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AQA Level 2 Further Maths revision notes

Geometry

Section G
10 specification points

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

Checked against AQA 8365 section G

Checked against AQA 8365 section G. Review basis: the qualification registry sourced from the AQA Level 2 Certificate in Further Mathematics (8365) specification; registry verification recorded 11 July 2026.

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G1

Knowledge of perimeter, area, surface area and volume of standard shapes; angle properties of parallel/intersecting lines, triangles, quadrilaterals and polygons; and understand and use circle theorems

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Standard areas include rectangle lwlw, triangle 12bh\frac12bh, parallelogram bhbh, trapezium 12(a+b)h\frac12(a+b)h and circle πr2\pi r^2; circumference is 2πr2\pi r. Prism volume is cross-sectional area times length; cylinder volume is πr2h\pi r^2h, curved area 2πrh2\pi rh and total area 2πrh+2πr22\pi rh+2\pi r^2.
  • Sphere volume and area are 43πr3\frac43\pi r^3 and 4πr24\pi r^2; cone volume and total area are 13πr2h\frac13\pi r^2h and πrl+πr2\pi rl+\pi r^2; pyramid volume is 13×base area×perpendicular height\frac13\times\text{base area}\times\text{perpendicular height}. Prism and pyramid surface areas sum all exposed faces.
  • Angle properties cover parallel and intersecting lines, triangles, all special quadrilaterals and polygons.
  • Circle facts include centre angle twice circumference angle, same-segment angles, semicircle right angle, opposite cyclic angles, radius–tangent perpendicularity, alternate segment, centre-perpendicular bisecting a chord, and equal tangents from an external point.
  • Examiners expect compatible units and named reasons.
An angle at the circumference is subtended by a chord joining its two endpoints.
Worked example

Opposite angles of a cyclic quadrilateral are 112112^\circ and xx^\circ. Work out xx and give a reason.

  1. 1.All four vertices lie on one circle, so the quadrilateral is cyclic.
  2. 2.Opposite angles in a cyclic quadrilateral sum to 180180^\circ.
  3. 3.x=180112=68x=180^\circ-112^\circ=68^\circ.

Answer: x=68x=68^\circ, because opposite angles in a cyclic quadrilateral sum to 180180^\circ.

Common mistakes

  • Don't use 360112360^\circ-112^\circ for one opposite angle instead of the cyclic total 180180^\circ.
  • Don't give the correct angle but omit the circle theorem when a reason is required.
  • Don't include a shared internal face when calculating the external surface area of a composite solid.

Exam tip

For a reasoning angle question, write the named theorem on the same line as the equation it justifies.

Tier 1 · Easy

ORIGINAL

1

Work out the interior angle of a regular 1212-sided polygon.

[2 marks]

Tier 2 · Standard

ORIGINAL

1

A closed cylinder has radius 33 cm and height 88 cm. Work out its total surface area in terms of π\pi.

[3 marks]

Tier 3 · Hard

ORIGINAL

1

Two radii of a circle of radius 1010 cm enclose an angle of 120120^\circ. Work out the exact area of the minor sector with the triangle formed by the two radii removed.

[4 marks]

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G2

Understand and construct geometrical proofs using formal arguments

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • A geometrical proof is a connected chain of statements, each justified by a definition, theorem or established result. Precise labels such as ABC\angle ABC identify the vertex and arms unambiguously.
  • Congruence can prove corresponding sides or angles equal; valid tests include SSS, SAS, ASA and RHS. Similarity, parallel-line facts and circle theorems may also supply steps.
  • A diagram must not be assumed to scale, and the desired conclusion cannot be used as a premise.
  • For circle proofs, the relevant chord, radius, tangent or arc should be identified.
  • Examiners expect reasons beside key statements and a final sentence that explicitly establishes the proposition.
A diagonal divides parallelogram ABCD into two triangles for a congruence proof.
Worked example

In parallelogram ABCDABCD, diagonal ACAC is drawn. Prove that triangles ABCABC and CDACDA are congruent.

  1. 1.AB=CDAB=CD and BC=ADBC=AD because opposite sides of a parallelogram are equal.
  2. 2.ACAC is common to both triangles.
  3. 3.The three corresponding side pairs are equal, so the triangles are congruent by SSS.

Answer: ABCCDA\triangle ABC\cong\triangle CDA by SSS.

Common mistakes

  • Don't claim the triangles are congruent because they look equal on the diagram.
  • Don't name SSS without identifying all three corresponding side pairs.
  • Don't list triangle vertices in an order that mismatches corresponding points.

Exam tip

For congruence, state each matching side or angle pair before naming the test.

Tier 1 · Easy

ORIGINAL

1

In isosceles triangle ABCABC, AB=ACAB=AC. Point DD is the midpoint of BCBC. Prove that ADAD is perpendicular to BCBC.

[2 marks]

Tier 2 · Standard

ORIGINAL

1

In a circle with centre OO, the minor angle AOBAOB is 2t2t, where 0<t<900^\circ<t<90^\circ. A tangent is drawn at AA. Prove that the acute angle between the tangent and ABAB is tt.

[3 marks]

Tier 3 · Hard

ORIGINAL

1

From a point PP outside a circle with centre OO, tangents touch the circle at AA and BB. Prove that OPOP is the perpendicular bisector of ABAB.

[4 marks]

G3

Sine and cosine rules in scalene triangles; area of a triangle = 1/2 ab sinC

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Right-angled triangles use sin=oppositehypotenuse\sin=\frac{\text{opposite}}{\text{hypotenuse}}, cos=adjacenthypotenuse\cos=\frac{\text{adjacent}}{\text{hypotenuse}} and tan=oppositeadjacent\tan=\frac{\text{opposite}}{\text{adjacent}}; inverse ratios recover an unknown acute angle from side lengths. For a scalene triangle, the sine rule asinA=bsinB=csinC\frac a{\sin A}=\frac b{\sin B}=\frac c{\sin C} needs an opposite side–angle pair.
  • The cosine rule c2=a2+b22abcosCc^2=a^2+b^2-2ab\cos C uses three sides or two sides and their included angle.
  • Area is 12absinC\frac12ab\sin C.
  • A sine-rule angle may have an alternative because sinθ=sin(180θ)\sin\theta=\sin(180^\circ-\theta); the triangle sum decides validity.
  • Examiners expect correctly paired labels, unrounded working and final rounding only as requested.
In a scalene triangle, each lowercase side is opposite its matching uppercase angle.
Worked example

Two sides of a triangle are 77 cm and 1010 cm and their included angle is 4848^\circ. Work out the third side to 33 significant figures.

  1. 1.Use the cosine rule: c2=72+1022(7)(10)cos48c^2=7^2+10^2-2(7)(10)\cos48^\circ.
  2. 2.Evaluate c255.3217c^2\approx55.3217.
  3. 3.Take the positive square root and round only the final value.

Answer: c=7.44 cmc=7.44\text{ cm} to 33 significant figures.

Common mistakes

  • Don't use the sine rule when no opposite side–angle pair is known.
  • Don't pair the 4848^\circ angle with a non-included pair of sides in the cosine rule.
  • Don't round cos48\cos48^\circ early and change the final length.

Exam tip

Mark the included angle and its opposite side on the diagram before selecting the cosine-rule form.

Tier 1 · Easy

ORIGINAL

1

Two sides of a triangle are 88 cm and 55 cm, and their included angle is 3030^\circ. Work out the area.

[2 marks]

Tier 2 · Standard

ORIGINAL

1

Two sides of a triangle are 77 cm and 99 cm, and their included angle is 6060^\circ. Work out the exact length of the third side.

[3 marks]

Tier 3 · Hard

ORIGINAL

1

In triangle ABCABC, A=30A=30^\circ, a=6a=6 cm and b=10b=10 cm. Work out both possible values of angle CC. Give each answer to 11 decimal place.

[4 marks]

G4

Use of Pythagoras' theorem in 2D and 3D; recognise Pythagorean triples

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • In a right-angled triangle, a2+b2=c2a^2+b^2=c^2, where cc is the hypotenuse opposite the right angle. The theorem can find a missing side or verify a right angle.
  • Required familiar triples include 3,4,53,4,5, 5,12,135,12,13, 8,15,178,15,17 and 7,24,257,24,25, together with their multiples.
  • In three dimensions, a face diagonal can be found first and used in a second right-angled triangle; for a cuboid this combines to d2=l2+w2+h2d^2=l^2+w^2+h^2.
  • Pythagoras is valid only when the right angle is established, and a calculated length must be positive.
  • Examiners expect the longest side to be treated as the hypotenuse and each 3D right-angled cross-section to be identified.
A cuboid’s body diagonal satisfies d² = l² + w² + h².
Worked example

A cuboid has side lengths 33 cm, 44 cm and 1212 cm. Work out its body diagonal.

  1. 1.Apply three-dimensional Pythagoras: d2=32+42+122d^2=3^2+4^2+12^2.
  2. 2.d2=9+16+144=169d^2=9+16+144=169.
  3. 3.Take the positive square root.

Answer: d=13 cmd=13\text{ cm}.

Common mistakes

  • Don't add the three side lengths instead of their squares.
  • Don't use only two dimensions and report the face diagonal 55 cm.
  • Don't use Pythagoras on a triangle without identifying a right angle.

Exam tip

In 3D, mark the face projection and body diagonal before writing the Pythagorean relation.

Tier 1 · Easy

ORIGINAL

1

A right-angled triangle has perpendicular sides 55 cm and 1212 cm. Work out the hypotenuse.

[1 mark]

Tier 2 · Standard

ORIGINAL

1

A cuboid has side lengths 66 cm, 66 cm and 77 cm. Work out the length of its body diagonal.

[3 marks]

Tier 3 · Hard

ORIGINAL

1

A cuboid has base dimensions 88 cm by 99 cm and body diagonal 1717 cm. Work out its volume.

[4 marks]

G5

Apply trigonometry and Pythagoras' theorem to 2 and 3 dimensional problems, including the angle between a line and a plane and between two planes

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • A 3D trigonometry problem should be reduced to a labelled right-angled cross-section.
  • The angle between a line and a plane is the angle between the line and its perpendicular projection onto that plane, not an angle with an arbitrary edge.
  • For the angle between two planes, take a cross-section perpendicular to their line of intersection; the angle between the two cross-section lines is the dihedral angle.
  • Pythagoras often supplies a face diagonal or projection before sine, cosine or tangent is applied.
  • Examiners expect the relevant projection, right angle and required angle to be identified, because using a visually convenient but incorrect triangle gives the wrong ratio.
The angle between a line and a plane is measured against the line’s perpendicular projection in the plane.
Worked example

A cuboid has base dimensions 66 cm by 88 cm and height 55 cm. Work out the angle between its body diagonal and the base, to 11 decimal place.

  1. 1.The body diagonal projects onto the base diagonal, whose length is 62+82=10\sqrt{6^2+8^2}=10 cm.
  2. 2.In the right-angled cross-section, tanθ=510=0.5\tan\theta=\frac{5}{10}=0.5.
  3. 3.θ=tan1(0.5)\theta=\tan^{-1}(0.5).

Answer: θ=26.6\theta=26.6^\circ to 11 decimal place.

Common mistakes

  • Don't measure the body diagonal’s angle against a 66 cm base edge instead of its 1010 cm projection.
  • Don't use the body diagonal as the adjacent side in the tangent ratio.
  • Don't find the complementary angle with the vertical height.

Exam tip

For a line–plane angle, identify the line’s projection on the plane before choosing a trigonometric ratio.

Tier 1 · Easy

ORIGINAL

1

A straight ramp is 1010 m long and rises vertically by 66 m. Work out the angle the ramp makes with the horizontal, to 11 decimal place.

[2 marks]

Tier 2 · Standard

ORIGINAL

1

A square-based pyramid has base side 1010 cm. Its apex is vertically above the centre of the base, and a sloping edge from the apex to a base vertex is 1313 cm. Work out the angle that this edge makes with the base, to 11 decimal place.

[3 marks]

Tier 3 · Hard

ORIGINAL

1

A cuboid ABCDEFGHABCDEFGH has AB=12AB=12 cm, BC=5BC=5 cm and vertical edge BF=8BF=8 cm. The base is ABCDABCD and EFGHEFGH is the top face. Work out the acute angle between plane ABGHABGH and the base plane ABCDABCD, to 11 decimal place.

[4 marks]

G6

Sketch and use graphs of y = sin x, y = cos x and y = tan x for angles of any size

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • In degrees, y=sinxy=\sin x and $y=\cos x$ have period 360360^\circ and range 1y1-1\leq y\leq1.
  • Sine passes through 00 at multiples of 180180^\circ; cosine begins at 11 when x=0x=0^\circ.
  • The tangent graph has period 180180^\circ, zeros at multiples of 180180^\circ and vertical asymptotes at 90+180n90^\circ+180^\circ n.
  • Graphs extend to angles of any size by repeating these periods in both directions.
  • Examiners expect smooth curves through exact key points, separate tangent branches that never cross their asymptotes, and correct use of graphs to read signs, intersections or approximate solutions.
Sine and cosine repeat every 360°; tangent repeats every 180° between vertical asymptotes.
Worked example

State the zeros and maximum point of y=sinxy=\sin x for 0x3600^\circ\leq x\leq360^\circ.

  1. 1.Sine is zero at the start, halfway point and end of one period.
  2. 2.Its maximum value 11 occurs one quarter of the way through the period.
  3. 3.Attach the corresponding angle coordinates.

Answer: Zeros at (0,0)(0^\circ,0), (180,0)(180^\circ,0) and (360,0)(360^\circ,0); maximum at (90,1)(90^\circ,1).

Common mistakes

  • Don't use period 180180^\circ for sine or cosine.
  • Don't draw tangent continuously through a vertical asymptote.
  • Don't join trigonometric key points with straight line segments.

Exam tip

Mark zeros, extrema and asymptotes first, then draw smooth periodic branches through them.

Tier 1 · Easy

ORIGINAL

1

Write down the coordinates of the minimum point of y=cosxy=\cos x for 0x3600^\circ\leq x\leq360^\circ.

[1 mark]

Tier 2 · Standard

ORIGINAL

1

Sketch y=tanxy=\tan x for 0x3600^\circ\leq x\leq360^\circ. Mark its zeros and vertical asymptotes.

[3 marks]

Tier 3 · Hard

ORIGINAL

1

Use the graphs of y=sinxy=\sin x and y=cosxy=\cos x to state the range of xx for which both values are positive and sinx>cosx\sin x>\cos x, where 0x3600^\circ\leq x\leq360^\circ.

[3 marks]

G7

Use the definitions of sin, cos and tan for any positive angle up to 360 degrees (measured in degrees only)

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • For any positive angle up to 360360^\circ, measured anticlockwise from the positive xx-axis, a point (x,y)(x,y) at distance rr from the origin gives cosθ=xr\cos\theta=\frac xr, sinθ=yr\sin\theta=\frac yr and tanθ=yx\tan\theta=\frac yx.
  • Sine is positive in quadrants I and II, cosine in I and IV, and tangent in I and III.
  • A reference angle gives the magnitude, while the quadrant fixes the sign and full angle.
  • Inverse trigonometric functions often return only a principal value, so quadrant information must be applied.
  • Examiners expect degree mode and an angle within the stated quadrant or interval.
Coordinate definitions and sign patterns for sine, cosine and tangent in four quadrants.
Worked example

The terminal point of angle θ\theta is (5,12)(-5,12), with 0<θ<1800^\circ<\theta<180^\circ. Work out sinθ\sin\theta, cosθ\cos\theta and tanθ\tan\theta exactly.

  1. 1.r=(5)2+122=13r=\sqrt{(-5)^2+12^2}=13.
  2. 2.sinθ=yr=1213\sin\theta=\frac{y}{r}=\frac{12}{13} and cosθ=xr=513\cos\theta=\frac{x}{r}=-\frac5{13}.
  3. 3.tanθ=yx=125=125\tan\theta=\frac{y}{x}=\frac{12}{-5}=-\frac{12}{5}.

Answer: sinθ=1213\sin\theta=\frac{12}{13}, cosθ=513\cos\theta=-\frac5{13} and tanθ=125\tan\theta=-\frac{12}{5}.

Common mistakes

  • Don't use distance r=17r=17 by adding the coordinate magnitudes instead of applying Pythagoras.
  • Don't make cosine positive even though the point lies in quadrant II.
  • Don't measure the angle clockwise from the positive xx-axis.

Exam tip

Write the quadrant sign and calculate rr before forming the three coordinate ratios.

Tier 1 · Easy

ORIGINAL

1

Work out the exact value of cos120\cos120^\circ.

[1 mark]

Tier 2 · Standard

ORIGINAL

1

180<θ<270180^\circ<\theta<270^\circ and sinθ=1213\sin\theta=-\frac{12}{13}. Work out the exact values of cosθ\cos\theta and tanθ\tan\theta.

[3 marks]

Tier 3 · Hard

ORIGINAL

1

0<θ<3600^\circ<\theta<360^\circ, tanθ=724\tan\theta=-\frac7{24} and cosθ>0\cos\theta>0. Work out sinθ\sin\theta exactly and θ\theta to 11 decimal place.

[4 marks]

G8

Knowledge and use of 30, 60, 90 triangles and 45, 45, 90 triangles

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • A 3030^\circ6060^\circ9090^\circ triangle has side ratio 1:3:21:\sqrt3:2, opposite 3030^\circ, 6060^\circ and 9090^\circ respectively. A 4545^\circ4545^\circ9090^\circ triangle has ratio 1:1:21:1:\sqrt2.
  • These ratios follow by halving an equilateral triangle or bisecting a square, and they generate exact trigonometric values such as sin30=12\sin30^\circ=\frac12, cos30=32\cos30^\circ=\frac{\sqrt3}{2} and sin45=22\sin45^\circ=\frac{\sqrt2}{2}.
  • They also produce exact lengths in regular polygons and composite geometry.
  • Scale every side by the same factor and retain exact surds unless a decimal is requested.
  • Examiners expect each ratio length to be matched to the angle opposite it.
The side ratios are 1:√3:2 for a 30°–60°–90° triangle and 1:1:√2 for a 45°–45°–90° triangle.
Worked example

A 3030^\circ6060^\circ9090^\circ triangle has hypotenuse 1010 cm. Work out the other two side lengths exactly.

  1. 1.The side ratio is 1:3:21:\sqrt3:2.
  2. 2.The scale factor is 10÷2=510\div2=5.
  3. 3.Multiply the two leg ratios by 55.

Answer: The shorter leg is 55 cm and the longer leg is 535\sqrt3 cm.

Common mistakes

  • Don't place the side of ratio 11 opposite 6060^\circ instead of 3030^\circ.
  • Don't scale the hypotenuse to 1010 but leave the other ratio lengths unscaled.
  • Don't round 535\sqrt3 when an exact length is requested.

Exam tip

Write the ratio beside the three corresponding sides before applying the common scale factor.

Tier 1 · Easy

ORIGINAL

1

Write down the exact value of sin30\sin30^\circ.

[1 mark]

Tier 2 · Standard

ORIGINAL

1

A right-angled isosceles triangle has hypotenuse 1414 cm. Work out the exact length of each equal side.

[2 marks]

Tier 3 · Hard

ORIGINAL

1

A regular hexagon has side length 88 cm. Work out the exact distance between a pair of opposite sides and hence the exact area of the hexagon.

[4 marks]

G9

Know and use tan = sin / cos and sin^2 + cos^2 = 1

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • The identities tanx=sinxcosx\tan x=\frac{\sin x}{\cos x} and sin2x+cos2x=1\sin^2x+\cos^2x=1 allow one trigonometric form to be replaced by another. Useful rearrangements are 1sin2x=cos2x1-\sin^2x=\cos^2x and 1cos2x=sin2x1-\cos^2x=\sin^2x.
  • They also solve equations by replacing squared terms or tangent with sine and cosine before factorising. When finding a missing value, the quadrant determines the sign of a square root.
  • In an identity proof, start from one side and transform it through valid equal expressions until the other is reached.
  • A common denominator often exposes an identity.
  • Examiners expect restrictions to be respected whenever a cancelled denominator could be zero.
Worked example

180<x<270180^\circ<x<270^\circ and cosx=1213\cos x=-\frac{12}{13}. Work out sinx\sin x and tanx\tan x exactly.

  1. 1.sin2x=1cos2x=1144169=25169\sin^2x=1-\cos^2x=1-\frac{144}{169}=\frac{25}{169}.
  2. 2.Quadrant III makes sine negative, so sinx=513\sin x=-\frac5{13}.
  3. 3.tanx=sinxcosx=5/1312/13=512\tan x=\frac{\sin x}{\cos x}=\frac{-5/13}{-12/13}=\frac5{12}.

Answer: sinx=513\sin x=-\frac5{13} and tanx=512\tan x=\frac5{12}.

Common mistakes

  • Don't take only the positive square root for sine despite the quadrant III restriction.
  • Don't write tanx=cosxsinx\tan x=\frac{\cos x}{\sin x} and obtain the reciprocal.
  • Don't cancel a trigonometric factor without noting values where the original denominator is zero.

Exam tip

After using sin2x+cos2x=1\sin^2x+\cos^2x=1, apply the quadrant sign before taking the final square root.

Tier 1 · Easy

ORIGINAL

1

xx is acute and sinx=35\sin x=\frac35. Work out the exact value of tanx\tan x.

[2 marks]

Tier 2 · Standard

ORIGINAL

1

Simplify 1cos2xsinx\frac{1-\cos^2x}{\sin x}.

[2 marks]

Tier 3 · Hard

ORIGINAL

1

Prove that sinx1+cosx+1+cosxsinx=2sinx\frac{\sin x}{1+\cos x}+\frac{1+\cos x}{\sin x}=\frac{2}{\sin x} for values of xx where both sides are defined.

[4 marks]

G10

Solution of simple trigonometric equations in given intervals

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • To solve a simple trigonometric equation, first isolate the trigonometric function and find a reference angle. Use its sign to choose the correct quadrants, then list every solution in the stated interval.
  • Sine and cosine repeat every 360360^\circ; tangent repeats every 180180^\circ.
  • A squared equation may produce positive and negative function values, so factor or take both square-root possibilities.
  • Interval endpoints must be included or excluded exactly as stated, and angles are measured in degrees for this specification.
  • Examiners expect all solutions, normally in ascending order, with substitution checks when factorisation or cancellation could introduce or remove a value.
The equation cos x = 1/2 has solutions in quadrants I and IV over one full turn.
Worked example

Solve 2cosx=12\cos x=1 for 0x<3600^\circ\leq x<360^\circ.

  1. 1.Isolate cosine: cosx=12\cos x=\frac12.
  2. 2.The reference angle is 6060^\circ.
  3. 3.Cosine is positive in quadrants I and IV, giving 6060^\circ and 36060360^\circ-60^\circ.

Answer: x=60x=60^\circ or x=300x=300^\circ.

Common mistakes

  • Don't report only the principal calculator value 6060^\circ.
  • Don't choose quadrant II because sine, not cosine, is positive there.
  • Don't include 360360^\circ when the interval endpoint is strict.

Exam tip

Write the reference angle and permitted quadrants before listing every solution in the interval.

Tier 1 · Easy

ORIGINAL

1

Solve sinx=1\sin x=1 for 0x3600^\circ\leq x\leq360^\circ.

[1 mark]

Tier 2 · Standard

ORIGINAL

1

Solve cosx=32\cos x=-\frac{\sqrt3}{2} for 0x3600^\circ\leq x\leq360^\circ.

[2 marks]

Tier 3 · Hard

ORIGINAL

1

Solve 2sin2x=3cosx2\sin^2x=3\cos x for 0x<3600^\circ\leq x<360^\circ.

[4 marks]

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