Flashcards · IB Diploma Programme Mathematics: Analysis and Approaches · Topic 3

Geometry and trigonometry flashcards

87 cards on Topic 3, subtopics 3.1 to 3.18. Pick a subtopic or study them all.

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3.1 3D geometry: distance, midpoint, volume and surface area

Distance between two points in 3D, and their midpoint: d = √((x₁ − x₂)² + (y₁ − y₂)² + (z₁ − z₂)²) Midpoint ((x₁ + x₂)/2, (y₁ + y₂)/2, (z₁ + z₂)/2)
A(0, 3, −1) and B(4, 5, 3). Find AB and the midpoint of AB.: AB = √(16 + 4 + 16) = √36 = 6. Midpoint (2, 4, 1).
Right cone: volume, curved surface area and slant height: V = (1/3)πr²h; curved surface area = πrl; l² = r² + h². Example: r = 5, h = 12 gives l = 13, curved area 65π, V = 100π.
Sphere and hemisphere: volume and surface area: Sphere: V = (4/3)πr³, S = 4πr². Hemisphere: V = (2/3)πr³; curved surface 2πr²; total with the flat face 3πr².
Volume of a right pyramid: V = (1/3)Ah, where A is the base area and h the perpendicular height (apex directly above the centre of the base).
Angle between a line and a plane: The angle between the line and its projection onto the plane. Drop a perpendicular from a point on the line to the plane, join to where the line meets the plane, and use the right-angled triangle.

3.2 The sine rule, cosine rule and area of a triangle

Sine rule: a/sin A = b/sin B = c/sin C Use it when you know a side and its opposite angle.
Cosine rule, for a side and for an angle: c² = a² + b² − 2ab cos C cos C = (a² + b² − c²)/(2ab)
Area of a triangle from two sides and the included angle: Area = ½ab sin C, where C is the angle between sides a and b.
How do you choose a rule for a non-right-angled triangle?: Side and its opposite angle known: sine rule. Two sides and the included angle: cosine rule for the third side. All three sides: cosine rule for an angle. Area: ½ab sin C.
A triangle has sides 3, 5 and 7. Find its largest angle.: Largest angle is opposite 7: cos C = (9 + 25 − 49)/30 = −1/2, so C = 120°. (The cosine rule handles an obtuse angle safely.)

3.3 Applications of right and non-right-angled trigonometry

Angle of elevation vs angle of depression: Both are measured from the horizontal: elevation upwards, depression downwards to the line of sight. They are equal for the same two points.
Bearing: A three-figure angle measured clockwise from north, such as 007° or 245°.
A cliff is 80 m high. The angle of depression to a boat is 14°. How far is the boat from the foot of the cliff?: tan 14° = 80/d, so d = 80/tan 14° = 321 m (3 s.f.).
How do you set up a bearings problem?: 1. Draw a north line at every point used. 2. Mark each bearing clockwise from north. 3. Use back bearing = bearing ± 180° for angles inside the triangle. 4. Solve the triangle, then convert back to a bearing.
3D trigonometry: what does SL set, compared with HL?: The guide says SL examinations set only right-angled trigonometry on 3D shapes. Find a right-angled triangle inside the solid and use SOH CAH TOA or Pythagoras.

3.4 The circle: radian measure, arc length and area of a sector

Radian: The angle at the centre of a circle subtended by an arc equal in length to the radius. 2π rad = 360°, so π rad = 180° and 1 rad ≈ 57.3°.
Arc length and sector area (θ in radians): Arc length l = rθ Sector area A = ½r²θ
Sector perimeter vs arc length: Arc alone: rθ. Sector perimeter: rθ + 2r (include the two radii).
Area of a segment cut off by a chord: Segment = sector − triangle = ½r²θ − ½r² sin θ (θ in radians).
Convert 135° to radians, and 2 radians to degrees: 135 × π/180 = 3π/4 2 × 180/π = 115° (3 s.f.)

3.5 Definitions of cosθ, sinθ and tanθ using the unit circle

Unit circle definitions of cos θ, sin θ and tan θ: The point at angle θ, anticlockwise from the positive x-axis, is (cos θ, sin θ). tan θ = sin θ/cos θ.
Exact values of sin and cos at π/6, π/4 and π/3: sin: 1/2, √2/2, √3/2 cos: √3/2, √2/2, 1/2 (tan: √3/3, 1, √3)
Signs of sin, cos and tan in each quadrant: First: all positive. Second: sin only. Third: tan only. Fourth: cos only.
Ambiguous case of the sine rule: how do you find both possible angles?: 1. Find sin of the unknown angle with the sine rule. 2. Take θ = arcsin(value) and also 180° − θ. 3. Keep 180° − θ only if it plus the given angle is less than 180°.
Equation of the line through the origin at 120° to the positive x-axis: y = x tan θ, and tan 120° = −√3, so y = −√3 x.
Exact values of sin(4π/3) and cos(−5π/4): sin(4π/3) = −√3/2 (third quadrant). cos(−5π/4) = −√2/2 (second quadrant).

3.6 The Pythagorean identity and double angle identities

Pythagorean identity: cos²θ + sin²θ = 1
Double angle identities for sin 2θ and cos 2θ: sin 2θ = 2 sin θ cos θ cos 2θ = cos²θ − sin²θ = 2cos²θ − 1 = 1 − 2sin²θ
cos θ = −1/3 and θ is in the second quadrant. Find sin θ and tan θ exactly.: sin²θ = 1 − 1/9 = 8/9, and sin is positive in the second quadrant, so sin θ = 2√2/3. tan θ = (2√2/3) ÷ (−1/3) = −2√2
Given sin x = 1/3, find cos 2x: cos 2x = 1 − 2sin²x = 1 − 2/9 = 7/9

3.7 The circular functions sin x, cos x and tan x

f(x) = a sin(b(x + c)) + d: amplitude, period, principal axis: Amplitude |a|; maximum d + |a|, minimum d − |a|. Period 2π/b (b is not the period). Principal axis y = d. Horizontal shift −c.
Period of tan(bx), and of y = tan(x/2): Period of tan(bx) is π/b, so tan(x/2) has period π ÷ (1/2) = 2π.
Amplitude, period and range of f(x) = 5 cos 4x − 2: Amplitude 5, period 2π/4 = π/2, range −7 ≤ f(x) ≤ 3.
Horizontal shift of y = sin(2x − π/2): Factorise b first: sin(2(x − π/4)), so a shift of π/4 to the right (not π/2).
Symmetry results for the circular functions: cos(−x), sin(−x), sin(π + x): cos(−x) = cos x (cos is even) sin(−x) = −sin x (sin is odd) sin(π + x) = −sin x
A wheel of radius 6 m, centre 8 m above the ground, turns once every 30 s, starting at the bottom. Model the height.: h(t) = 8 − 6 cos(πt/15), since b = 2π/30 = π/15. Minimum 2 m, maximum 14 m.

3.8 Solving trigonometric equations

How do you solve sin(bx + c) = k in a given interval?: 1. Isolate the trig function. 2. Change the interval to match bx + c. 3. List every angle in that interval with the right sign. 4. Solve for x and check each value is in the original interval.
Solve 2sin²x = sin x for 0 ≤ x ≤ 2π: Factorise, do not divide: sin x(2 sin x − 1) = 0. sin x = 0: x = 0, π, 2π. sin x = 1/2: x = π/6, 5π/6.
How do you solve a quadratic in one trig ratio?: 1. Use cos²x = 1 − sin²x (or a cos 2x form) so only one ratio remains. 2. Rearrange to = 0 and factorise. 3. Reject any value of sin or cos outside [−1, 1]. 4. Solve each remaining equation in the interval.
Solve 2 cos x = −1 for 0 ≤ x ≤ 2π: cos x = −1/2, so x = 2π/3 or x = 4π/3.
Solve tan²x = 3 for 0 ≤ x ≤ π: tan x = ±√3, so x = π/3 or x = 2π/3.

3.9 Reciprocal trigonometric ratios and inverse circular functions (AHL only)

Reciprocal trigonometric ratiosHL: sec θ = 1/cos θ cosec θ = 1/sin θ cot θ = 1/tan θ
Pythagorean identities with tan and cotHL: 1 + tan²θ = sec²θ 1 + cot²θ = cosec²θ
Domains and ranges of arcsin, arccos and arctanHL: arcsin x: domain [−1, 1], range [−π/2, π/2] arccos x: domain [−1, 1], range [0, π] arctan x: domain ℝ, range ]−π/2, π/2[
Find arccos(−1/2)HL: 2π/3, not −π/3, because the range of arccos is [0, π].
Given tan θ = 2, find sec²θHL: sec²θ = 1 + tan²θ = 1 + 4 = 5

3.10 Compound angle identities (AHL only)

Compound angle identities for sin(A ± B)HL: sin(A ± B) = sin A cos B ± cos A sin B
Compound angle identities for cos(A ± B)HL: cos(A ± B) = cos A cos B ∓ sin A sin B The sign in the middle is the opposite: cos(A + B) has a minus.
Compound angle identity for tan(A ± B)HL: tan(A ± B) = (tan A ± tan B)/(1 ∓ tan A tan B)
Double angle identity for tan 2θ, and tan 2A when tan A = 3HL: tan 2θ = 2 tan θ/(1 − tan²θ) (put B = A in tan(A + B)). tan 2A = 6/(1 − 9) = −3/4
Find the exact value of cos 15°HL: cos(45° − 30°) = cos 45° cos 30° + sin 45° sin 30° = (√2/2)(√3/2) + (√2/2)(1/2) = (√6 + √2)/4

3.11 Relationships between trigonometric functions and symmetry of their graphs (AHL only)

sin(π − θ), cos(π − θ) and tan(π − θ)HL: sin(π − θ) = sin θ cos(π − θ) = −cos θ tan(π − θ) = −tan θ
sin x = 0.4 has the solution α in [0, π/2]. Find the other solution in [0, π].HL: π − α, because sin(π − θ) = sin θ.
sin(π/2 − θ) and cos(π/2 − θ)HL: sin(π/2 − θ) = cos θ and cos(π/2 − θ) = sin θ. The graph of cos x is the graph of sin x translated π/2 to the left.
What symmetry of the graph of y = sin x does sin(π − x) = sin x show?HL: The graph is symmetric about the line x = π/2: points at equal distances either side of π/2 have the same height.

3.12 Vectors: concept, components and algebraic/geometric operations (AHL only)

Displacement vectorHL: AB = b − a ("end minus start").
Base vectorsHL: i, j, k are unit vectors along the axes; (v₁, v₂, v₃) = v₁i + v₂j + v₃k.
ParallelHL: w = kv for some scalar k (non-zero vectors).
MagnitudeHL: |v| = √(v₁² + v₂² + v₃²). Distance AB = |AB|.
Find the unit vector in the direction of (2, −6, 3)HL: |v| = √(4 + 36 + 9) = 7, so the unit vector is v/|v| = (2/7, −6/7, 3/7).
How do you write a vector proof?HL: 1. Name the given vectors, for example OA = a, OC = c. 2. Write other vectors in terms of them, using AB = b − a. 3. Show the fact: a scalar multiple (parallel), zero scalar product (perpendicular) or equal magnitudes. 4. State the result in words.

3.13 The scalar product of two vectors (AHL only)

Scalar product: component form and geometric formHL: v · w = v₁w₁ + v₂w₂ + v₃w₃ = |v||w| cos θ, where θ is the angle between v and w. The result is a number.
Find t so that (t, 2, −1) is perpendicular to (3, t, 4)HL: Perpendicular ⇔ scalar product 0: 3t + 2t − 4 = 0, so t = 4/5.
Find the angle between lines with directions (1, 2, 2) and (2, −1, 2)HL: cos θ = |2 − 2 + 4|/(3 × 3) = 4/9, so θ = 63.6°. For lines take the acute angle, using the modulus.
Properties of the scalar product to quoteHL: v · w = w · v; v · v = |v|²; v · (w + u) = v · w + v · u; (kv) · w = k(v · w). For non-zero vectors, v · w = 0 means perpendicular.

3.14 Vector equation of a line in two and three dimensions (AHL only)

Vector equation of a lineHL: r = a + λb, where a is the position vector of a point on the line and b is a direction vector. For a line through A and B, use b = AB.
Parametric and Cartesian forms of a line through (x₀, y₀, z₀) with direction (l, m, n)HL: Parametric: x = x₀ + λl, y = y₀ + λm, z = z₀ + λn Cartesian: (x − x₀)/l = (y − y₀)/m = (z − z₀)/n
Write r = (3, −1, 0) + λ(2, 5, −4) in Cartesian formHL: (x − 3)/2 = (y + 1)/5 = z/(−4)
An object has position r = (4, 1) + t(6, −8). What are its velocity and speed?HL: Velocity (6, −8) (the direction vector); speed |(6, −8)| = 10 units per unit time. The start position is (4, 1).

3.15 Coincident, parallel, intersecting and skew lines (AHL only)

Skew linesHL: Non-parallel lines that do not intersect (three dimensions only).
How do you decide whether two lines are coincident, parallel, intersecting or skew?HL: 1. Directions multiples: parallel or coincident; test one point of L₁ in L₂. 2. Otherwise set a + λb = c + μd (different parameters). 3. Solve two component equations for λ and μ. 4. Third equation true: intersect. False: skew.
Two lines have directions (1, −2, 3) and (−2, 4, −6). Can they be skew?HL: No. The second direction is −2 times the first, so the lines are parallel or coincident.
Two objects move along lines with time t. Collision vs paths crossing: what is the difference in method?HL: Collision: use the same t for both objects (same place at the same time). Paths crossing: use different parameters for the two lines.

3.16 The vector product of two vectors (AHL only)

Vector product of v and w in componentsHL: v × w = (v₂w₃ − v₃w₂, v₃w₁ − v₁w₃, v₁w₂ − v₂w₁) The result is a vector perpendicular to both v and w. Check with scalar products.
Magnitude of v × w, and the areas it givesHL: |v × w| = |v||w| sin θ. Parallelogram with sides v and w: area |v × w|. Triangle ABC: area ½|AB × AC|.
Find (1, 0, 2) × (0, 1, 3) and the area of the parallelogram with these sidesHL: (0 × 3 − 2 × 1, 2 × 0 − 1 × 3, 1 × 1 − 0 × 0) = (−2, −3, 1) Area = √(4 + 9 + 1) = √14
Properties of the vector productHL: v × w = −w × v; v × v = 0; it distributes over addition; (kv) × w = k(v × w). A zero vector product means the vectors are parallel.

3.17 Vector equations of a plane (AHL only)

Normal to a planeHL: A vector perpendicular to every direction in the plane.
Three forms of the equation of a planeHL: Vector: r = a + λb + μc Normal (scalar product): r · n = a · n Cartesian: ax + by + cz = d, with normal (a, b, c)
How do you find the plane through three points A, B and C?HL: 1. Find AB and AC. 2. n = AB × AC (simplify by a common factor). 3. d = a · n; write ax + by + cz = d. 4. Check the other two points.
Find the Cartesian equation of the plane through (1, −2, 4) with normal (3, 1, −2)HL: d = 3(1) + 1(−2) − 2(4) = −7, so 3x + y − 2z = −7.

3.18 Intersections of lines and planes (AHL only)

How do you find where a line meets a plane?HL: 1. Write the line in parametric form. 2. Substitute x, y, z into the plane equation. 3. Solve for the parameter and substitute back for the point.
How do you find the line of intersection of two planes?HL: 1. Direction = n₁ × n₂. 2. Set one coordinate to 0 and solve the two equations for a point. 3. Write r = point + t(direction). Check the point in both planes.
Angle between a line and a plane, and between two planesHL: Line–plane: sin φ = |b · n|/(|b||n|) (sin, because n is at 90° to the plane). Plane–plane: cos θ = |n₁ · n₂|/(|n₁||n₂|).
Three planes: what can the solution set be?HL: A single point, a line, a plane, or no solution. With no solution, either at least two planes are parallel or they form a triangular prism: check the normals.
Find the angle between planes with normals (1, 1, 0) and (0, 1, 1)HL: cos θ = 1/(√2 × √2) = 1/2, so θ = 60°.