MATH AI / UNIT 7 / HL

UNIT 7 · HL

Unit circle and rotation

Describe a direction. Track a rotation. Keep the magnitude.

You will connect the vectors and transformations you already know to sine, cosine and tangent. Then you will use those connections to explain the rotation previously associated with the number i.

01   Directions on a circle02   Coordinates and graphs03   Rotation matrices

Work in pairs unless a task says “individually”. Write predictions, labelled sketches and explanations in your notebook. Use radians on your GDC. Attempt a question before opening its worked answer.

SL students: open the independent geometry investigations.

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TRANSFER · PAIRS · 6 MINUTES

Separate length from direction

A controller describes an arrow by the vector v = (3, 4). It needs a second arrow pointing in exactly the same direction, but with magnitude 1.

Find the new vector. Explain why your operation changes its length without changing its direction.

Now consider w = (−6, 8). Apply the same idea without finding an angle. Write a rule that works for any non-zero vector (a, b).

After recording your answer, select Show unit vector to compare the first pair.

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UNIT-CIRCLE DEFINITION · 2 MINUTES

The coordinates name the direction

For an acute angle θ, your unit vector has components cos θ and sin θ. The familiar triangle ratios now give the coordinates directly because its length is 1.

A unit circle has centre (0, 0) and radius 1. We extend the meaning of sine and cosine to every directed angle by defining

P = (cos θ, sin θ).

Measure θ from the positive x-axis: anticlockwise is positive; clockwise is negative. A coordinate is signed even though a triangle side length is not.

Move P beyond π/2. Explain what the negative horizontal coordinate means.

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DIRECTED ANGLES · PAIRS · 3 MINUTES

The same endpoint can record different turns

A unit vector starts at (1, 0). Three instructions rotate it through π/3, 7π/3 and −5π/3 radians.

Predict whether the final arrows coincide. Explain how the journeys differ.

On a circle of radius 1, one radian sweeps an arc of length 1. A full turn is 2π radians, so adding 2π changes the turn count but not the final direction.

Enter each instruction in the radians box. You may type pi/3, 7pi/3 or −5pi/3.

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INVESTIGATION 1 · PAIRS · 6 MINUTES

Construct two directions with the same height

A sensor reports the vertical coordinate of a point on the unit circle as exactly 0.6. It does not report the horizontal coordinate.

Construct every possible point. Explain why your list is complete.

Call the acute direction α. Express the other direction in terms of α without using inverse trigonometry. Then state what extra information would distinguish the two points.

Use the circle to test your predictions. The dashed line marks y = 0.6; it is not a second circle.

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INVESTIGATION 1 · GENERALISE · 5 MINUTES

Transform a direction without recalculating

Let the pinned unit vector be P = (a, b) at an acute angle α. Predict the image of P under each transformation before pressing its button.

Reflect P in the y-axis.

Rotate P anticlockwise through π.

Reflect P in the x-axis.

Rotate P anticlockwise through π/2.

Write each new coordinate pair and its directed angle. Which moves swap the components?

Every button acts on the purple pinned starting point, not on the last image. You can move P and pin a different starting point to test a general rule.

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TANGENT · PAIRS · 5 MINUTES

Tangent records a gradient

The line through O and a unit-circle point P = (x, y) has gradient y/x, provided x ≠ 0. Substituting the coordinate definitions gives

tan θ = sin θ / cos θ = y/x.

Find two different unit directions whose gradient is −3/4. Does knowing tangent identify one direction?

Sketch both arrows before using Opposite direction. Explain why a half-turn leaves the gradient unchanged even though the vector reverses.

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TANGENT · EXPLAIN · 3 MINUTES

What happens at a vertical direction?

Compare directions just before π/2, exactly at π/2 and just after π/2. Predict the gradient before selecting each position.

Explain why tangent is undefined at π/2, rather than zero or an ordinary very large number.

The vertical coordinate approaches 1 while the horizontal coordinate approaches 0. The sign of that small horizontal coordinate changes as the arrow passes the vertical.

Give every vertical direction using π and an integer k.

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INDIVIDUAL CHECK · 3 MINUTES

Use both coordinates to check a claim

A rotating arrow has length 1. A programmer makes these statements about its direction.

A. If two positions have the same sine, they are the same point.

B. If two positions have the same sine and the same cosine, they are the same point.

C. If two positions have the same tangent, their arrows point in the same direction.

Which statement is always true? Give a diagram or counterexample for the other two.

Your evidence

Write one choice and two counterexamples in your notebook. Compare only after everyone has committed.

A counterexample must obey the original condition: all proposed points must actually lie on the unit circle.

INVESTIGATION 2 · PAIRS · 6 MINUTES

Draw a graph from the rotating point

Sketch the vertical coordinate of a unit vector over one full turn, starting at (1,0). Mark the zeros and extrema. Select Sine, then play one quarter-turn at a time to test your sketch. The graph records (θ, sin θ); its horizontal axis is angle, not P’s spatial x-coordinate.

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INVESTIGATION 2 · COMPARE · 5 MINUTES

Change the coordinate that is recorded

Sketch the horizontal coordinate of the same rotating unit vector before selecting Cosine. Use Both and Full curves to compare the signals. Explain their different starting values and find the horizontal shift that makes them coincide. Relate that shift to a quarter-turn.

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INVESTIGATION 2 · GENERALISE · 3 MINUTES

Explain the graph using the circle

A graphing window shows only part of a function. Use the rotating vector to explain the whole functions, including negative angles and repeated turns.

Why are sine and cosine defined for every real angle, but restricted to values from −1 to 1?

For each function, state the period and identify the zeros and extrema over 0 ≤ θ ≤ 2π. Explain at least one feature using a point on the circle, not a memorised graph.

Two views of one point

Circle: P = (cos θ, sin θ)
Sine graph: (θ, sin θ)
Cosine graph: (θ, cos θ)

The horizontal coordinate of a graph point is θ. It is not the horizontal coordinate of P.

Use this distinction when drawing or interpreting the traces.

INDIVIDUAL TRANSFER · 3 MINUTES

Recover a sensor signal

At a particular angle θ, a rotating unit vector has sin θ = −0.96 and points to the right of the y-axis. The readings are exact.

Find its horizontal coordinate. Predict the sine and cosine readings after an anticlockwise quarter-turn.

Explain how your predictions would appear on the two coordinate graphs. Use the vector transformation rather than calculating θ.

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MAGNITUDE · PAIRS · 4 MINUTES

Use the fixed magnitude to obtain an identity

Every point P = (x, y) on the unit circle has distance 1 from O, regardless of the signs of its coordinates.

Write a relationship between x and y that rotation cannot change. Translate it into sine and cosine.

Justify the relationship for every point, including points on the axes. A table of a few correct readings is evidence, but it does not explain every possible angle.

After writing your argument, select Check magnitude and move the point to test it.

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IDENTITY · APPLY · 4 MINUTES

One component is not a complete direction

A unit vector points into quadrant III, so π < θ < 3π/2. Its vertical component is sin θ = −5/13.

Find cos θ and tan θ. Then find the vector of magnitude 26 in this same direction.

Show the squared quantity before choosing a sign. Explain how the interval for θ removes the ambiguity created when you take a square root.

Keep the sign and the magnitude separate

x² + y² = 1
x = ±√(1 − y²)

The square root determines a component’s magnitude. The position on the circle determines its sign.

For a vector of magnitude r > 0, multiply both unit components by r.

IDENTITY · CRITIQUE · 2 MINUTES

Can these readings describe a pure rotation?

A device reports the exact unit-direction components cos θ = −0.7 and sin θ = −0.8 for the same angle.

Can both readings be correct? Give a test that does not require finding θ.

A second device reports the vector (−7, −8) but does not claim it has length 1. Explain why that vector is valid, and how you would recover its unit direction.

A useful distinction

Coordinates describe the actual endpoint.

Unit coordinates describe its direction independently of its length.

For rounded sensor data, a tolerance would be needed before deciding whether a discrepancy is significant. No tolerance is needed for the exact claim here.

INVESTIGATION 3 · PAIRS · 5 MINUTES

Build a rotation from two familiar directions

A rotation through θ about O sends e₁ = (1, 0) to (cos θ, sin θ). The second basis vector, e₂ = (0, 1), begins a quarter-turn anticlockwise from e₁.

Where must e₂ go under the same rotation? Use a coordinate sketch before selecting “Show second image”.

The two transformed arrows must keep length 1, stay perpendicular and preserve their anticlockwise order. Test your prediction by changing θ.

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INVESTIGATION 3 · EXPLAIN · 4 MINUTES

Read the matrix from the images

The columns of a transformation matrix are the images of e₁ and e₂. Place the two vectors from your diagram in the correct columns.

Rθ = cos θ−sin θsin θcos θ

Explain why each column has this form, then multiply Rθ by a general vector (x, y).

Use column vectors for the matrix multiplication. The bracketed matrix acts on any vector, not only a unit vector.

Check a familiar rotation

Set θ = π/2. Use the axis point (0,1) to identify cosine and sine, rather than a table of special-angle values.

Rπ/2 = 0−110

The result sends (x, y) to (−y, x), exactly as in your quarter-turn construction.

INVESTIGATION 3 · APPLY · 4 MINUTES

Audit a rotation command

A controller uses the matrix

A = −0.6−0.80.8−0.6

It acts on the vector (5, 0). All entries are exact.

Find the image and verify its length. Identify the rotation angle in 0 ≤ θ < 2π, to three significant figures.

Explain which matrix entries give cosine and sine. Do not use an inverse tangent result without checking its direction.

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ROTATION · JUSTIFY · 3 MINUTES

Why does every length survive?

A few successful numerical tests do not establish that a matrix preserves every vector’s length. Use the general image

x′ = x cos θ − y sin θ
y′ = x sin θ + y cos θ

Show that x′² + y′² = x² + y².

Expand the two squares. Identify which terms cancel and where the unit-circle identity is needed. Then explain the result in a sentence.

The identity has a geometric job

The equation cos²θ + sin²θ = 1 does more than fill in a missing trig value.

It guarantees that this matrix changes direction without stretching the vector.

Both the algebra and the moving circle express the same invariant.

COMPLEX-NUMBER CONNECTION · 3 MINUTES

Return to the rotation represented by i

You previously met the idea that multiplying by the imaginary number i rotates a point through π/2 anticlockwise. Its coordinate matrix is

J = 0−110 = Rπ/2.

Starting from (1,0), predict the position after one, two, three and four applications of J.

Use Apply J to check each prediction. Explain why two applications give the negative of the starting vector, while four restore it.

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TRANSFER TO THE NEXT TOPIC · 3 MINUTES

Combine magnitude and direction

A vector of magnitude r > 0 points at directed angle θ. Start from (1,0), rotate through θ, then scale by r.

v = r(cos θ, sin θ).

Would scaling first and rotating second give the same result? Explain using components or matrices.

A vector of magnitude 5 points in the direction of (−3,4). After a quarter-turn, state its coordinates, its magnitude and the change in direction.

Two independent pieces of information

Magnitude tells us how far the endpoint is from the origin.

Direction tells us where it lies around the origin.

The unit circle separates these two pieces: first describe direction on radius 1, then rescale.

INDEPENDENT EXIT · 8 MINUTES

Show what you can transfer

Work independently in your notebook. Use a labelled diagram and a reason, not just a final value. Set your GDC to radians.

1 · Components and direction

A unit vector has cos θ = −8/17 and π < θ < 3π/2. Find sin θ and tan θ. Give the coordinates of the vector of magnitude 34 in that direction.

2 · Circle and graph

Explain why the sine graph is negative for π < θ < 2π. Identify where cosine is zero in that interval and state what happens to tangent there.

3 · A quarter-turn

Rotate your magnitude-34 vector from question 1 anticlockwise through π/2. State its new coordinates and explain why its length stays 34.

Check only after completing all three. Keep your first attempt visible so that your correction shows what changed.

REFERENCE

Keep these connections together

Coordinates and angle

P = (cos θ, sin θ)

θ is measured from the positive x-axis. Positive turns are anticlockwise. Adding 2π gives the same point.

Gradient and magnitude

tan θ = sin θ / cos θ
cos²θ + sin²θ = 1

Tangent is undefined when cos θ = 0. Both coordinates, not tangent alone, identify a unit direction.

From rotation to graphs

Sine records the vertical coordinate; cosine records the horizontal coordinate. Both have range [−1,1] and period 2π.

cos θ = sin(θ + π/2)

From direction to a vector

v = r(cos θ, sin θ),   r > 0

Uniform scaling changes magnitude. Rotation changes direction. A quarter-turn sends (x,y) to (−y,x).

The unit circle gives the geometric foundation. Graphical trigonometric equations are a separate next step.

Worked answers

Check the normalization

The magnitudes are 5 and 10. Divide every component by the positive magnitude:

u = (3/5, 4/5),   ŵ = (−3/5, 4/5).

For a non-zero vector, the unit vector in its direction is

(a, b) / √(a² + b²).

Both components have the same positive scale factor, so their ratio and direction are unchanged. The zero vector has no direction to preserve.

Check the three rotations

All three end in the same direction. The second makes one extra anticlockwise turn. The third reaches that direction by rotating clockwise through 5π/3.

cos(θ + 2πk) = cos θ
sin(θ + 2πk) = sin θ,   k ∈ ℤ.

The endpoint alone cannot tell you how many complete turns occurred.

Compare the possible directions

The points are (0.8, 0.6) and (−0.8, 0.6). A horizontal line at height 0.6 cuts the circle at exactly these two points. Reflection in the y-axis changes α to π − α and preserves height.

sin(π − α) = sin α
cos(π − α) = −cos α.

The quadrant, or the sign of the horizontal coordinate, distinguishes the two positions. Extra full turns do not produce new points.

Check the coordinate rules
TransformationImageAngle
Reflect in y-axis(−a, b)π − α
Half-turn(−a, −b)α + π
Reflect in x-axis(a, −b)−α, or 2π − α
Anticlockwise quarter-turn(−b, a)α + π/2

In particular, cos(−α) = cos α and sin(−α) = −sin α. A quarter-turn swaps the component roles and changes one sign.

Check the two directions

The unit points are (4/5, −3/5) and (−4/5, 3/5). Both have gradient −3/4, but their arrows point in opposite directions.

tan(θ + π) = (−sin θ)/(−cos θ) = tan θ.

Tangent has period π. Sine and cosine together distinguish the two directions; tangent alone does not.

Check the limiting behaviour

Just before π/2, y/x is positive and very large. Just after it, y/x is negative with large magnitude. At π/2 itself, division by zero is undefined.

cos θ = 0 when θ = π/2 + kπ,   k ∈ ℤ.

These are all the angles where tangent is undefined. They become the vertical asymptotes in the tangent graph; zero is not an appropriate value.

Check after committing your answer

B is true: sine and cosine together specify both coordinates. The turn count may still differ.

A is false: (0.8, 0.6) and (−0.8, 0.6) have the same sine but are different points.

C is false: (0.8, −0.6) and (−0.8, 0.6) have equal tangent but opposite directions. This is why gradient alone will not be enough to describe a complex-number direction.

Check the sine trace

The trace is y = sin θ. Over 0 ≤ θ ≤ 2π it passes through (0,0), (π/2,1), (π,0), (3π/2,−1) and (2π,0).

The curve rises and falls continuously because it records the moving point’s vertical coordinate. A second complete turn repeats the same pattern.

Compare the two graphs

The second signal is y = cos θ. Its main points over one turn are (0,1), (π/2,0), (π,−1), (3π/2,0), (2π,1).

cos θ = sin(θ + π/2).

Shifting the sine graph π/2 to the left gives the cosine graph. Both have amplitude 1 and period 2π. This is a change of which component is observed, not a change in the vector’s length.

Check the graph features

Any real directed angle gives a position, so both domains are ℝ. A unit vector cannot have either component outside [−1,1]. Both ranges are [−1,1] and both periods are 2π.

On [0, 2π]SineCosine
Zeros0, π, 2ππ/2, 3π/2
Maximum 1π/20, 2π
Minimum −13π/2π

For example, sine reaches 1 at the top of the circle because the vertical coordinate is 1 there.

Check the sensor prediction

The squared magnitude is 1, so x² = 1 − 0.96² = 0.0784. The point is to the right, so x = 0.28.

A quarter-turn maps (0.28, −0.96) to (0.96, 0.28). The new cosine reading is 0.96; the new sine reading is 0.28.

sin(θ + π/2) = cos θ
cos(θ + π/2) = −sin θ.

The new vertical signal equals the old horizontal signal. The new horizontal signal is the negative of the old vertical signal.

Reveal the geometric argument
|OP|² = x² + y² = 1
cos²θ + sin²θ = 1.

This follows from the distance formula. It remains valid when a coordinate is negative or zero. The notation sin²θ means (sin θ)², not sin(θ²).

An identity holds for every angle in its domain. Here the fixed radius explains why the identity is true for all real θ.

Check the components and rescaling
cos²θ = 1 − 25/169 = 144/169.

Quadrant III has a negative horizontal coordinate, so cos θ = −12/13. The quotient gives tan θ = 5/12.

26(cos θ, sin θ) = (−24, −10).

Its magnitude is √(576 + 100) = 26. A positive rescaling changes magnitude but not direction. Choosing +12/13 would put the vector in quadrant IV, contrary to the given interval.

Check the distinction

The first claim is impossible: (−0.7)² + (−0.8)² = 1.13, not 1. The conclusion uses the stated exact readings.

The second vector has magnitude √113 and a valid direction. Its unit direction is (−7/√113, −8/√113).

A point need not be on the unit circle to describe a vector. Its coordinates equal cosine and sine only after normalization.

Check the second image

The image of e₂ remains a quarter-turn anticlockwise from the image of e₁. Applying (a, b) → (−b, a) gives

Rθe₁ = (cos θ, sin θ)
Rθe₂ = (−sin θ, cos θ).

Their dot product is zero. Both squared magnitudes equal cos²θ + sin²θ = 1.

Check the general image
x′ = x cos θ − y sin θ
y′ = x sin θ + y cos θ.

The first column is the image of (1,0). The second is the image of (0,1). Since (x,y) = x e₁ + y e₂, the output is x times the first image plus y times the second.

Check the rotation audit

The image is (−3,4), which has length 5. Comparing A with Rθ gives cos θ = −0.6 and sin θ = 0.8, so the direction is in quadrant II.

θ = cos⁻¹(−0.6) ≈ 2.21 radians.

The angle is approximately 126.9°. Using tan⁻¹(0.8/−0.6) alone returns an angle in the wrong quadrant. Both components determine the correct direction.

Check the proof
x′² + y′²
= x²cos²θ − 2xy cos θ sin θ + y²sin²θ
+ x²sin²θ + 2xy sin θ cos θ + y²cos²θ
= x²(cos²θ + sin²θ) + y²(sin²θ + cos²θ)
= x² + y².

The mixed terms cancel and the identity replaces each bracket by 1. Therefore the squared magnitude, and hence the magnitude, is preserved for every vector.

Check the rotation cycle
(1,0) → (0,1) → (−1,0) → (0,−1) → (1,0).

One application makes a quarter-turn. Two make a half-turn, so J² = −I. Four make a full turn, so J⁴ = I.

Here I is the identity matrix, while i is the imaginary number. The matrix J represents the coordinate effect of multiplication by i. This is the geometric connection behind i² = −1.

Check magnitude and rotation

Yes. Uniform scaling by r commutes with rotation: rRθv = Rθ(rv). Both components receive the same scale factor.

The starting vector is (−3,4), already of magnitude 5. A quarter-turn maps it to (−4,−3). Its magnitude remains 5, and its direction increases by π/2.

Describing a point by magnitude and directed angle will be useful when we represent complex numbers geometrically.

Check the exit after completing it

1. sin²θ = 1 − 64/289 = 225/289. In quadrant III, sin θ = −15/17; tan θ = 15/8. The rescaled vector is (−16,−30).

2. The point is below the x-axis throughout (π,2π), so its vertical coordinate is negative. At θ = 3π/2, cosine is 0. The direction is vertical and tangent is undefined.

3. The image is (30,−16). Its squared magnitude is 900 + 256 = 1156 = 34². A quarter-turn maps (x,y) to (−y,x), preserving x²+y².