5 Trigonometry — revision question pack

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

Checked against Edexcel 9MA0 section 5. Review basis: the qualification registry sourced from the Pearson Edexcel Level 3 Advanced GCE in Mathematics (9MA0) specification; registry verification recorded 11 July 2026.

How this checking works

5.1 · Understand and use the definitions of sine, cosine and tangent for all arguments; the sine and cosine rules; the area of a triangle in the form ½ab sin C; work with radian measure, including use for arc length and area of sector.

Explanation

For a unit-circle angle $\theta$ , $\cos\theta$ and $\sin\theta$ are the point's horizontal and vertical coordinates, while $\tan\theta=\sin\theta/\cos\theta$ where $\cos\theta\neq0$ ; this extends the ratios beyond acute angles.
For a triangle, use $a/\sin A=b/\sin B=c/\sin C$ when an opposite side-angle pair is known, $a^2=b^2+c^2-2bc\cos A$ for three sides or an included angle, and $\tfrac12ab\sin C$ for area.
In the ambiguous sine-rule case, test the supplementary angle because $\sin A=\sin(\pi-A)$ , then reject any triangle inconsistent with the given sides and angles.
Radian measure makes arc and sector formulae direct: an angle $\theta$ radians in a circle of radius $r$ gives arc length $s=r\theta$ and sector area $A=\tfrac12r^2\theta$ .
Keep the calculator in the required angle mode and label sides opposite their matching angles; using degrees in $s=r\theta$ or pairing the wrong side and angle is a common error.

A labelled triangle: each lower-case side is opposite its matching upper-case angle.

Worked example

Two sides of a triangular sail are $8\,\text{m}$ and $11\,\text{m}$ , with included angle $0.9$ radians. Calculate the third side and the area of the sail, giving each answer to $3$ significant figures.

1.The cosine rule gives $c^2=8^2+11^2-2(8)(11)\cos(0.9)=75.597\ldots$ , so $c=8.694\ldots\,\text{m}=8.69\,\text{m}$ .
2.The area is $\tfrac12(8)(11)\sin(0.9)=34.466\ldots\,\text{m}^2$ , hence $34.5\,\text{m}^2$ .

Answer: Third side $=8.69\,\text{m}$ Area $=34.5\,\text{m}^2$

Common mistakes

Don't use degrees in $s=r\theta$ or $A=\tfrac12r^2\theta$ , even though these formulae require radians.
Don't pair a side with a non-opposite angle in the sine rule.

Exam tip

Label each side opposite its matching angle and check the calculator angle mode before substituting.

Tier 1 · Easy

1.
A circular arc has radius $7.5\,\text{cm}$ and subtends $1.2$ radians at the centre. Find its length.
(2)
(Total for Question 1 is 2 marks)
2.
Two sides of a triangle have lengths $10\,\text{cm}$ and $13\,\text{cm}$ . The included angle is $30^\circ$ . Find the area of the triangle.
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1.
A sector of a circle has radius $7\,\text{cm}$ and arc length $11.2\,\text{cm}$ . Find the angle of the sector in radians and its area.
(3)
(Total for Question 1 is 3 marks)
2.
In triangle $ABC$ , side $a$ has length $6\,\text{cm}$ , $A=30^\circ$ and $B=45^\circ$ . Find the exact perimeter of the triangle.
(4)
(Total for Question 2 is 4 marks)
3.
In triangle $ABC$ , sides $a$ and $b$ are opposite angles $A$ and $B$ respectively. Given that $a=7\,\text{cm}$ , $b=10\,\text{cm}$ and $A=30^\circ$ , find the two possible values of angle $B$ and the corresponding areas of triangle $ABC$ . Give the angles to $1$ decimal place and the areas to $3$ significant figures. Use unrounded values in your working.
(5)
(Total for Question 3 is 5 marks)

Tier 3 · Hard

1.
A minor segment is cut from a circle of radius $6\,\text{cm}$ by a chord whose endpoints subtend $1.4$ radians at the centre. Determine the perimeter and area of the segment, giving both to $3$ significant figures.
(5)
(Total for Question 1 is 5 marks)
2.
Two concentric circles and two radial lines enclose an annular sector of angle $1.2$ radians. The outer radius is $3\,\text{cm}$ greater than the inner radius, and the area of the annular sector is $54\,\text{cm}^2$ . Find both radii and the perimeter of the annular sector.
(5)
(Total for Question 2 is 5 marks)
3.
A sector of a circle has perimeter $20\,\text{cm}$ and area $24\,\text{cm}^2$ . Find all possible values of its radius and its angle in radians.
(5)
(Total for Question 3 is 5 marks)
4.
Two sides of a triangle have lengths $8\,\text{cm}$ and $13\,\text{cm}$ , and its area is $26\,\text{cm}^2$ . Find the two possible included angles. For each angle, find the exact length of the third side.
(5)
(Total for Question 4 is 5 marks)
5.
A chord of a circle has length $16\,\text{cm}$ . The perpendicular distance from the centre of the circle to the chord is $6\,\text{cm}$ . Calculate the perimeter and area of the major segment cut off by the chord, giving each answer to $3$ significant figures. Use unrounded values in your working.
(6)
(Total for Question 5 is 6 marks)

5.2 · Understand and use the standard small angle approximations of sine, cosine and tangent: sin θ ≈ θ, cos θ ≈ 1 − θ²/2, tan θ ≈ θ.

Explanation

For $|\theta|$ close to zero and measured in radians, $\sin\theta\approx\theta$ , $\tan\theta\approx\theta$ and $\cos\theta\approx1-\theta^2/2$ . Replace each trigonometric function by its stated approximation, simplify algebraically, and retain only a solution whose magnitude is small enough for the approximation to be credible.
For example, $1-\cos\theta\approx\theta^2/2$ , so $(1-\cos\theta)/\theta^2\approx1/2$ for a small non-zero $\theta$ .
The approximations are radian results, not degree results; another common error is to accept a large root created by the approximate polynomial.
The approximation sign matters: these are local approximations near zero, not identities.
Quote the approximation used, substitute the small radian angle, and reject any root whose magnitude is inconsistent with the small-angle assumption.

Worked example

Without using a calculator's trigonometric keys, estimate $\dfrac{1-\cos(0.08)}{(0.08)^2}$ by a small-angle approximation.

1.Use $\cos\theta\approx1-\theta^2/2$ .
2.Then $1-\cos(0.08)\approx(0.08)^2/2$ , so division by $(0.08)^2$ gives $1/2=0.5$ .

Answer: $0.5$

Common mistakes

Don't apply the small-angle approximations to an angle measured in degrees.
Don't accept a large algebraic root even though the approximation is valid only near zero.

Exam tip

State that the angle is in radians and reject any solution that is not small.

Tier 1 · Easy

1.
Use a standard small-angle approximation to estimate $\sin(0.064)$ .
(1)
(Total for Question 1 is 1 mark)
2.
For $x=0.06$ radians, estimate $\cos x$ using the appropriate small-angle approximation.
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1.
Use small-angle approximations to estimate the non-zero small positive solution of $\sin x=5(1-\cos x)$ .
(3)
(Total for Question 1 is 3 marks)
2.
A pendulum of length $2.5\,\text{m}$ is displaced through $0.12$ radians from the downward vertical. By considering the vertical component of the length, derive an expression for its vertical rise $h$ . Then use a standard small-angle approximation to estimate $h$ in centimetres, giving your answer to $3$ significant figures.
(3)
(Total for Question 2 is 3 marks)
3.
For $x=0.05$ radians, use standard small-angle approximations to estimate $\dfrac{3\sin(2x)-2\tan x}{1-\cos x}$ .
(3)
(Total for Question 3 is 3 marks)

Tier 3 · Hard

1.
A small positive angle $x$ satisfies $\sin x+\cos x=1.08$ . Use the standard small-angle approximations to estimate $x$ , giving $4$ decimal places, and explain which algebraic root is admissible.
(4)
(Total for Question 1 is 4 marks)
2.
A sensor hangs from a straight cable of length $24\,\text{m}$ . The cable makes a small positive angle $x$ radians with the downward vertical, and the sensor is $23.88\,\text{m}$ vertically below the support. Use standard small-angle approximations to estimate $x$ and the sensor's horizontal displacement from the support.
(4)
(Total for Question 2 is 4 marks)
3.
For small $x$ , $F(x)=a\sin x+b\cos x+\tan(2x)$ , where $a$ and $b$ are positive constants. It is known that $F(x)\approx5+7x-\dfrac52x^2$ . Use standard small-angle approximations to find $a$ and $b$ . Hence estimate the smaller positive solution of $F(x)=5.24$ , giving your answer to $3$ significant figures.
(5)
(Total for Question 3 is 5 marks)
4.
A rigid boom of length $26\,\text{m}$ is raised at an angle of $0.18$ radians above the horizontal. Use standard small-angle approximations to estimate the horizontal and vertical coordinates of its end relative to the pivot, giving each coordinate to $4$ significant figures. Calculate the corresponding coordinates using trigonometric functions, and show that the distance between the approximate and calculated positions is less than $2.6\,\text{cm}$ .
(5)
(Total for Question 4 is 5 marks)
5.
A shallow circular arc has chord length $30\,\text{m}$ and maximum height $0.60\,\text{m}$ above the chord. The chord subtends a small angle $2u$ radians at the centre. By resolving the radius along and perpendicular to the chord, show that $15=r\sin u$ and $0.60=r(1-\cos u)$ . Use standard small-angle approximations to obtain the values of $u$ , $r$ and $2u$ predicted by the approximation model; in particular, report the approximate value of $r$ , not the value from exact circle geometry.
(6)
(Total for Question 5 is 6 marks)

5.3 · Understand and use the sine, cosine and tangent functions; their graphs, symmetries and periodicity; know and use exact values of sin, cos and tan for standard angles and their multiples.

Explanation

$\sin x$ and $\cos x$ have range $[-1,1]$ and period $2\pi$ , while $\tan x$ has range $\mathbb{R}$ , period $\pi$ and vertical asymptotes at $x=\pi/2+k\pi$ .
Use $\sin(-x)=-\sin x$ , $\cos(-x)=\cos x$ and $\tan(-x)=-\tan x$ , then reduce an angle by a whole period before using its reference angle and quadrant.
Know exact sine and cosine values at $0$ , $\pi/6$ , $\pi/4$ , $\pi/3$ , $\pi/2$ and $\pi$ , and exact tangent values where defined.
Transformations such as $y=\cos(x+\pi/6)$ shift the graph, while $y=\tan2x$ halves its period.
In $a\cos(bx)+c$ , the amplitude is $|a|$ , period $2\pi/|b|$ , and $c$ moves the midline.

Two cycles of

y=\cos x

, showing range

[-1,1]

and period

2\pi

Worked example

For $y=3\cos(2x)-1$ , state the amplitude, period, maximum value and minimum value.

1.The coefficient outside cosine gives amplitude $|3|=3$ .
2.The factor $2$ inside gives period $2\pi/2=\pi$ .
3.Since $-1\leq\cos(2x)\leq1$ , multiplying by $3$ and subtracting $1$ gives $-4\leq y\leq2$ .

Answer: Amplitude $3$ Period $\pi$ Maximum $2$ Minimum $-4$

Common mistakes

Don't use the coefficient outside the trigonometric function to calculate the period instead of the amplitude.
Don't give tangent a period of $2\pi$ or omit its vertical asymptotes.

Exam tip

On a graph question, state amplitude, period, midline and range separately before sketching one complete cycle.

Tier 1 · Easy

1.
Find the exact value of $\tan(5\pi/6)$ .
(2)
(Total for Question 1 is 2 marks)
2.
Find the exact value of $\cos(-7\pi/3)$ .
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1.
Find the exact values of $\sin\left(\dfrac{7\pi}{6}\right)$ , $\cos\left(\dfrac{5\pi}{3}\right)$ and $\tan\left(\dfrac{3\pi}{4}\right)$ .
(3)
(Total for Question 1 is 3 marks)
2.
For $y=2\sin(3x)-4$ , state the amplitude and period, and find the exact coordinates of every maximum point for $0\leq x\leq2\pi$ .
(4)
(Total for Question 2 is 4 marks)
3.
For $y=\tan(2x-\pi/4)$ , state the period and find all zeros and vertical asymptotes for $0\leq x\leq\pi$ .
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
Evaluate exactly $\sin(-11\pi/6)$ , $\cos(13\pi/3)$ and $\tan(7\pi/4)$ .
(4)
(Total for Question 1 is 4 marks)
2.
The function $f(x)=a\cos(bx)+c$ , where $b>0$ , has maximum value $7$ , minimum value $-3$ and least positive period $4\pi/3$ . Given that $f(0)=-3$ , find $a$ , $b$ and $c$ .
(5)
(Total for Question 2 is 5 marks)
3.
Let $f(x)=\sin^2x$ . Without using a double-angle identity, show that $f$ is even and has period $\pi$ . State its range and give the exact coordinates of every maximum and minimum point for $0\leq x\leq2\pi$ .
(5)
(Total for Question 3 is 5 marks)
4.
For $f(x)=|2\cos x-1|$ , state the least positive period and the range. Find the exact coordinates of every zero and every global maximum point for $0\leq x\leq2\pi$ .
(5)
(Total for Question 4 is 5 marks)
5.
For integers $n\geq0$ , define $u_n=2\cos(n\pi/3)+\sin(n\pi/2)$ . Make a table of the exact values of $u_n$ for $0\leq n\leq11$ . Hence state the least positive integer period of the sequence and find how many integers $n$ with $0\leq n\leq59$ satisfy $u_n\leq-1$ .
(6)
(Total for Question 5 is 6 marks)

5.4 · Understand and use the definitions of secant, cosecant and cotangent and of arcsin, arccos and arctan; their relationships to sine, cosine and tangent; understanding of their graphs; their ranges and domains.

Explanation

The reciprocal functions are $\sec x=1/\cos x$ , $\cosec x=1/\sin x$ and $\cot x=\cos x/\sin x$ ; their graphs inherit zeros of the denominator as vertical asymptotes. Secant and cosecant have range $(-\infty,-1]\cup[1,\infty)$ and period $2\pi$ , while cotangent has range $\mathbb{R}$ and period $\pi$ .
The inverse graphs are restrictions reflected in $y=x$ .
Their principal ranges are $-\pi/2\leq\arcsin x\leq\pi/2$ , $0\leq\arccos x\leq\pi$ and $-\pi/2<\arctan x<\pi/2$ ; $\arcsin x$ and $\arccos x$ require $-1\leq x\leq1$ .
Angles may be in degrees or radians.
The notation $\sin^{-1}x$ means $\arcsin x$ , not the reciprocal $\cosec x$ .

Worked example

Give the principal values, in radians, of $\arcsin(-1/2)$ and $\arctan(-1)$ .

1.Within the principal sine-inverse range, $\sin(-\pi/6)=-1/2$ , so $\arcsin(-1/2)=-\pi/6$ .
2.Within the principal tangent-inverse range, $\tan(-\pi/4)=-1$ , so $\arctan(-1)=-\pi/4$ .

Answer: $\arcsin(-1/2)=-\pi/6$ $\arctan(-1)=-\pi/4$

Common mistakes

Don't read $\sin^{-1}x$ as the reciprocal $1/\sin x$ instead of the inverse function $\arcsin x$ .
Don't give an inverse-trigonometric answer outside the stated principal range.

Exam tip

Check the input domain and principal output range before giving an inverse-trigonometric value.

Tier 1 · Easy

1.
Given $\cos\theta=-4/5$ , write down $\sec\theta$ .
(1)
(Total for Question 1 is 1 mark)
2.
Find the principal value, in radians, of $\arccos(-\sqrt3/2)$ .
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1.
For $0\leq x\leq2\pi$ , state where $\sec x$ is undefined and solve $\sec x=-2$ .
(4)
(Total for Question 1 is 4 marks)
2.
Let $f(x)=\arcsin(2x-1)$ . State the domain and range of $f$ , and solve $f(x)=\pi/6$ .
(4)
(Total for Question 2 is 4 marks)
3.
Let $f(x)=\arccos(1/x)$ . State the domain and range of $f$ , and solve $f(x)=2\pi/3$ .
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
For $y=2\sec x-1$ , state the period and range, then give all vertical asymptotes in $-\pi\leq x\leq\pi$ .
(5)
(Total for Question 1 is 5 marks)
2.
Show that $\arcsin x+\arccos x=\pi/2$ for $-1\leq x\leq1$ . Hence solve $3\arcsin x-\arccos x=0$ exactly.
(5)
(Total for Question 2 is 5 marks)
3.
For $0\leq x\leq3\pi$ , express $f(x)=\arcsin(\sin x)$ as a piecewise function. State the range of $f$ on this interval and hence solve $f(x)=\pi/6$ .
(6)
(Total for Question 3 is 6 marks)
4.
For $x\neq0$ , determine the value of $\arctan x+\arctan(1/x)$ separately for $x>0$ and $x<0$ . Hence solve $\arctan x+\arctan(1/x)=-\pi/2$ .
(5)
(Total for Question 4 is 5 marks)
5.
For $f(x)=3-2\cosec x$ , find the exact coordinates of the maximum point on $0<x<\pi$ and the minimum point on $\pi<x<2\pi$ . Hence solve $f(x)=7$ for $0<x<2\pi$ .
(5)
(Total for Question 5 is 5 marks)

5.5 · Understand and use tan θ = sin θ / cos θ; understand and use sin²θ + cos²θ = 1, sec²θ = 1 + tan²θ and cosec²θ = 1 + cot²θ.

Explanation

The quotient identity is $\tan\theta=\sin\theta/\cos\theta$ , and the three Pythagorean identities are $\sin^2\theta+\cos^2\theta=1$ , $\sec^2\theta=1+\tan^2\theta$ and $\cosec^2\theta=1+\cot^2\theta$ .
Choose an identity containing the known and required functions, rearrange it, and use the stated quadrant to select the correct sign after taking a square root.
If $\tan\theta=-7/24$ in quadrant II, a reference triangle has side magnitudes $7$ , $24$ and $25$ , so $\sin\theta=7/25$ and $\cos\theta=-24/25$ .
From $\sin^2\theta+\cos^2\theta=1$ one obtains two possible signs; ignoring the quadrant or silently choosing the positive root can change every reciprocal ratio that follows.

Worked example

Given that $\tan\theta=-7/24$ and $\pi/2<\theta<\pi$ , determine $\sin\theta$ , $\cos\theta$ and $\cosec\theta$ .

1.Use $\sec^2\theta=1+\tan^2\theta=1+49/576=625/576$ .
2.In quadrant II cosine and secant are negative, so $\sec\theta=-25/24$ and $\cos\theta=-24/25$ .
3.Then $\sin\theta=\tan\theta\cos\theta=(-7/24)(-24/25)=7/25$ , giving $\cosec\theta=25/7$ .

Answer: $\sin\theta=7/25$ $\cos\theta=-24/25$ $\cosec\theta=25/7$

Common mistakes

Don't treat $\cosec\theta$ as $\sin\theta$ rather than its reciprocal.
Don't use $\tan\theta=\cos\theta/\sin\theta$ instead of $\sin\theta/\cos\theta$ .

Exam tip

Use the interval to fix the signs of sine and cosine before taking reciprocals.

Tier 1 · Easy

1.
An acute angle $\theta$ satisfies $\sin\theta=5/13$ . Find $\tan\theta$ and $\sec\theta$ exactly.
(3)
(Total for Question 1 is 3 marks)
2.
Given that $\cot\theta=-3/4$ and $3\pi/2<\theta<2\pi$ , find $\cosec\theta$ exactly.
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1.
Prove that $\dfrac{1}{1-\sin x}+\dfrac{1}{1+\sin x}=2\sec^2x$ for values of $x$ for which both sides are defined.
(4)
(Total for Question 1 is 4 marks)
2.
Simplify $\dfrac{\sec x-\cos x}{\tan x}$ , for values of $x$ for which the original expression is defined.
(3)
(Total for Question 2 is 3 marks)
3.
Given that $\tan\theta=-2$ and $3\pi/2<\theta<2\pi$ , find the exact value of $\dfrac{3\sin\theta+\cos\theta}{\sin\theta-2\cos\theta}$ without finding $\theta$ .
(3)
(Total for Question 3 is 3 marks)

Tier 3 · Hard

1.
An angle $\theta$ lies in the first quadrant and satisfies $\sec\theta+\tan\theta=5$ . Find $\sin\theta$ exactly.
(5)
(Total for Question 1 is 5 marks)
2.
Given that $\sin\theta+\cos\theta=7/5$ and $0<\theta<\pi/4$ , find $\tan\theta$ exactly.
(5)
(Total for Question 2 is 5 marks)
3.
An acute angle $\theta$ satisfies $2\sec\theta+\tan\theta=4$ . Find $\sin\theta$ exactly.
(5)
(Total for Question 3 is 5 marks)
4.
An angle $\theta$ satisfies $\pi/2<\theta<\pi$ and $\dfrac{2\sin\theta+\cos\theta}{\sin\theta-2\cos\theta}=\dfrac{2}{11}$ . Find $\cosec\theta-\cot\theta$ exactly.
(5)
(Total for Question 4 is 5 marks)
5.
Prove that $\sec^2x+\cosec^2x\geq4$ wherever the expression is defined. Find all values of $x$ for which equality holds in the interval $0\leq x<2\pi$ .
(5)
(Total for Question 5 is 5 marks)

5.6 · Understand and use double angle formulae; formulae for sin(A ± B), cos(A ± B), tan(A ± B) with geometrical proofs; express a cos θ + b sin θ in the form r cos(θ ± α) or r sin(θ ± α).

Explanation

The compound-angle formulae are $\sin(A\pm B)=\sin A\cos B\pm\cos A\sin B$ , $\cos(A\pm B)=\cos A\cos B\mp\sin A\sin B$ and $\tan(A\pm B)=(\tan A\pm\tan B)/(1\mp\tan A\tan B)$ ; a geometrical proof can calculate the same unit-circle chord by coordinates and by the cosine rule.
Setting $A=B=\theta$ gives $\sin2\theta=2\sin\theta\cos\theta$ , $\cos2\theta=\cos^2\theta-\sin^2\theta=1-2\sin^2\theta=2\cos^2\theta-1$ and $\tan2\theta=2\tan\theta/(1-\tan^2\theta)$ .
Rearranging a double-angle identity also gives half-angle results, for example $\sin^2\theta=(1-\cos2\theta)/2$ .
To write $a\cos\theta+b\sin\theta=R\cos(\theta-\alpha)$ , compare coefficients to obtain $R\cos\alpha=a$ , $R\sin\alpha=b$ , hence $R=\sqrt{a^2+b^2}$ with the quadrant of $\alpha$ set by the signs.
For a compound-angle expression, the sign in the cosine formula reverses; for an $R$ -form, expanding the proposed form before choosing $\alpha$ prevents a wrong sign.

Worked example

Express $5\cos\theta-12\sin\theta$ as $R\cos(\theta+\alpha)$ , where $R>0$ and $0<\alpha<\pi/2$ . Hence state its maximum and minimum values.

1.Expand $R\cos(\theta+\alpha)=R\cos\theta\cos\alpha-R\sin\theta\sin\alpha$ .
2.Comparing coefficients gives $R\cos\alpha=5$ and $R\sin\alpha=12$ , so $R=\sqrt{25+144}=13$ and $\tan\alpha=12/5$ .
3.Since cosine ranges from $-1$ to $1$ , the expression ranges from $-13$ to $13$ .

Answer: $13\cos(\theta+\alpha)$ , where $\alpha=\arctan(12/5)$ Maximum $13$ Minimum $-13$

Common mistakes

Don't use the same sign in $\cos(A\pm B)$ instead of reversing it in the expansion.
Don't choose the sign of $\alpha$ in an $R$ -form without expanding and comparing both coefficients.

Exam tip

Expand the proposed $R$ -form, compare coefficients, then state $R>0$ and the required range for $\alpha$ .

Tier 1 · Easy

1.
Use a compound-angle formula to find the exact value of $\sin(75^\circ)$ .
(3)
(Total for Question 1 is 3 marks)
2.
By writing $15^\circ$ as the difference of two standard angles, find $\tan15^\circ$ exactly.
(3)
(Total for Question 2 is 3 marks)

Tier 2 · Standard

1.
Given that $\sin\theta=\dfrac35$ and $0<\theta<\dfrac{\pi}{2}$ , find the exact values of $\sin2\theta$ , $\cos2\theta$ and $\tan2\theta$ .
(4)
(Total for Question 1 is 4 marks)
2.
Write $7\cos\theta+24\sin\theta$ as the single sine expression $R\sin(\theta+\alpha)$ , choosing $R$ positive and $\alpha$ acute. Hence find the exact range of $f(\theta)=\dfrac{1}{26+7\cos\theta+24\sin\theta}$ .
(4)
(Total for Question 2 is 4 marks)
3.
Express $5\cos^2x-2\sin^2x$ in the form $a+b\cos(2x)$ , where $a$ and $b$ are constants. Hence state its maximum and minimum values.
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
Use two points on the unit circle and the cosine rule to prove geometrically that $\cos(A-B)=\cos A\cos B+\sin A\sin B$ .
(5)
(Total for Question 1 is 5 marks)
2.
Starting from the addition formulae for sine and cosine, derive $\tan(A+B)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$ for values of $A$ and $B$ for which the expressions are defined.
(4)
(Total for Question 2 is 4 marks)
3.
Using double-angle formulae, prove that $\tan(\theta/2)=\dfrac{\sin\theta}{1+\cos\theta}$ for values of $\theta$ for which both sides are defined. Hence find $\tan(\pi/8)$ exactly.
(5)
(Total for Question 3 is 5 marks)
4.
Given that $\cos(2\theta)=-7/25$ and $\pi/2<\theta<3\pi/4$ , find $\sin\theta$ and $\cos\theta$ exactly. Hence find the exact value of $\sin(\theta+\pi/4)$ .
(5)
(Total for Question 4 is 5 marks)
5.
Show, by repeated use of the double-angle formula for sine, that $\sin(8x)=8\sin x\cos x\cos(2x)\cos(4x)$ . Hence find the exact value of $\cos20^\circ\cos40^\circ\cos80^\circ$ .
(5)
(Total for Question 5 is 5 marks)

5.7 · Solve simple trigonometric equations in a given interval, including quadratic equations in sin, cos and tan and equations involving multiples of the unknown angle.

Explanation

Solve first for the trigonometric ratio, use a reference angle and the signs in each quadrant, then list only solutions in the stated interval.
For a quadratic in one trigonometric function, substitute a temporary variable, factorise or use the quadratic formula, and reject any ratio outside its possible range before solving each remaining branch.
When the equation involves $kx$ , transform the given interval for $x$ into the corresponding interval for $kx$ , find every solution there, and divide only at the end.
Inverse-trigonometric buttons return a principal value rather than the full solution set; endpoints, excluded endpoints and degree-versus-radian mode must all be checked explicitly.

Worked example

Determine all $x$ satisfying $2\cos^2x-3\cos x+1=0$ for $0\leq x<2\pi$ .

1.Factorise to $(2\cos x-1)(\cos x-1)=0$ .
2.Hence $\cos x=1/2$ or $\cos x=1$ .
3.In the interval, $\cos x=1/2$ at $x=\pi/3$ and $5\pi/3$ , while $\cos x=1$ at $x=0$ ; $2\pi$ is excluded.

Answer: $x=0,\ \pi/3,\ 5\pi/3$

Common mistakes

Don't give only the principal value returned by the inverse-trigonometric button.
Don't include an excluded endpoint or miss solutions created by the multiple angle.

Exam tip

Solve for the trig value first, then use symmetry and periodicity to list every solution in the stated interval.

Tier 1 · Easy

1.
Solve $\sin\theta=0.4$ for $0\leq\theta\leq2\pi$ , giving solutions to $3$ decimal places.
(3)
(Total for Question 1 is 3 marks)
2.
Solve $\cos x=-\dfrac{\sqrt2}{2}$ for $0\leq x<2\pi$ .
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1.
Solve $\sin(2x)=\dfrac{\sqrt3}{2}$ for $0\leq x\leq2\pi$ .
(4)
(Total for Question 1 is 4 marks)
2.
Solve $2\sin^2x+\sin x-1=0$ for $0\leq x<2\pi$ .
(4)
(Total for Question 2 is 4 marks)
3.
Solve $2\sin x\cos x=\sin x$ for $0\leq x<2\pi$ .
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
Find every solution of $\tan(2x)=-\sqrt3$ in the interval $-\pi/2\leq x\leq\pi$ .
(5)
(Total for Question 1 is 5 marks)
2.
Solve $4\sin^2(3x)-1=0$ for $-\pi/2\leq x\leq2\pi/3$ . Show the complete set of values of $3x$ in the corresponding interval.
(6)
(Total for Question 2 is 6 marks)
3.
Determine the positive constant $R$ and acute angle $\alpha$ such that $3\sin x+4\cos x=R\sin(x+\alpha)$ . Hence solve $3\sin x+4\cos x=2$ for $-\pi\leq x\leq\pi$ , giving solutions to $3$ decimal places. Use unrounded values in your working.
(5)
(Total for Question 3 is 5 marks)
4.
Solve $\cos(2x)=2\sin x$ for $0\leq x<2\pi$ , giving all solutions exactly.
(5)
(Total for Question 4 is 5 marks)
5.
Determine all real values of $k$ for which $\sin^2x-\sin x=k$ has exactly three distinct solutions in the interval $0\leq x<2\pi$ .
(6)
(Total for Question 5 is 6 marks)

5.8 · Construct proofs involving trigonometric functions and identities.

Explanation

A trigonometric identity is true for every value in its domain, so a proof transforms one side into the other using exact algebra and established identities rather than testing selected angles.
Usually begin with the more complicated side, replace secant, cosecant, cotangent or tangent by sine and cosine when helpful, and factor or take a common denominator before cancelling.
Multiplying numerator and denominator by a conjugate can expose $1-\sin^2x=\cos^2x$ or $1-\cos^2x=\sin^2x$ and complete the proof cleanly.
Never cancel terms across addition, and record domain restrictions: algebra such as division by $\sin x$ is valid only where that denominator is non-zero.

Worked example

Prove that $\dfrac{1-\cos(2x)}{\sin(2x)}=\tan x$ for values at which both sides are defined.

1.Apply the double-angle forms to the left-hand side: $(1-\cos2x)/\sin2x=(2\sin^2x)/(2\sin x\cos x)=\sin x/\cos x=\tan x$ , with cancellation only where the original expressions are defined.

Answer: Use $1-\cos(2x)=2\sin^2x$ and $\sin(2x)=2\sin x\cos x$ .

Common mistakes

Don't cancel a term across addition while manipulating a trigonometric fraction.
Don't divide by $\sin x$ or $\cos x$ without recording where that factor is zero.

Exam tip

For “prove”, transform one side only with named identities until it exactly matches the other side.

Tier 1 · Easy

1.
Prove that $\sin x\cot x=\cos x$ wherever the left-hand side is defined.
(2)
(Total for Question 1 is 2 marks)
2.
Prove that $(1+\cot^2x)\sin^2x=1$ wherever the left-hand side is defined.
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1.
Prove that $\cos(A+B)\cos(A-B)=\cos^2A-\sin^2B$ .
(4)
(Total for Question 1 is 4 marks)
2.
Prove that $\dfrac{\tan x}{\sec x-1}=\dfrac{1+\cos x}{\sin x}$ wherever both sides are defined.
(4)
(Total for Question 2 is 4 marks)
3.
Prove that $\dfrac{1+\sin x}{\cos x}+\dfrac{\cos x}{1+\sin x}=2\sec x$ wherever both sides are defined.
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
Prove that $\dfrac{1}{1-\sin x}-\dfrac{1}{1+\sin x}=2\tan x\sec x$ wherever the expressions exist.
(5)
(Total for Question 1 is 5 marks)
2.
Prove that $(\sin x+\cos x)^4-(\sin x-\cos x)^4=4\sin2x$ .
(4)
(Total for Question 2 is 4 marks)
3.
Prove that $\sec^2x+\cosec^2x=(\tan x+\cot x)^2=4\cosec^2(2x)$ wherever the expressions are defined.
(5)
(Total for Question 3 is 5 marks)
4.
Prove that $\sin^6x+\cos^6x=1-\dfrac34\sin^2(2x)$ . Hence find the exact value of $\sin^6(\pi/8)+\cos^6(\pi/8)$ .
(5)
(Total for Question 4 is 5 marks)
5.
Prove that $(\sec x+\tan x)^2=\dfrac{1+\sin x}{1-\sin x}$ wherever both sides are defined. Hence solve $\sec x+\tan x=\sqrt3$ for $-\pi<x<\pi$ .
(6)
(Total for Question 5 is 6 marks)

5.9 · Use trigonometric functions to solve problems in context, including problems involving vectors, kinematics and forces.

Explanation

Resolve a vector of magnitude $V$ at angle $\theta$ to the positive horizontal into components $V\cos\theta$ and $V\sin\theta$ , changing signs to match its actual direction.
For resultant or equilibrium problems, form separate equations in two perpendicular directions; a zero resultant requires both component sums to equal zero.
Trigonometric models can describe wave motion, a point on a vertical circular wheel or changing hours of sunlight.
Interpret every solution using the stated time interval, units and physical constraints.
A calculator angle without a quadrant check can point in the opposite direction; draw and label a diagram, then state bearings or directions in the form the context requests.

Perpendicular components combine to give a resultant whose direction is measured from east.

Worked example

Two horizontal forces act on a crate: $8\,\text{N}$ due east and $11\,\text{N}$ at $60^\circ$ north of east. Find the magnitude and direction of their resultant, to $3$ significant figures and the nearest degree respectively.

1.The east component is $8+11\cos60^\circ=13.5\,\text{N}$ and the north component is $11\sin60^\circ=9.526\ldots\,\text{N}$ .
2.Hence $R=\sqrt{13.5^2+9.526^2}=16.523\ldots\,\text{N}$ .
3.Its direction is $\arctan(9.526/13.5)=35.21\ldots^\circ$ north of east.

Answer: Magnitude $=16.5\,\text{N}$ Direction $35^\circ$ north of east

Common mistakes

Don't use an inverse-tangent calculator value without checking the resultant's quadrant.
Don't state a bare angle without the bearing or directional wording required by the context.

Exam tip

Draw and label component directions, then give the final magnitude with units and the direction in contextual form.

Tier 1 · Easy

1.
A drone travels at $12\,\text{m s}^{-1}$ on a path $35^\circ$ above the horizontal. Calculate its horizontal and vertical velocity components to $3$ significant figures.
(3)
(Total for Question 1 is 3 marks)
2.
The height of a marker above the floor is modelled by $h=5+3\sin(\pi t/6)$ metres, where $t$ is measured in seconds. Find the exact height when $t=2$ .
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1.
A hiker walks $24\,\text{km}$ on a bearing of $040^\circ$ and then $18\,\text{km}$ on a bearing of $130^\circ$ . Find the hiker's distance and bearing from the starting point, giving the bearing to the nearest degree.
(5)
(Total for Question 1 is 5 marks)
2.
The position vector of a particle is $\mathbf r=(6\cos t)\mathbf i+(8\sin t)\mathbf j$ metres. When $t=\pi/3$ , find the particle's distance from the origin exactly and the direction of $\mathbf r$ , in degrees above the positive $x$ -axis, to $1$ decimal place.
(4)
(Total for Question 2 is 4 marks)
3.
Points $A$ and $B$ are $10\,\text{km}$ apart, with $B$ due east of $A$ . A buoy $C$ is on a bearing of $040^\circ$ from $A$ and a bearing of $310^\circ$ from $B$ . Find the distances $AC$ and $BC$ , giving each answer to $3$ significant figures.
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
A ring is in equilibrium under its weight of $18\,\text{N}$ , a tension $P$ directed $25^\circ$ above the horizontal, and a horizontal tension $Q$ acting oppositely. Calculate $P$ and $Q$ to $3$ significant figures.
(5)
(Total for Question 1 is 5 marks)
2.
An aircraft is required to have a ground velocity of $120\,\text{km h}^{-1}$ due north. A steady wind has velocity $30\,\text{km h}^{-1}$ due east. Find the constant airspeed and bearing that the pilot must use, giving the airspeed to $3$ significant figures and the bearing to the nearest degree.
(5)
(Total for Question 2 is 5 marks)
3.
A sign of weight $W$ newtons is held in equilibrium by two light cables. The left cable runs upwards to the left at $30^\circ$ above the horizontal and can withstand a maximum tension of $80\,\text{N}$ . The right cable runs upwards to the right at $60^\circ$ above the horizontal and can withstand a maximum tension of $96\,\text{N}$ . The sign is modelled as a particle. Find the greatest possible value of $W$ , giving your answer to $3$ significant figures, and state which cable is at its limiting tension. Use unrounded values in your working.
(6)
(Total for Question 3 is 6 marks)
4.
Three horizontal tugs act on a floating platform in equilibrium. One tug has magnitude $14\,\text{N}$ and bearing $030^\circ$ ; a second has magnitude $20\,\text{N}$ and bearing $150^\circ$ . Find the exact magnitude of the third tug and its bearing to the nearest degree.
(6)
(Total for Question 4 is 6 marks)
5.
During an $18$ -second test, a sensor reading is modelled by $S=70+30\cos\left(\dfrac{\pi(t-1)}{4}\right)$ for $0\leq t\leq18$ , where $t$ is measured in seconds. Find every time interval for which $S\geq85$ , and find the total time for which this condition holds.
(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.

5.1 · Understand and use the definitions of sine, cosine and tangent for all arguments; the sine and cosine rules; the area of a triangle in the form ½ab sin C; work with radian measure, including use for arc length and area of sector.

Tier 1 · Easy

Mark scheme for 5.1 Tier 1 · Easy
Question	Scheme	Marks
1	$9.0\,\text{cm}$	2
(2 marks)		2
Notes
Use $s=r\theta$ with the angle already in radians: $s=7.5(1.2)=9.0\,\text{cm}$ .
2	$32.5\,\text{cm}^2$	2
(2 marks)		2
Notes
Use $A=\tfrac12ab\sin C$ : $A=\tfrac12(10)(13)\sin30^\circ=65(1/2)=32.5\,\text{cm}^2$ .

Tier 2 · Standard

Mark scheme for 5.1 Tier 2 · Standard
Question	Scheme	Marks
1	Angle $=1.6$ radians Area $=39.2\,\text{cm}^2$	3
(3 marks)		3
Notes
Using $s=r\theta$ , the angle is $\theta=11.2/7=1.6$ radians. The sector area is $\tfrac12r^2\theta=\tfrac12(7^2)(1.6)=39.2\,\text{cm}^2$ .
2	$6+9\sqrt2+3\sqrt6\,\text{cm}$	4
(4 marks)		4
Notes
First $C=180^\circ-30^\circ-45^\circ=105^\circ$ . By the sine rule, $b=6\sin45^\circ/\sin30^\circ=6\sqrt2$ and $c=6\sin105^\circ/\sin30^\circ=3(\sqrt6+\sqrt2)$ . Therefore the perimeter is $6+6\sqrt2+3(\sqrt6+\sqrt2)=6+9\sqrt2+3\sqrt6\,\text{cm}$ .
3	$B=45.6^\circ$ , area $=33.9\,\text{cm}^2$ $B=134.4^\circ$ , area $=9.40\,\text{cm}^2$	5
(5 marks)		5
Notes
By the sine rule, $\sin B=10\sin30^\circ/7=5/7$ . Hence $B=45.584\ldots^\circ$ or $180^\circ-45.584\ldots^\circ=134.415\ldots^\circ$ . The corresponding values of $C$ are $104.415\ldots^\circ$ and $15.584\ldots^\circ$ . Using $\tfrac12ab\sin C$ , the areas are $\tfrac12(7)(10)\sin(104.415\ldots^\circ)=33.898\ldots\,\text{cm}^2$ and $\tfrac12(7)(10)\sin(15.584\ldots^\circ)=9.403\ldots\,\text{cm}^2$ .

Tier 3 · Hard

Mark scheme for 5.1 Tier 3 · Hard
Question	Scheme	Marks
1	Perimeter $=16.1\,\text{cm}$ Area $=7.46\,\text{cm}^2$	5
(5 marks)		5
Notes
The arc length is $6(1.4)=8.4\,\text{cm}$ . Splitting the isosceles triangle in half gives chord length $2(6)\sin(0.7)=7.730\ldots\,\text{cm}$ , so the perimeter is $16.130\ldots\,\text{cm}$ . The sector area is $\tfrac12(6^2)(1.4)=25.2\,\text{cm}^2$ and the triangle area is $\tfrac12(6^2)\sin(1.4)=17.738\ldots\,\text{cm}^2$ . Their difference is $7.461\ldots\,\text{cm}^2$ , giving the stated answers.
2	Inner radius $=13.5\,\text{cm}$ and outer radius $=16.5\,\text{cm}$ Perimeter $=42\,\text{cm}$	5
(5 marks)		5
Notes
Let the outer and inner radii be $R$ and $r$ . Then $R-r=3$ . The area difference is $\tfrac12(1.2)(R^2-r^2)=0.6(R-r)(R+r)=54$ , so $1.8(R+r)=54$ and $R+r=30$ . Solving gives $R=16.5$ and $r=13.5$ . The perimeter consists of both arcs and two radial lengths: $1.2R+1.2r+2(R-r)=1.2(30)+2(3)=42\,\text{cm}$ .
3	Radius $4\,\text{cm}$ and angle $3$ radians Radius $6\,\text{cm}$ and angle $4/3$ radians	5
(5 marks)		5
Notes
Let the radius be $r$ cm and the angle be $\theta$ radians. The perimeter gives $r\theta+2r=20$ , so $\theta=20/r-2$ . Substitution into $\tfrac12r^2\theta=24$ gives $10r-r^2=24$ , hence $(r-4)(r-6)=0$ . If $r=4$ , then $\theta=3$ ; if $r=6$ , then $\theta=4/3$ . Both radii and angles are positive, so both sectors are valid.
4	Included angle $30^\circ$ , third side $=\sqrt{233-104\sqrt3}\,\text{cm}$ Included angle $150^\circ$ , third side $=\sqrt{233+104\sqrt3}\,\text{cm}$	5
(5 marks)		5
Notes
If the included angle is $C$ , then $26=\tfrac12(8)(13)\sin C$ , so $\sin C=1/2$ . The two possible angles in a triangle are $C=30^\circ$ and $C=150^\circ$ . By the cosine rule, the third side $c$ satisfies $c^2=8^2+13^2-2(8)(13)\cos C$ . This gives $c^2=233-104\sqrt3$ when $C=30^\circ$ and $c^2=233+104\sqrt3$ when $C=150^\circ$ . Both values are positive and each angle, together with the two given sides, constructs a valid triangle.
5	Perimeter $=60.3\,\text{cm}$ Area $=269\,\text{cm}^2$	6
(6 marks)		6
Notes
The perpendicular from the centre bisects the chord, so the radius is $r=\sqrt{8^2+6^2}=10\,\text{cm}$ . If the minor central angle is $\theta$ , then $\theta=2\arctan(8/6)=1.854590\ldots$ radians. The major angle is $2\pi-\theta=4.428594\ldots$ radians, so the major-segment perimeter is $10(2\pi-\theta)+16=60.285948\ldots\,\text{cm}$ . The minor triangle has area $\tfrac12(10^2)\sin\theta=48\,\text{cm}^2$ . Hence the major-segment area is $\tfrac12(10^2)(2\pi-\theta)+48=269.429743\ldots\,\text{cm}^2$ , giving the stated answers.

5.2 · Understand and use the standard small angle approximations of sine, cosine and tangent: sin θ ≈ θ, cos θ ≈ 1 − θ²/2, tan θ ≈ θ.

Tier 1 · Easy

Mark scheme for 5.2 Tier 1 · Easy
Question	Scheme	Marks
1	$0.064$	1
(1 mark)		1
Notes
Since $0.064$ is a small angle in radians, use $\sin\theta\approx\theta$ to obtain $\sin(0.064)\approx0.064$ .
2	$0.9982$	2
(2 marks)		2
Notes
For a small angle in radians, $\cos x\approx1-x^2/2$ . Hence $\cos(0.06)\approx1-(0.06)^2/2=0.9982$ .

Tier 2 · Standard

Mark scheme for 5.2 Tier 2 · Standard
Question	Scheme	Marks
1	$x\approx0.4$ radians	3
(3 marks)		3
Notes
For small $x$ , use $\sin x\approx x$ and $1-\cos x\approx x^2/2$ . The equation becomes $x\approx5x^2/2$ . Since the required solution is non-zero, divide by $x$ to obtain $1\approx5x/2$ , hence $x\approx2/5=0.4$ radians.
2	$h=2.5(1-\cos(0.12))\,\text{m}$ $h\approx1.80\,\text{cm}$	3
(3 marks)		3
Notes
Initially the bob is $2.5\,\text{m}$ vertically below the pivot. After the displacement its vertical distance below the pivot is $2.5\cos(0.12)\,\text{m}$ , so $h=2.5-2.5\cos(0.12)=2.5(1-\cos(0.12))\,\text{m}$ . For a small angle in radians, $1-\cos x\approx x^2/2$ . Hence $h\approx2.5(0.12)^2/2=0.018\,\text{m}=1.80\,\text{cm}$ to $3$ significant figures.
3	$160$	3
(3 marks)		3
Notes
For small $x$ in radians, $\sin(2x)\approx2x$ , $\tan x\approx x$ and $1-\cos x\approx x^2/2$ . The expression is therefore approximately $\dfrac{6x-2x}{x^2/2}=\dfrac{8}{x}$ . At $x=0.05$ , this is $8/0.05=160$ .

Tier 3 · Hard

Mark scheme for 5.2 Tier 3 · Hard
Question	Scheme	Marks
1	$x\approx0.0835$ radians The other root is not a small angle.	4
(4 marks)		4
Notes
Substitution gives $x+1-x^2/2\approx1.08$ , hence $x^2-2x+0.16\approx0$ . Therefore $x\approx1\pm\sqrt{0.84}$ , giving $0.08348\ldots$ or $1.9165\ldots$ . Only $0.08348\ldots$ is small, so $x\approx0.0835$ radians; the larger root lies outside the approximation's intended range.
2	$x\approx0.1$ radians Horizontal displacement $\approx2.4\,\text{m}$	4
(4 marks)		4
Notes
The vertical component gives $24\cos x=23.88$ , so $\cos x=0.995$ . Using $\cos x\approx1-x^2/2$ gives $1-x^2/2\approx0.995$ , hence $x^2\approx0.01$ and the positive angle is $x\approx0.1$ radians. The horizontal displacement is $24\sin x\approx24x=2.4\,\text{m}$ .
3	$a=5$ and $b=5$ $x\approx0.0347$ radians	5
(5 marks)		5
Notes
Using $\sin x\approx x$ , $\cos x\approx1-x^2/2$ and $\tan(2x)\approx2x$ gives $F(x)\approx b+(a+2)x-bx^2/2$ . Comparing coefficients with $5+7x-5x^2/2$ gives $b=5$ and $a=5$ . The equation is then $5+7x-2.5x^2\approx5.24$ , or $2.5x^2-7x+0.24\approx0$ . Its roots are $0.034716\ldots$ and $2.76528\ldots$ ; the smaller positive solution is $x\approx0.0347$ radians.
4	Approximate coordinates $(25.58,\ 4.680)\,\text{m}$ Calculated coordinates $(25.58,\ 4.655)\,\text{m}$ Position error $<2.6\,\text{cm}$	5
(5 marks)		5
Notes
Using $\cos x\approx1-x^2/2$ and $\sin x\approx x$ , the approximate coordinates are $\bigl(26(1-0.18^2/2),26(0.18)\bigr)=(25.5788,4.68)$ , giving $(25.58,4.680)\,\text{m}$ to $4$ significant figures. Direct calculation gives $(26\cos0.18,26\sin0.18)=(25.579936\ldots,4.654768\ldots)$ , giving $(25.58,4.655)\,\text{m}$ . Using the unrounded coordinates, the distance between the positions is $\sqrt{(25.5788-25.579936\ldots)^2+(4.68-4.654768\ldots)^2}=0.0252566\ldots\,\text{m}=2.52566\ldots\,\text{cm}<2.6\,\text{cm}$ .
5	$15=r\sin u$ and $0.60=r(1-\cos u)$ $u\approx0.080$ radians $r\approx187.5\,\text{m}$ Angle subtended by the chord $\approx0.160$ radians	6
(6 marks)		6
Notes
The perpendicular from the centre bisects the chord, so the half-chord gives $15=r\sin u$ . The distance from the centre to the chord is $r\cos u$ , hence the height is $r-r\cos u=0.60$ . For small $u$ , use $\sin u\approx u$ and $1-\cos u\approx u^2/2$ . Thus $15\approx ru$ and $0.60\approx ru^2/2$ . Dividing the second relation by the first gives $0.60/15\approx u/2$ , so the approximation model gives $u\approx0.080$ . It then gives $r\approx15/0.080=187.5\,\text{m}$ and $2u\approx0.160$ radians. (Exact circle geometry would give $r=187.8\,\text{m}$ , but that is not the value requested.)

5.3 · Understand and use the sine, cosine and tangent functions; their graphs, symmetries and periodicity; know and use exact values of sin, cos and tan for standard angles and their multiples.

Tier 1 · Easy

Mark scheme for 5.3 Tier 1 · Easy
Question	Scheme	Marks
1	$-\dfrac{1}{\sqrt3}$ $-\dfrac{\sqrt3}{3}$	2
(2 marks)		2
Notes
The reference angle is $\pi/6$ . Tangent is negative in the second quadrant, so $\tan(5\pi/6)=-\tan(\pi/6)=-1/\sqrt3=-\sqrt3/3$ .
2	$\dfrac12$	2
(2 marks)		2
Notes
Cosine is even and has period $2\pi$ . Thus $\cos(-7\pi/3)=\cos(7\pi/3)=\cos(\pi/3)=1/2$ .

Tier 2 · Standard

Mark scheme for 5.3 Tier 2 · Standard
Question	Scheme	Marks
1	$-\dfrac12$ $\dfrac12$ $-1$	3
(3 marks)		3
Notes
The reference angles are $\pi/6$ , $\pi/3$ and $\pi/4$ . The angle $7\pi/6$ is in quadrant III, so its sine is $-1/2$ . The angle $5\pi/3$ is in quadrant IV, so its cosine is $1/2$ . The angle $3\pi/4$ is in quadrant II, so its tangent is $-1$ .
2	Amplitude $2$ and period $2\pi/3$ Maximum points $(\pi/6,-2)$ , $(5\pi/6,-2)$ and $(3\pi/2,-2)$	4
(4 marks)		4
Notes
The amplitude is $2$ and the period is $2\pi/3$ . A maximum occurs when $\sin(3x)=1$ , so $3x=\pi/2+2k\pi$ . Hence $x=\pi/6+2k\pi/3$ , giving $x=\pi/6,5\pi/6,3\pi/2$ in the interval. At each, $y=2(1)-4=-2$ .
3	Period $=\pi/2$ Zeros at $x=\pi/8$ and $x=5\pi/8$ Vertical asymptotes at $x=3\pi/8$ and $x=7\pi/8$	4
(4 marks)		4
Notes
The factor $2$ gives period $\pi/2$ . Zeros occur when $2x-\pi/4=k\pi$ , so $x=\pi/8+k\pi/2$ ; the values in the interval are $\pi/8$ and $5\pi/8$ . Vertical asymptotes occur when $2x-\pi/4=\pi/2+k\pi$ , so $x=3\pi/8+k\pi/2$ ; the values in the interval are $3\pi/8$ and $7\pi/8$ .

Tier 3 · Hard

Mark scheme for 5.3 Tier 3 · Hard
Question	Scheme	Marks
1	$\sin(-11\pi/6)=\dfrac12$ $\cos(13\pi/3)=\dfrac12$ $\tan(7\pi/4)=-1$	4
(4 marks)		4
Notes
Add $2\pi$ to $-11\pi/6$ to get $\pi/6$ , so its sine is $1/2$ . Subtract $4\pi$ from $13\pi/3$ to get $\pi/3$ , so its cosine is $1/2$ . The angle $7\pi/4$ has reference angle $\pi/4$ in quadrant IV, where tangent is negative, giving $-1$ .
2	$a=-5$ , $b=\dfrac32$ , $c=2$	5
(5 marks)		5
Notes
The midline is $c=(7+(-3))/2=2$ and the amplitude is $\|a\|=(7-(-3))/2=5$ . Since $f(0)=a+c=-3$ , $a=-5$ . The period gives $2\pi/b=4\pi/3$ , so $b=3/2$ .
3	$f(-x)=f(x)$ and $f(x+\pi)=f(x)$ Range $[0,1]$ Maximum points $(\pi/2,1)$ and $(3\pi/2,1)$ Minimum points $(0,0)$ , $(\pi,0)$ and $(2\pi,0)$	5
(5 marks)		5
Notes
Since $\sin(-x)=-\sin x$ , $f(-x)=(-\sin x)^2=\sin^2x=f(x)$ , so $f$ is even. Also $\sin(x+\pi)=-\sin x$ , hence $f(x+\pi)=f(x)$ and $\pi$ is a period. No smaller positive period exists: $f(x)=0$ only at integer multiples of $\pi$ , so any period must map $0$ to another zero of $f$ and is therefore at least $\pi$ . From $-1\leq\sin x\leq1$ , the range of $\sin^2x$ is $[0,1]$ . The value $1$ occurs at $x=\pi/2,3\pi/2$ , and the value $0$ occurs at $x=0,\pi,2\pi$ in the stated interval.
4	Least positive period $=2\pi$ and range $[0,3]$ Zeros $(\pi/3,0)$ and $(5\pi/3,0)$ Global maximum point $(\pi,3)$	5
(5 marks)		5
Notes
Since $-1\leq\cos x\leq1$ , the expression $2\cos x-1$ ranges from $-3$ to $1$ , so $0\leq f(x)\leq3$ . The zeros satisfy $\cos x=1/2$ , giving $x=\pi/3,5\pi/3$ . The maximum value $3$ occurs when $\cos x=-1$ , at $x=\pi$ . The absolute value does not halve the period here: the gaps between consecutive zeros alternate between $4\pi/3$ and $2\pi/3$ , so the zero pattern first repeats after $2\pi$ . Hence the least positive period is $2\pi$ .
5	$(u_0,u_1,\ldots,u_{11})=(2,2,-1,-3,-1,2,2,0,-1,-1,-1,0)$ Least positive integer period $12$ $30$ integers	6
(6 marks)		6
Notes
Using exact values at multiples of $\pi/3$ and $\pi/2$ gives the stated $12$ -term table. Both component sequences repeat after $12$ integer steps, so $12$ is a period; the table has no repeated initial block with a smaller length dividing $12$ , so it is the least positive integer period. In one period, $u_n\leq-1$ for $n=2,3,4,8,9,10$ , giving $6$ qualifying terms. The $60$ integers from $0$ to $59$ form five complete periods, so the required number is $5\times6=30$ .

5.4 · Understand and use the definitions of secant, cosecant and cotangent and of arcsin, arccos and arctan; their relationships to sine, cosine and tangent; understanding of their graphs; their ranges and domains.

Tier 1 · Easy

Mark scheme for 5.4 Tier 1 · Easy
Question	Scheme	Marks
1	$\sec\theta=-\dfrac54$	1
(1 mark)		1
Notes
Secant is the reciprocal of cosine, so $\sec\theta=1/(-4/5)=-5/4$ .
2	$\dfrac{5\pi}{6}$	2
(2 marks)		2
Notes
The principal range of arccos is $[0,\pi]$ . In this range, $\cos(5\pi/6)=-\sqrt3/2$ , so the principal value is $5\pi/6$ .

Tier 2 · Standard

Mark scheme for 5.4 Tier 2 · Standard
Question	Scheme	Marks
1	Undefined at $x=\dfrac{\pi}{2},\dfrac{3\pi}{2}$ $x=\dfrac{2\pi}{3},\dfrac{4\pi}{3}$	4
(4 marks)		4
Notes
Since $\sec x=1/\cos x$ , it is undefined when $\cos x=0$ , namely at $x=\pi/2$ and $3\pi/2$ . Also $\sec x=-2$ is equivalent to $\cos x=-1/2$ , which occurs in the interval at $x=2\pi/3$ and $4\pi/3$ .
2	Domain $0\leq x\leq1$ and range $-\pi/2\leq f(x)\leq\pi/2$ $x=\dfrac34$	4
(4 marks)		4
Notes
For arcsin, the input must satisfy $-1\leq2x-1\leq1$ , giving $0\leq x\leq1$ . Its principal range is $[-\pi/2,\pi/2]$ . If $\arcsin(2x-1)=\pi/6$ , then $2x-1=\sin(\pi/6)=1/2$ , so $x=3/4$ .
3	Domain $(-\infty,-1]\cup[1,\infty)$ Range $[0,\pi/2)\cup(\pi/2,\pi]$ $x=-2$	4
(4 marks)		4
Notes
The input to arccos must satisfy $-1\leq1/x\leq1$ , which gives $x\leq-1$ or $x\geq1$ . On this domain, $1/x$ takes every value in $[-1,0)\cup(0,1]$ , so applying the decreasing principal arccos function gives the range $[0,\pi/2)\cup(\pi/2,\pi]$ ; $\pi/2$ is excluded because $1/x$ cannot equal zero. Finally, $\arccos(1/x)=2\pi/3$ gives $1/x=\cos(2\pi/3)=-1/2$ , so $x=-2$ .

Tier 3 · Hard

Mark scheme for 5.4 Tier 3 · Hard
Question	Scheme	Marks
1	Period $2\pi$ Range $y\leq-3$ or $y\geq1$ Vertical asymptotes $x=-\pi/2$ and $x=\pi/2$	5
(5 marks)		5
Notes
Secant has period $2\pi$ , so the transformation does not change the period. Since $\sec x\leq-1$ or $\sec x\geq1$ , multiplying by $2$ and subtracting $1$ gives $y\leq-3$ or $y\geq1$ . Vertical asymptotes occur where $\cos x=0$ , namely $x=\pi/2+k\pi$ ; in the stated interval these are $x=-\pi/2$ and $x=\pi/2$ .
2	$\arcsin x+\arccos x=\pi/2$ $x=\dfrac{\sqrt{2-\sqrt2}}{2}$	5
(5 marks)		5
Notes
Let $y=\arcsin x$ , so $-\pi/2\leq y\leq\pi/2$ and $\sin y=x$ . Since $\cos(\pi/2-y)=\sin y=x$ and $0\leq\pi/2-y\leq\pi$ , the principal value is $\arccos x=\pi/2-y$ . Hence $\arcsin x+\arccos x=\pi/2$ . The equation becomes $3\arcsin x-(\pi/2-\arcsin x)=0$ , so $\arcsin x=\pi/8$ . Therefore $x=\sin(\pi/8)=\sqrt{2-\sqrt2}/2$ .
3	$f(x)=x$ for $0\leq x\leq\pi/2$ ; $f(x)=\pi-x$ for $\pi/2\leq x\leq3\pi/2$ ; $f(x)=x-2\pi$ for $3\pi/2\leq x\leq5\pi/2$ ; $f(x)=3\pi-x$ for $5\pi/2\leq x\leq3\pi$ Range $[-\pi/2,\pi/2]$ $x=\pi/6,\ 5\pi/6,\ 13\pi/6,\ 17\pi/6$	6
(6 marks)		6
Notes
The principal range of arcsin is $[-\pi/2,\pi/2]$ . On the first quarter-cycle the principal angle is $x$ ; reflection across $\pi/2$ gives $\pi-x$ until $3\pi/2$ . Over the next principal branch it is $x-2\pi$ , followed by the reflection $3\pi-x$ . These branches attain every value from $-\pi/2$ to $\pi/2$ . For $f(x)=\pi/6$ , equivalently $\sin x=1/2$ with principal output $\pi/6$ , the four values in the stated interval are $\pi/6,5\pi/6,13\pi/6,17\pi/6$ .
4	$\arctan x+\arctan(1/x)=\pi/2$ for $x>0$ and $-\pi/2$ for $x<0$ The solution is every $x<0$ .	5
(5 marks)		5
Notes
For $x>0$ , let $\alpha=\arctan x$ , so $0<\alpha<\pi/2$ . The angle $\pi/2-\alpha$ is also in the principal range of arctan and has tangent $1/x$ , hence $\arctan(1/x)=\pi/2-\alpha$ . The sum is therefore $\pi/2$ . If $x<0$ , apply the oddness of arctan to $-x>0$ : both inverse-tangent terms change sign, so the sum is $-\pi/2$ . Consequently the equation holds exactly when $x<0$ .
5	Maximum point $(\pi/2,1)$ on $0<x<\pi$ Minimum point $(3\pi/2,5)$ on $\pi<x<2\pi$ $x=7\pi/6,\ 11\pi/6$	5
(5 marks)		5
Notes
On $0<x<\pi$ , $0<\sin x\leq1$ , so $\cosec x\geq1$ and $f(x)\leq1$ . Equality occurs only when $\sin x=1$ , giving the maximum point $(\pi/2,1)$ . On $\pi<x<2\pi$ , $-1\leq\sin x<0$ , so $\cosec x\leq-1$ and $f(x)\geq5$ . Equality occurs only when $\sin x=-1$ , giving the minimum point $(3\pi/2,5)$ . Finally, $f(x)=7$ gives $\cosec x=-2$ , or $\sin x=-1/2$ . The complete solution set in the stated interval is $x=7\pi/6,11\pi/6$ .

5.5 · Understand and use tan θ = sin θ / cos θ; understand and use sin²θ + cos²θ = 1, sec²θ = 1 + tan²θ and cosec²θ = 1 + cot²θ.

Tier 1 · Easy

Mark scheme for 5.5 Tier 1 · Easy
Question	Scheme	Marks
1	$\tan\theta=5/12$ $\sec\theta=13/12$	3
(3 marks)		3
Notes
Because $\theta$ is acute, $\cos\theta$ is positive. From $\cos^2\theta=1-25/169=144/169$ , $\cos\theta=12/13$ . Hence $\tan\theta=(5/13)/(12/13)=5/12$ and $\sec\theta=1/(12/13)=13/12$ .
2	$\cosec\theta=-\dfrac54$	2
(2 marks)		2
Notes
Using a $3$ - $4$ - $5$ reference triangle, $\|\sin\theta\|=4/5$ . The angle is in quadrant IV, where sine is negative, so $\sin\theta=-4/5$ and $\cosec\theta=-5/4$ .

Tier 2 · Standard

Mark scheme for 5.5 Tier 2 · Standard
Question	Scheme	Marks
1	Both sides simplify to $\dfrac{2}{\cos^2x}$ .	4
(4 marks)		4
Notes
Combine the fractions on the left: $\dfrac{(1+\sin x)+(1-\sin x)}{(1-\sin x)(1+\sin x)}=\dfrac{2}{1-\sin^2x}$ . Using $1-\sin^2x=\cos^2x$ , this is $2/\cos^2x=2\sec^2x$ . The manipulation is valid wherever the original expressions are defined.
2	$\sin x$	3
(3 marks)		3
Notes
$\sec x-\cos x=1/\cos x-\cos x=(1-\cos^2x)/\cos x=\sin^2x/\cos x$ . Dividing this by $\tan x=\sin x/\cos x$ gives $\sin x$ wherever the original expression is defined.
3	$5/4$	3
(3 marks)		3
Notes
In the stated quadrant $\cos\theta\neq0$ , so divide the numerator and denominator by $\cos\theta$ . The expression becomes $\dfrac{3\tan\theta+1}{\tan\theta-2}$ . Substituting $\tan\theta=-2$ gives $(-6+1)/(-2-2)=5/4$ .

Tier 3 · Hard

Mark scheme for 5.5 Tier 3 · Hard
Question	Scheme	Marks
1	$\sin\theta=12/13$	5
(5 marks)		5
Notes
Since $(\sec\theta+\tan\theta)(\sec\theta-\tan\theta)=\sec^2\theta-\tan^2\theta=1$ , it follows that $\sec\theta-\tan\theta=1/5$ . Adding the two equations gives $2\sec\theta=26/5$ , so $\sec\theta=13/5$ and $\tan\theta=12/5$ . Thus $\cos\theta=5/13$ and $\sin\theta=\tan\theta\cos\theta=(12/5)(5/13)=12/13$ .
2	$\tan\theta=\dfrac34$	5
(5 marks)		5
Notes
Squaring the given relation gives $1+2\sin\theta\cos\theta=49/25$ , so $\sin\theta\cos\theta=12/25$ . Therefore $(\cos\theta-\sin\theta)^2=1-2\sin\theta\cos\theta=1/25$ . In the stated interval $\cos\theta>\sin\theta$ , so $\cos\theta-\sin\theta=1/5$ . Solving this with $\sin\theta+\cos\theta=7/5$ gives $\sin\theta=3/5$ and $\cos\theta=4/5$ . Hence $\tan\theta=3/4$ .
3	$\sin\theta=\dfrac{4\sqrt{13}-2}{17}$	5
(5 marks)		5
Notes
Let $s=\sec\theta$ and $t=\tan\theta$ . Then $t=4-2s$ and $s^2=1+t^2$ , so $s^2=1+(4-2s)^2$ . Hence $3s^2-16s+17=0$ , giving $s=(8\pm\sqrt{13})/3$ . Since $\theta$ is acute, $t>0$ ; the plus sign would give $t=4-2s<0$ , so $s=(8-\sqrt{13})/3$ . Thus $\cos\theta=1/s=(8+\sqrt{13})/17$ , and $\sin\theta=\tan\theta\cos\theta=(4-2s)/s=4\cos\theta-2=(4\sqrt{13}-2)/17$ .
4	$\cosec\theta-\cot\theta=3$	5
(5 marks)		5
Notes
In the stated interval $\cos\theta\neq0$ , so divide the numerator and denominator by $\cos\theta$ . With $t=\tan\theta$ , the equation becomes $(2t+1)/(t-2)=2/11$ . Cross-multiplying gives $22t+11=2t-4$ , hence $t=-3/4$ . Since $\theta$ is in quadrant II, a $3$ - $4$ - $5$ reference triangle gives $\sin\theta=3/5$ and $\cos\theta=-4/5$ . Therefore $\cosec\theta-\cot\theta=5/3-(-4/3)=3$ .
5	$\sec^2x+\cosec^2x\geq4$ $x=\pi/4,\ 3\pi/4,\ 5\pi/4,\ 7\pi/4$	5
(5 marks)		5
Notes
Where the expression is defined, $\sin x\cos x\neq0$ and $\sec^2x+\cosec^2x=(\sin^2x+\cos^2x)/(\sin^2x\cos^2x)=1/(\sin^2x\cos^2x)$ . Also $(\sin^2x-\cos^2x)^2\geq0$ , so $(\sin^2x+\cos^2x)^2\geq4\sin^2x\cos^2x$ . Thus $\sin^2x\cos^2x\leq1/4$ , proving the inequality. Equality requires $\sin^2x=\cos^2x$ , which gives the four stated values in the interval.

5.6 · Understand and use double angle formulae; formulae for sin(A ± B), cos(A ± B), tan(A ± B) with geometrical proofs; express a cos θ + b sin θ in the form r cos(θ ± α) or r sin(θ ± α).

Tier 1 · Easy

Mark scheme for 5.6 Tier 1 · Easy
Question	Scheme	Marks
1	$\dfrac{\sqrt6+\sqrt2}{4}$	3
(3 marks)		3
Notes
Write $75^\circ=45^\circ+30^\circ$ . Then $\sin75^\circ=\sin45^\circ\cos30^\circ+\cos45^\circ\sin30^\circ=(1/\sqrt2)(\sqrt3/2)+(1/\sqrt2)(1/2)=(\sqrt6+\sqrt2)/4$ .
2	$2-\sqrt3$	3
(3 marks)		3
Notes
Write $15^\circ=45^\circ-30^\circ$ . Then $\tan15^\circ=(1-1/\sqrt3)/(1+1/\sqrt3)=(\sqrt3-1)/(\sqrt3+1)=2-\sqrt3$ .

Tier 2 · Standard

Mark scheme for 5.6 Tier 2 · Standard
Question	Scheme	Marks
1	$\sin2\theta=\dfrac{24}{25}$ $\cos2\theta=\dfrac{7}{25}$ $\tan2\theta=\dfrac{24}{7}$	4
(4 marks)		4
Notes
Because $\theta$ is acute, $\cos\theta=4/5$ . Hence $\sin2\theta=2\sin\theta\cos\theta=2(3/5)(4/5)=24/25$ . Also $\cos2\theta=\cos^2\theta-\sin^2\theta=16/25-9/25=7/25$ . Therefore $\tan2\theta=(\sin2\theta)/(\cos2\theta)=24/7$ .
2	$25\sin(\theta+\alpha)$ , where $\alpha=\arctan(7/24)$ $\dfrac{1}{51}\leq f(\theta)\leq1$	4
(4 marks)		4
Notes
Expanding gives $R\sin(\theta+\alpha)=R\sin\theta\cos\alpha+R\cos\theta\sin\alpha$ . Thus $R\cos\alpha=24$ and $R\sin\alpha=7$ , so $R=\sqrt{24^2+7^2}=25$ and $\tan\alpha=7/24$ . Therefore $1\leq26+25\sin(\theta+\alpha)\leq51$ . All values are positive, so taking reciprocals reverses the bounds and gives $1/51\leq f(\theta)\leq1$ .
3	$a=3/2$ and $b=7/2$ Maximum $=5$ and minimum $=-2$	4
(4 marks)		4
Notes
Use $\cos^2x=(1+\cos2x)/2$ and $\sin^2x=(1-\cos2x)/2$ . Then $5\cos^2x-2\sin^2x=\tfrac52(1+\cos2x)-(1-\cos2x)=\tfrac32+\tfrac72\cos2x$ . Since $-1\leq\cos2x\leq1$ , the maximum is $3/2+7/2=5$ and the minimum is $3/2-7/2=-2$ .

Tier 3 · Hard

Mark scheme for 5.6 Tier 3 · Hard
Question	Scheme	Marks
1	Equate the coordinate and cosine-rule expressions for the squared chord joining $(\cos A,\sin A)$ and $(\cos B,\sin B)$ .	5
(5 marks)		5
Notes
Let $P=(\cos A,\sin A)$ and $Q=(\cos B,\sin B)$ on the unit circle. By coordinates, $PQ^2=(\cos A-\cos B)^2+(\sin A-\sin B)^2=2-2(\cos A\cos B+\sin A\sin B)$ . In triangle $OPQ$ , $OP=OQ=1$ . If $\phi\in[0,\pi]$ is the smaller central angle, then $\cos\phi=\cos(A-B)$ , so the cosine rule gives $PQ^2=2-2\cos(A-B)$ . Equating the two expressions for $PQ^2$ and dividing by $-2$ proves $\cos(A-B)=\cos A\cos B+\sin A\sin B$ .
2	Divide the expansions of $\sin(A+B)$ and $\cos(A+B)$ by $\cos A\cos B$ .	4
(4 marks)		4
Notes
Using the addition formulae, $\tan(A+B)=\dfrac{\sin A\cos B+\cos A\sin B}{\cos A\cos B-\sin A\sin B}$ . Divide numerator and denominator by $\cos A\cos B$ . The numerator becomes $\tan A+\tan B$ and the denominator becomes $1-\tan A\tan B$ , giving the required formula wherever all the displayed expressions are defined.
3	$\tan(\theta/2)=\dfrac{\sin\theta}{1+\cos\theta}$ $\tan(\pi/8)=\sqrt2-1$	5
(5 marks)		5
Notes
Write $u=\theta/2$ . Then $\sin\theta=2\sin u\cos u$ and $1+\cos\theta=1+2\cos^2u-1=2\cos^2u$ . Where both sides are defined, $\cos u\neq0$ , so division gives $\sin\theta/(1+\cos\theta)=\tan u=\tan(\theta/2)$ . Taking $\theta=\pi/4$ gives $\tan(\pi/8)=\dfrac{\sqrt2/2}{1+\sqrt2/2}=\dfrac{\sqrt2}{2+\sqrt2}=\sqrt2-1$ .
4	$\sin\theta=4/5$ and $\cos\theta=-3/5$ $\sin(\theta+\pi/4)=\sqrt2/10$	5
(5 marks)		5
Notes
From $\cos(2\theta)=2\cos^2\theta-1=-7/25$ , we obtain $\cos^2\theta=9/25$ . The stated interval is in quadrant II, so $\cos\theta=-3/5$ and $\sin\theta=4/5$ . Using the sine addition formula, $\sin(\theta+\pi/4)=\sin\theta\cos(\pi/4)+\cos\theta\sin(\pi/4)=(4/5-3/5)/\sqrt2=\sqrt2/10$ .
5	$\sin(8x)=8\sin x\cos x\cos(2x)\cos(4x)$ $\cos20^\circ\cos40^\circ\cos80^\circ=1/8$	5
(5 marks)		5
Notes
Apply $\sin(2u)=2\sin u\cos u$ three times: $\sin(8x)=2\sin(4x)\cos(4x)=4\sin(2x)\cos(2x)\cos(4x)=8\sin x\cos x\cos(2x)\cos(4x)$ . Set $x=20^\circ$ . Since $\sin160^\circ=\sin20^\circ$ and $\sin20^\circ\neq0$ , the identity gives $\sin20^\circ=8\sin20^\circ\cos20^\circ\cos40^\circ\cos80^\circ$ . Cancelling $\sin20^\circ$ gives the exact product $1/8$ .

5.7 · Solve simple trigonometric equations in a given interval, including quadratic equations in sin, cos and tan and equations involving multiples of the unknown angle.

Tier 1 · Easy

Mark scheme for 5.7 Tier 1 · Easy
Question	Scheme	Marks
1	$\theta=0.412$ or $2.730$	3
(3 marks)		3
Notes
The principal value is $\arcsin(0.4)=0.4115\ldots$ . Sine is positive in quadrants I and II, so the second solution is $\pi-0.4115\ldots=2.7301\ldots$ . Thus $\theta=0.412$ or $2.730$ .
2	$x=\dfrac{3\pi}{4},\ \dfrac{5\pi}{4}$	2
(2 marks)		2
Notes
The reference angle is $\pi/4$ . Cosine is negative in quadrants II and III, giving $x=3\pi/4$ and $x=5\pi/4$ .

Tier 2 · Standard

Mark scheme for 5.7 Tier 2 · Standard
Question	Scheme	Marks
1	$x=\dfrac{\pi}{6},\ \dfrac{\pi}{3},\ \dfrac{7\pi}{6},\ \dfrac{4\pi}{3}$	4
(4 marks)		4
Notes
Since $0\leq2x\leq4\pi$ , solve over two complete periods. The solutions for $2x$ are $\pi/3$ , $2\pi/3$ , $7\pi/3$ and $8\pi/3$ . Dividing each by $2$ gives $x=\pi/6,\pi/3,7\pi/6,4\pi/3$ .
2	$x=\dfrac{\pi}{6},\ \dfrac{5\pi}{6},\ \dfrac{3\pi}{2}$	4
(4 marks)		4
Notes
Factorise to $(2\sin x-1)(\sin x+1)=0$ . Thus $\sin x=1/2$ or $\sin x=-1$ . In the stated interval these give $x=\pi/6,5\pi/6$ and $3\pi/2$ respectively.
3	$x=0,\ \pi/3,\ \pi,\ 5\pi/3$	4
(4 marks)		4
Notes
Bring all terms to one side and factor without dividing by $\sin x$ : $\sin x(2\cos x-1)=0$ . Thus $\sin x=0$ or $\cos x=1/2$ . In $0\leq x<2\pi$ , the first branch gives $x=0,\pi$ and the second gives $x=\pi/3,5\pi/3$ . These four values form the complete solution set.

Tier 3 · Hard

Mark scheme for 5.7 Tier 3 · Hard
Question	Scheme	Marks
1	$x=-\pi/6,\ \pi/3,\ 5\pi/6$	5
(5 marks)		5
Notes
The transformed interval is $-\pi\leq2x\leq2\pi$ . Since tangent has period $\pi$ and reference angle $\pi/3$ , the solutions for $2x$ in this interval are $-\pi/3$ , $2\pi/3$ and $5\pi/3$ . Dividing each by $2$ gives $x=-\pi/6$ , $\pi/3$ and $5\pi/6$ .
2	$3x=-\dfrac{7\pi}{6},\ -\dfrac{5\pi}{6},\ -\dfrac{\pi}{6},\ \dfrac{\pi}{6},\ \dfrac{5\pi}{6},\ \dfrac{7\pi}{6},\ \dfrac{11\pi}{6}$ (for $-3\pi/2\leq3x\leq2\pi$ ) $x=-\dfrac{7\pi}{18},\ -\dfrac{5\pi}{18},\ -\dfrac{\pi}{18},\ \dfrac{\pi}{18},\ \dfrac{5\pi}{18},\ \dfrac{7\pi}{18},\ \dfrac{11\pi}{18}$	6
(6 marks)		6
Notes
Let $y=3x$ , so $-3\pi/2\leq y\leq2\pi$ . Factorising gives $(2\sin y-1)(2\sin y+1)=0$ , so $\sin y=1/2$ or $\sin y=-1/2$ . For $\sin y=1/2$ , the values in the expanded interval are $y=-7\pi/6,\pi/6,5\pi/6$ . For $\sin y=-1/2$ , they are $y=-5\pi/6,-\pi/6,7\pi/6,11\pi/6$ . Dividing all seven values by $3$ and ordering them gives $x=-7\pi/18,-5\pi/18,-\pi/18,\pi/18,5\pi/18,7\pi/18,11\pi/18$ .
3	$3\sin x+4\cos x=5\sin(x+\alpha)$ , where $\alpha=\arctan(4/3)$ $x=-0.516,\ 1.803$	5
(5 marks)		5
Notes
Expanding $R\sin(x+\alpha)$ gives $R\cos\alpha=3$ and $R\sin\alpha=4$ , so $R=5$ and $\alpha=\arctan(4/3)$ . The equation becomes $\sin(x+\alpha)=2/5$ . Let $\beta=\arcsin(2/5)$ . Since $-\pi+\alpha\leq x+\alpha\leq\pi+\alpha$ , the only values in the transformed interval are $x+\alpha=\beta$ and $x+\alpha=\pi-\beta$ . Hence $x=\beta-\alpha=-0.515778\ldots$ or $x=\pi-\beta-\alpha=1.802780\ldots$ , giving $x=-0.516,1.803$ .
4	$x=\arcsin\left(\dfrac{\sqrt3-1}{2}\right),\ \pi-\arcsin\left(\dfrac{\sqrt3-1}{2}\right)$	5
(5 marks)		5
Notes
Use $\cos(2x)=1-2\sin^2x$ . With $s=\sin x$ , the equation becomes $1-2s^2=2s$ , or $2s^2+2s-1=0$ . Thus $s=(-1\pm\sqrt3)/2$ . The value $(-1-\sqrt3)/2$ is less than $-1$ and is rejected, leaving $\sin x=(\sqrt3-1)/2$ . This value is positive, so the complete solution set in the interval consists of the quadrant-I value and its quadrant-II partner, as stated.
5	$k=0$	6
(6 marks)		6
Notes
Let $s=\sin x$ , so $-1\leq s\leq1$ and $s^2-s=k$ . Each root strictly between $-1$ and $1$ produces two values of $x$ in the interval, while either endpoint $s=1$ or $s=-1$ produces one value. A total of three solutions therefore requires one endpoint root and one interior root. If $s=1$ , then $k=0$ and the other root is $s=0$ , giving one solution from $s=1$ and two from $s=0$ . If $s=-1$ , then $k=2$ and the other root is $s=2$ , which is outside the sine range, so there is only one solution. A repeated interior root produces two solutions, and two interior roots produce four. Hence only $k=0$ gives exactly three distinct solutions.

5.8 · Construct proofs involving trigonometric functions and identities.

Tier 1 · Easy

Mark scheme for 5.8 Tier 1 · Easy
Question	Scheme	Marks
1	Replace $\cot x$ by $\cos x/\sin x$ and cancel $\sin x$ .	2
(2 marks)		2
Notes
$\sin x\cot x=\sin x(\cos x/\sin x)=\cos x$ . The cancellation is valid wherever $\cot x$ is defined, namely where $\sin x\neq0$ .
2	Use $1+\cot^2x=\cosec^2x$ .	2
(2 marks)		2
Notes
By the Pythagorean identity, $(1+\cot^2x)\sin^2x=\cosec^2x\sin^2x=(1/\sin^2x)\sin^2x=1$ wherever $\sin x\neq0$ .

Tier 2 · Standard

Mark scheme for 5.8 Tier 2 · Standard
Question	Scheme	Marks
1	Expand both cosine factors and simplify using $\sin^2A+\cos^2A=1$ .	4
(4 marks)		4
Notes
Expand the left-hand side: $(\cos A\cos B-\sin A\sin B)(\cos A\cos B+\sin A\sin B)=\cos^2A\cos^2B-\sin^2A\sin^2B$ . Now use $\cos^2B=1-\sin^2B$ : this becomes $\cos^2A-\sin^2B(\cos^2A+\sin^2A)=\cos^2A-\sin^2B$ .
2	Multiply the left-hand side by $(\sec x+1)/(\sec x+1)$ .	4
(4 marks)		4
Notes
Multiply the left-hand side by the conjugate: $\dfrac{\tan x(\sec x+1)}{\sec^2x-1}=\dfrac{\tan x(\sec x+1)}{\tan^2x}=\dfrac{\sec x+1}{\tan x}$ . Replacing secant and tangent by sine and cosine gives $\dfrac{1/\cos x+1}{\sin x/\cos x}=\dfrac{1+\cos x}{\sin x}$ , as required. The working multiplies by $\sec x+1$ and divides by $\tan^2x$ ; both are nonzero wherever both sides are defined, since $\tan x=0$ or $\sec x=-1$ would make a side undefined.
3	Combine the left-hand side over $\cos x(1+\sin x)$ and use $\sin^2x+\cos^2x=1$ .	4
(4 marks)		4
Notes
The left-hand side is $\dfrac{(1+\sin x)^2+\cos^2x}{\cos x(1+\sin x)}$ . Its numerator simplifies to $1+2\sin x+\sin^2x+\cos^2x=2(1+\sin x)$ . Since the original expressions require $\cos x\neq0$ and $1+\sin x\neq0$ , cancellation is valid, leaving $2/\cos x=2\sec x$ .

Tier 3 · Hard

Mark scheme for 5.8 Tier 3 · Hard
Question	Scheme	Marks
1	The left-hand side simplifies to $2\sin x/\cos^2x=2\tan x\sec x$ .	5
(5 marks)		5
Notes
Combine the fractions: the numerator is $(1+\sin x)-(1-\sin x)=2\sin x$ and the denominator is $(1-\sin x)(1+\sin x)=1-\sin^2x=\cos^2x$ . Thus the left-hand side is $2\sin x/\cos^2x=2(\sin x/\cos x)(1/\cos x)=2\tan x\sec x$ .
2	Factor the difference of fourth powers, then use $\sin^2x+\cos^2x=1$ and $\sin2x=2\sin x\cos x$ .	4
(4 marks)		4
Notes
Let $a=\sin x+\cos x$ and $b=\sin x-\cos x$ . Then $a^4-b^4=(a-b)(a+b)(a^2+b^2)$ . Here $a-b=2\cos x$ , $a+b=2\sin x$ , and $a^2+b^2=2(\sin^2x+\cos^2x)=2$ . Thus the left-hand side is $(2\cos x)(2\sin x)(2)=8\sin x\cos x=4\sin2x$ , as required.
3	Each expression simplifies to $1/(\sin^2x\cos^2x)$ .	5
(5 marks)		5
Notes
Where the expressions are defined, $\sin x\neq0$ and $\cos x\neq0$ . First, $\sec^2x+\cosec^2x=1/\cos^2x+1/\sin^2x=(\sin^2x+\cos^2x)/(\sin^2x\cos^2x)=1/(\sin^2x\cos^2x)$ . Also $\tan x+\cot x=\sin x/\cos x+\cos x/\sin x=1/(\sin x\cos x)$ , so its square is the same expression. Finally, $4\cosec^2(2x)=4/\sin^2(2x)=4/(4\sin^2x\cos^2x)$ , again giving the same result.
4	$\sin^6x+\cos^6x=1-\dfrac34\sin^2(2x)$ $5/8$	5
(5 marks)		5
Notes
Using $a^3+b^3=(a+b)^3-3ab(a+b)$ with $a=\sin^2x$ and $b=\cos^2x$ gives $\sin^6x+\cos^6x=1-3\sin^2x\cos^2x$ . Since $\sin(2x)=2\sin x\cos x$ , this is $1-3\sin^2(2x)/4$ . At $x=\pi/8$ , $\sin(2x)=\sin(\pi/4)=\sqrt2/2$ , so the value is $1-(3/4)(1/2)=5/8$ .
5	$(\sec x+\tan x)^2=\dfrac{1+\sin x}{1-\sin x}$ $x=\pi/6$	6
(6 marks)		6
Notes
Write $\sec x+\tan x=(1+\sin x)/\cos x$ . Squaring gives $(1+\sin x)^2/\cos^2x$ . Since $\cos^2x=(1-\sin x)(1+\sin x)$ , cancellation gives $(1+\sin x)/(1-\sin x)$ wherever the original expressions are defined. If $\sec x+\tan x=\sqrt3$ , the identity gives $(1+\sin x)/(1-\sin x)=3$ , so $\sin x=1/2$ . The candidates in the interval are $x=\pi/6$ and $x=5\pi/6$ . Substitution into the unsquared equation gives $\sqrt3$ at $\pi/6$ but $-\sqrt3$ at $5\pi/6$ , so only $x=\pi/6$ is valid.

5.9 · Use trigonometric functions to solve problems in context, including problems involving vectors, kinematics and forces.

Tier 1 · Easy

Mark scheme for 5.9 Tier 1 · Easy
Question	Scheme	Marks
1	Horizontal $=9.83\,\text{m s}^{-1}$ Vertical $=6.88\,\text{m s}^{-1}$	3
(3 marks)		3
Notes
Resolve the velocity vector: the horizontal component is $12\cos35^\circ=9.829\ldots\,\text{m s}^{-1}$ and the vertical component is $12\sin35^\circ=6.882\ldots\,\text{m s}^{-1}$ .
2	$5+\dfrac{3\sqrt3}{2}\,\text{m}$	2
(2 marks)		2
Notes
Substitute $t=2$ : $h=5+3\sin(\pi/3)=5+3(\sqrt3/2)=5+3\sqrt3/2\,\text{m}$ .

Tier 2 · Standard

Mark scheme for 5.9 Tier 2 · Standard
Question	Scheme	Marks
1	Distance $=30\,\text{km}$ Bearing $077^\circ$	5
(5 marks)		5
Notes
Resolve each displacement into east and north components. The totals are $E=24\sin40^\circ+18\sin130^\circ=29.2157\ldots$ and $N=24\cos40^\circ+18\cos130^\circ=6.81489\ldots$ . Thus the distance is $\sqrt{E^2+N^2}=30\,\text{km}$ . The bearing is measured clockwise from north, so it is $\arctan(E/N)=76.869\ldots^\circ$ , giving $077^\circ$ .
2	Distance $=\sqrt{57}\,\text{m}$ Direction $=66.6^\circ$	4
(4 marks)		4
Notes
At $t=\pi/3$ , $\mathbf r=3\mathbf i+4\sqrt3\mathbf j$ . Its magnitude is $\sqrt{3^2+(4\sqrt3)^2}=\sqrt{57}\,\text{m}$ . Both components are positive, so the direction is $\arctan(4\sqrt3/3)=66.586\ldots^\circ$ , giving $66.6^\circ$ above the positive $x$ -axis.
3	$AC=6.43\,\text{km}$ $BC=7.66\,\text{km}$	4
(4 marks)		4
Notes
The bearing of $B$ from $A$ is $090^\circ$ , so $\angle BAC=50^\circ$ . The bearing of $A$ from $B$ is $270^\circ$ , so $\angle ABC=40^\circ$ . Hence $\angle ACB=90^\circ$ . By the sine rule, $AC=10\sin40^\circ=6.427\ldots\,\text{km}$ and $BC=10\sin50^\circ=7.660\ldots\,\text{km}$ , giving the stated answers.

Tier 3 · Hard

Mark scheme for 5.9 Tier 3 · Hard
Question	Scheme	Marks
1	$P=42.6\,\text{N}$ $Q=38.6\,\text{N}$	5
(5 marks)		5
Notes
Vertical equilibrium gives $P\sin25^\circ=18$ , so $P=18/\sin25^\circ=42.592\ldots\,\text{N}$ . Horizontal equilibrium then gives $Q=P\cos25^\circ=38.601\ldots\,\text{N}$ . Therefore $P=42.6\,\text{N}$ and $Q=38.6\,\text{N}$ .
2	Airspeed $=124\,\text{km h}^{-1}$ Bearing $346^\circ$	5
(5 marks)		5
Notes
Use $\mathbf v_{\text{air}}+\mathbf v_{\text{wind}}=\mathbf v_{\text{ground}}$ . Taking east and north as positive component directions, $\mathbf v_{\text{air}}=(-30,120)$ . Hence the airspeed is $\sqrt{30^2+120^2}=30\sqrt{17}=123.693\ldots\,\text{km h}^{-1}$ , giving $124\,\text{km h}^{-1}$ . The direction is $\arctan(30/120)=14.036\ldots^\circ$ west of north, so the bearing measured clockwise from north is $360^\circ-14.036\ldots^\circ=345.963\ldots^\circ$ , giving $346^\circ$ .
3	$W=111\,\text{N}$ The right cable is at its limiting tension.	6
(6 marks)		6
Notes
Let the left and right tensions be $L$ and $R$ . Horizontal equilibrium gives $L\cos30^\circ=R\cos60^\circ$ , so $R=\sqrt3L$ . Vertical equilibrium gives $W=L\sin30^\circ+R\sin60^\circ=L/2+(\sqrt3L)(\sqrt3/2)=2L$ . The left-cable constraint gives $W\leq160\,\text{N}$ . The right-cable limit gives $R=96$ , so $L=96/\sqrt3$ and $W=192/\sqrt3=64\sqrt3=110.851\ldots\,\text{N}$ . This is the lower limit, so the greatest possible weight is $111\,\text{N}$ and the right cable is limiting.
4	Magnitude $=2\sqrt{79}\,\text{N}$ Bearing $287^\circ$	6
(6 marks)		6
Notes
Take east and north as the positive component directions. The first two forces have total east component $14\sin30^\circ+20\sin150^\circ=17$ and total north component $14\cos30^\circ+20\cos150^\circ=-3\sqrt3$ . For equilibrium, the third force must therefore have components $(-17,3\sqrt3)$ . Its magnitude is $\sqrt{17^2+(3\sqrt3)^2}=\sqrt{316}=2\sqrt{79}\,\text{N}$ . It points west and north. Its angle west of north is $\arctan(17/(3\sqrt3))=73.0039\ldots^\circ$ , so its bearing is $360^\circ-73.0039\ldots^\circ=286.996\ldots^\circ$ , which rounds to $287^\circ$ .
5	$0\leq t\leq7/3$ , $23/3\leq t\leq31/3$ , and $47/3\leq t\leq18$ Total time $=22/3$ seconds	6
(6 marks)		6
Notes
The condition is $\cos u\geq1/2$ , where $u=\pi(t-1)/4$ . The time interval gives $-\pi/4\leq u\leq17\pi/4$ . In this transformed interval, $\cos u\geq1/2$ on $[-\pi/4,\pi/3]$ , $[5\pi/3,7\pi/3]$ and $[11\pi/3,17\pi/4]$ . Since $t=1+4u/\pi$ , these become the three stated time intervals. Their lengths are $7/3$ , $8/3$ and $7/3$ seconds, giving a total of $22/3$ seconds.