta}{2}$ Alternatively the angle in the alternate segment at $A$ is half the central angle $\theta$ Each $t$ therefore names one chord through $A$ and so one point of the circle other than $A$ itself, which is why the substitution is a bijection onto the circle minus the point $\theta=\pi$	6
(6 marks)		6
Notes
Write $P=(\cos\theta,\sin\theta)$ . Then the gradient of $AP$ is $\dfrac{\sin\theta}{1+\cos\theta}$ , and substituting the t-formulae gives $\dfrac{2t/(1+t^2)}{(1+t^2+1-t^2)/(1+t^2)}=\dfrac{2t}{2}=t$ . Independent check by circle geometry: the angle subtended by the arc at the circumference point $A$ is half the angle $\theta$ subtended at the centre, so the chord makes an angle $\tfrac{\theta}{2}$ with $AO$ , giving gradient $\tan\tfrac{\theta}{2}$ . As $t$ ranges over the real numbers the chord sweeps the whole circle except $A$ itself, which is the point $\theta=\pi$ excluded by the substitution.
5	$\dfrac{2t}{1-t^2}=\dfrac34$ gives $3-3t^2=8t$ $3t^2+8t-3=0$ , so $(3t-1)(t+3)=0$ $t=\dfrac13$ or $t=-3$ $0<\tfrac{\theta}{2}<\tfrac{\pi}{4}$ requires $0<t<1$ , so $t=\dfrac13$	5
(5 marks)		5
Notes
Substituting into $\tan\theta=\dfrac{2t}{1-t^2}$ gives $8t=3(1-t^2)$ , so $3t^2+8t-3=0$ , which factorises as $(3t-1)(t+3)=0$ . The restriction $0<\theta<\tfrac{\pi}{2}$ forces $0<\tfrac{\theta}{2}<\tfrac{\pi}{4}$ , where $0<t<1$ , so $t=\tfrac13$ . Independent check: $\arctan 0.75=0.6435$ radians and $\tan(0.32175)=0.33333$ .

FP1-1.2 · Applications of t-formulae to trigonometric identities.

Tier 1 · Easy

Mark scheme for FP1-1.2 Tier 1 · Easy
Question	Scheme	Marks
1	$\operatorname{cosec}\theta+\cot\theta=\dfrac{1+\cos\theta}{\sin\theta}$ $=\dfrac{2/(1+t^2)}{2t/(1+t^2)}=\dfrac1t$ $=\cot\tfrac{\theta}{2}$	3
(3 marks)		3
Notes
Combine over $\sin\theta$ to get $\dfrac{1+\cos\theta}{\sin\theta}$ , then substitute $1+\cos\theta=\dfrac{2}{1+t^2}$ and $\sin\theta=\dfrac{2t}{1+t^2}$ , so the quotient is $\dfrac1t=\cot\tfrac{\theta}{2}$ . Independent check at $\theta=\tfrac{\pi}{3}$ : $\operatorname{cosec}\theta+\cot\theta=1.1547+0.5774=1.7321$ and $\cot\tfrac{\pi}{6}=\sqrt3=1.7321$ .
2	$1-\cos\theta=\dfrac{2t^2}{1+t^2}$ and $\sin\theta=\dfrac{2t}{1+t^2}$ The quotient is $\dfrac{2t^2}{2t}=t$ $=\tan\tfrac{\theta}{2}$	3
(3 marks)		3
Notes
Substitute the two t-formulae; the common factor $\dfrac{2}{1+t^2}$ cancels, leaving $\dfrac{t^2}{t}=t$ . Independent check at $\theta=\tfrac{\pi}{2}$ : $\dfrac{1-0}{1}=1$ and $\tan\tfrac{\pi}{4}=1$ .

Tier 2 · Standard

Mark scheme for FP1-1.2 Tier 2 · Standard
Question	Scheme	Marks
1	$\cot\tfrac{\theta}{2}-\tan\tfrac{\theta}{2}=\dfrac1t-t=\dfrac{1-t^2}{t}$ $\cot\theta=\dfrac{\cos\theta}{\sin\theta}=\dfrac{1-t^2}{2t}$ Hence $\dfrac{1-t^2}{t}=2\cot\theta$	4
(4 marks)		4
Notes
Write the left side over the common denominator $t$ to get $\dfrac{1-t^2}{t}$ . Separately, $\cot\theta=\dfrac{(1-t^2)/(1+t^2)}{2t/(1+t^2)}=\dfrac{1-t^2}{2t}$ , so twice this is $\dfrac{1-t^2}{t}$ , matching the left side. Independent check at $\theta=\tfrac{\pi}{2}$ : $\cot\tfrac{\pi}{4}-\tan\tfrac{\pi}{4}=0$ and $2\cot\tfrac{\pi}{2}=0$ ; at $\theta=\tfrac{\pi}{3}$ both sides equal $1.1547$ .
2	$1-\cos\theta=\dfrac{2t^2}{1+t^2}$ $1+\cos\theta=\dfrac{2}{1+t^2}$ The quotient is $t^2=\tan^2\tfrac{\theta}{2}$	3
(3 marks)		3
Notes
Both numerator and denominator share the factor $\dfrac{2}{1+t^2}$ , so the quotient is simply $t^2$ . Independent check at $\theta=\tfrac{2\pi}{3}$ : $\dfrac{1+0.5}{1-0.5}=3$ and $\tan^2\tfrac{\pi}{3}=3$ .
3	Left side $=\dfrac{2t}{1-t^2}-t=\dfrac{2t-t(1-t^2)}{1-t^2}=\dfrac{t(1+t^2)}{1-t^2}$ $\sec\theta=\dfrac{1+t^2}{1-t^2}$ Right side $=t\cdot\dfrac{1+t^2}{1-t^2}$ , which equals the left side	4
(4 marks)		4
Notes
Write the left side over $1-t^2$ : the numerator is $2t-t+t^3=t+t^3=t(1+t^2)$ . The right side is $t\sec\theta$ and $\sec\theta$ is the reciprocal of $\dfrac{1-t^2}{1+t^2}$ , so it equals $\dfrac{t(1+t^2)}{1-t^2}$ as well. Independent check at $\theta=\tfrac{\pi}{3}$ : left $=1.7321-0.5774=1.1547$ ; right $=0.5774\times2=1.1547$ .

Tier 3 · Hard

Mark scheme for FP1-1.2 Tier 3 · Hard
Question	Scheme	Marks
1	Numerator $=\dfrac{(1+t^2)+2t-(1-t^2)}{1+t^2}=\dfrac{2t^2+2t}{1+t^2}$ Denominator $=\dfrac{(1+t^2)+2t+(1-t^2)}{1+t^2}=\dfrac{2+2t}{1+t^2}$ The quotient is $\dfrac{2t(t+1)}{2(1+t)}=t$ $=\tan\tfrac{\theta}{2}$ , provided $t\ne-1$	5
(5 marks)		5
Notes
Put every term over $1+t^2$ . The numerator collapses to $2t(t+1)$ and the denominator to $2(t+1)$ , so the common factor $2(t+1)$ cancels and the quotient is $t$ . The cancellation requires $t\ne-1$ , that is $\theta\ne-\tfrac{\pi}{2}$ modulo $2\pi$ , where the denominator would vanish. Independent check at $\theta=\tfrac{\pi}{2}$ : $\dfrac{1+1-0}{1+1+0}=1$ and $\tan\tfrac{\pi}{4}=1$ .
2	$\sec\theta+\tan\theta=\dfrac{1+t^2}{1-t^2}+\dfrac{2t}{1-t^2}=\dfrac{(1+t)^2}{(1-t)(1+t)}$ $=\dfrac{1+t}{1-t}$ $\tan\left(\tfrac{\pi}{4}+\tfrac{\theta}{2}\right)=\dfrac{\tan\frac{\pi}{4}+t}{1-\tan\frac{\pi}{4}\,t}=\dfrac{1+t}{1-t}$ The two expressions are therefore identical	5
(5 marks)		5
Notes
Over the common denominator $1-t^2$ the numerator is $1+2t+t^2=(1+t)^2$ , and $1-t^2=(1-t)(1+t)$ , so one factor of $1+t$ cancels. The compound-angle formula for $\tan(A+B)$ with $A=\tfrac{\pi}{4}$ gives the same fraction, which is the required deduction. Independent check at $\theta=\tfrac{\pi}{3}$ : $\sec\theta+\tan\theta=2+1.7321=3.7321$ and $\tan\left(\tfrac{\pi}{4}+\tfrac{\pi}{6}\right)=\tan 75^\circ=3.7321$ .
3	$t+\dfrac1t=\dfrac{1+t^2}{t}$ $\operatorname{cosec}\theta=\dfrac{1+t^2}{2t}$ Hence the left side equals $2\operatorname{cosec}\theta$ Undefined at $\theta=\pi$ , where $\tan\tfrac{\theta}{2}$ does not exist and $\sin\theta=0$	5
(5 marks)		5
Notes
Adding over the denominator $t$ gives $\dfrac{1+t^2}{t}$ , while $\operatorname{cosec}\theta$ is the reciprocal of $\dfrac{2t}{1+t^2}$ , namely $\dfrac{1+t^2}{2t}$ ; doubling it reproduces the left side. Within $0<\theta<2\pi$ the only failure is $\theta=\pi$ , where $t$ is undefined and $\operatorname{cosec}\theta$ does not exist either. Independent check at $\theta=\tfrac{\pi}{2}$ : $1+1=2$ and $2\operatorname{cosec}\tfrac{\pi}{2}=2$ .
4	The two fractions are $\dfrac{1}{t^2}$ and $t^2$ Their sum is $t^2+\dfrac1{t^2}=\left(\dfrac1t-t\right)^2+2$ $\dfrac1t-t=2\cot\theta$ , so the sum is $4\cot^2\theta+2$ $4\cot^2\theta+2=4(\operatorname{cosec}^2\theta-1)+2=4\operatorname{cosec}^2\theta-2$	6
(6 marks)		6
Notes
The t-formulae give $\dfrac{1-\cos\theta}{1+\cos\theta}=t^2$ , so its reciprocal is $\dfrac1{t^2}$ and the sum is $t^2+t^{-2}$ . Completing the square as $\left(t^{-1}-t\right)^2+2$ and using $t^{-1}-t=2\cot\theta$ gives $4\cot^2\theta+2$ , which the Pythagorean identity converts to $4\operatorname{cosec}^2\theta-2$ . Independent check at $\theta=\tfrac{\pi}{3}$ : the left side is $3+\tfrac13=3.3333$ and $4(1.1547)^2-2=3.3333$ .
5	$\operatorname{cosec}\theta=\dfrac{1+t^2}{2t}=\tfrac12\left(t+\dfrac1t\right)$ For $0<\theta<\pi$ we have $t>0$ $t+\dfrac1t-2=\dfrac{(t-1)^2}{t}\ge0$ , so $t+\dfrac1t\ge2$ Hence $\operatorname{cosec}\theta\ge1$ , with equality when $t=1$ , that is $\theta=\tfrac{\pi}{2}$	6
(6 marks)		6
Notes
Inverting $\sin\theta=\dfrac{2t}{1+t^2}$ gives $\operatorname{cosec}\theta=\dfrac{1+t^2}{2t}$ , which splits as $\tfrac12(t+t^{-1})$ . On $0<\theta<\pi$ the half-angle lies in $\left(0,\tfrac{\pi}{2}\right)$ so $t>0$ , and $t+t^{-1}-2=\dfrac{(t-1)^2}{t}$ is non-negative, giving $\operatorname{cosec}\theta\ge1$ with equality only at $t=1$ . Independent check: the minimum of $\operatorname{cosec}\theta$ on $(0,\pi)$ occurs where $\sin\theta$ is greatest, namely $\theta=\tfrac{\pi}{2}$ , where $\operatorname{cosec}\theta=1$ .

FP1-1.3 · Applications of t-formulae to solve trigonometric equations.

Tier 1 · Easy

Mark scheme for FP1-1.3 Tier 1 · Easy
Question	Scheme	Marks
1	$2t+1-t^2=1+t^2$ , so $2t^2-2t=0$ $t=0$ or $t=1$ $x=0$ or $x=\tfrac{\pi}{2}$ $x=\pi$ gives $0+(-1)=-1\ne1$ , so it is not a solution	4
(4 marks)		4
Notes
Multiplying through by $1+t^2$ gives $2t+(1-t^2)=1+t^2$ , hence $2t^2-2t=0$ and $t(t-1)=0$ . Then $x=2\arctan0=0$ and $x=2\arctan1=\tfrac{\pi}{2}$ . Independent check by the $R$ -method: $\sin x+\cos x=\sqrt2\sin\left(x+\tfrac{\pi}{4}\right)=1$ gives $x+\tfrac{\pi}{4}=\tfrac{\pi}{4}$ or $\tfrac{3\pi}{4}$ , so $x=0$ or $\tfrac{\pi}{2}$ .
2	$5(1-t^2)+24t=5(1+t^2)$ , so $10t^2-24t=0$ $t=0$ or $t=\tfrac{12}{5}$ $x=0$ or $x=2\arctan\tfrac{12}{5}=2.35$ $x=\pi$ gives $-5\ne5$ , so it is not a solution	4
(4 marks)		4
Notes
Substituting and clearing the denominator gives $5-5t^2+24t=5+5t^2$ , so $24t=10t^2$ and $t=0$ or $t=2.4$ . Hence $x=0$ or $x=2\arctan 2.4=2.3520$ , which is $2.35$ to $3$ significant figures. Independent check: $5\cos 2.3520+12\sin 2.3520=-3.5207+8.5207=5.0000$ .

Tier 2 · Standard

Mark scheme for FP1-1.3 Tier 2 · Standard
Question	Scheme	Marks
1	$2t-(1-t^2)=1+t^2$ reduces to $2t=2$ , so $t=1$ $x=\tfrac{\pi}{2}$ $t=\tan\tfrac{x}{2}$ is undefined at $x=\pi$ , so that value can never be a root of the quadratic Substituting $x=\pi$ into the original equation gives $0-(-1)=1$ , so $x=\pi$ is a solution $x=\tfrac{\pi}{2}$ and $x=\pi$	5
(5 marks)		5
Notes
The $t^2$ terms cancel, leaving the linear equation $2t=2$ and the single root $t=1$ , giving $x=\tfrac{\pi}{2}$ . Because the substitution requires $\tan\tfrac{x}{2}$ to exist, $x=\pi$ is excluded from the start and must be tested by hand; it satisfies the equation. Independent check by the $R$ -method: $\sin x-\cos x=\sqrt2\sin\left(x-\tfrac{\pi}{4}\right)=1$ gives $x-\tfrac{\pi}{4}=\tfrac{\pi}{4}$ or $\tfrac{3\pi}{4}$ , so $x=\tfrac{\pi}{2}$ or $x=\pi$ .
2	$1-t^2+2t=0$ , so $t^2-2t-1=0$ $t=1\pm\sqrt2$ $\tan\tfrac{3\pi}{8}=1+\sqrt2$ gives $x=\tfrac{3\pi}{4}$ $\tan\left(-\tfrac{\pi}{8}\right)=1-\sqrt2$ gives $x=\tfrac{7\pi}{4}$	5
(5 marks)		5
Notes
Clearing the denominator gives $1-t^2+2t=0$ , so $t^2-2t-1=0$ and $t=1\pm\sqrt2$ . Since $\tan\tfrac{\pi}{8}=\sqrt2-1$ and $\tan\tfrac{3\pi}{8}=\sqrt2+1$ , the two half-angles are $\tfrac{3\pi}{8}$ and $-\tfrac{\pi}{8}$ , so $x=\tfrac{3\pi}{4}$ or $x=2\pi-\tfrac{\pi}{4}=\tfrac{7\pi}{4}$ . Independent check: the equation is $\tan x=-1$ , whose roots in $[0,2\pi)$ are exactly $\tfrac{3\pi}{4}$ and $\tfrac{7\pi}{4}$ .
3	$6t+4(1-t^2)=4(1+t^2)$ , so $8t^2-6t=0$ $t=0$ or $t=\tfrac34$ $x=0$ or $x=2\arctan\tfrac34=1.29$ $x=\pi$ gives $-4\ne4$ , so it is rejected	5
(5 marks)		5
Notes
The equation becomes $6t+4-4t^2=4+4t^2$ , so $8t^2=6t$ and $t=0$ or $t=0.75$ . Hence $x=0$ or $x=2\arctan0.75=1.2870$ , which is $1.29$ to $3$ significant figures. Independent check: $3\sin1.2870+4\cos1.2870=2.8800+1.1200=4.0000$ .

Tier 3 · Hard

Mark scheme for FP1-1.3 Tier 3 · Hard
Question	Scheme	Marks
1	$2(1-t^2)-2t=1+t^2$ , so $3t^2+2t-1=0$ $(3t-1)(t+1)=0$ , so $t=\tfrac13$ or $t=-1$ $t=\tfrac13$ gives $x=2\arctan\tfrac13=0.644$ $t=-1$ gives $\tfrac{x}{2}=\tfrac{3\pi}{4}$ , so $x=\tfrac{3\pi}{2}=4.71$ $x=\pi$ gives $-2\ne1$ , so it is rejected	6
(6 marks)		6
Notes
Substituting gives $2-2t^2-2t=1+t^2$ , so $3t^2+2t-1=0$ with roots $t=\tfrac13$ and $t=-1$ . Then $x=2\arctan\tfrac13=0.6435$ and, taking the half-angle in $\left(\tfrac{\pi}{2},\pi\right)$ so that $x$ lies in $[0,2\pi)$ , $t=-1$ gives $\tfrac{x}{2}=\tfrac{3\pi}{4}$ and $x=\tfrac{3\pi}{2}$ . Independent check: $2\cos0.6435-\sin0.6435=1.6000-0.6000=1$ , and $2\cos\tfrac{3\pi}{2}-\sin\tfrac{3\pi}{2}=0+1=1$ .
2	$\operatorname{cosec}x+\cot x=\dfrac{1+\cos x}{\sin x}=\dfrac1t$ $\dfrac1t=2$ , so $t=\tfrac12$ $x=2\arctan\tfrac12=0.927$ There is no second root because $\dfrac1t=2$ is linear in $t$	5
(5 marks)		5
Notes
Using the t-formulae, $\operatorname{cosec}x+\cot x=\dfrac{1+\cos x}{\sin x}=\dfrac1t$ , so the equation is simply $t=\tfrac12$ and $x=2\arctan0.5=0.9273$ , which is $0.927$ to $3$ significant figures. Independent check: $\operatorname{cosec}0.9273+\cot0.9273=1.2500+0.7500=2.0000$ .
3	$a(1-t^2)+2bt=c(1+t^2)$ after multiplying by $1+t^2$ Rearranging gives $(a+c)t^2-2bt+(c-a)=0$ Discriminant $=4b^2-4(a+c)(c-a)=4b^2-4(c^2-a^2)$ $=4\left(a^2+b^2-c^2\right)$ This is negative when $c^2>a^2+b^2$ , and the excluded value $x=\pi$ would require $c=-a$ , whence $c^2=a^2\le a^2+b^2$ ; so no solution exists	6
(6 marks)		6
Notes
Substitute the t-formulae and multiply by $1+t^2$ to clear denominators, then collect powers of $t$ . The discriminant of the quadratic is $(-2b)^2-4(a+c)(c-a)=4b^2-4\left(c^2-a^2\right)=4\left(a^2+b^2-c^2\right)$ , which is negative precisely when $c^2>a^2+b^2$ . The only root the substitution cannot see is $x=\pi$ , and that satisfies the original equation only if $-a=c$ , which forces $c^2=a^2\le a^2+b^2$ and so cannot happen under the stated condition. Independent check with the $R$ -method: $a\cos x+b\sin x=\sqrt{a^2+b^2}\cos(x-\alpha)$ has range $\left[-\sqrt{a^2+b^2},\sqrt{a^2+b^2}\right]$ , so $\|c\|>\sqrt{a^2+b^2}$ makes it unsolvable.
4	$(7+2)t^2-12t+(2-7)=0$ , that is $9t^2-12t-5=0$ $(3t-5)(3t+1)=0$ , so $t=\tfrac53$ or $t=-\tfrac13$ $t=\tfrac53$ gives $x=2\arctan\tfrac53=2.06$ $t=-\tfrac13$ gives $x=2\pi-2\arctan\tfrac13=5.64$ $x=\pi$ gives $-7\ne2$ , so it is rejected	6
(6 marks)		6
Notes
Using the standard reduction $(a+c)t^2-2bt+(c-a)=0$ with $a=7$ , $b=6$ , $c=2$ gives $9t^2-12t-5=0$ , whose discriminant is $144+180=324=18^2$ , so $t=\dfrac{12\pm18}{18}$ , that is $t=\tfrac53$ or $t=-\tfrac13$ . Hence $x=2\arctan\tfrac53=2.0608$ and $x=2\pi-2\arctan\tfrac13=5.6397$ . Independent check: $7\cos2.0608+6\sin2.0608=-3.2941+5.2941=2.0000$ and $7\cos5.6397+6\sin5.6397=5.6000-3.6000=2.0000$ .
5	$(3+5)t^2-8t+(5-3)=0$ , that is $8t^2-8t+2=0$ $4t^2-4t+1=0$ , so $(2t-1)^2=0$ and $t=\tfrac12$ is a repeated root $x=2\arctan\tfrac12=0.927$ The discriminant is $4\left(3^2+4^2-5^2\right)=0$ , so the quadratic has one repeated root Geometrically $c=\sqrt{a^2+b^2}$ , so the line meets the maximum of $5\cos(x-\alpha)$ exactly once in the interval	6
(6 marks)		6
Notes
The reduction gives $8t^2-8t+2=0$ , that is $4t^2-4t+1=(2t-1)^2=0$ , so $t=\tfrac12$ twice and $x=2\arctan0.5=0.9273$ . The general discriminant $4\left(a^2+b^2-c^2\right)$ is zero here because $3^2+4^2=25=5^2$ , the boundary case in which $c$ equals the amplitude. Independent check with the $R$ -method: $3\cos x+4\sin x=5\cos(x-0.9273)$ , which attains the value $5$ only at $x=0.9273$ in $[0,2\pi)$ .