Geometry, trigonometry and functions

6 visual steps · 3 worked examples · 5 exercises with solutions

First, let’s see the idea.

How can an angle become a point and an identity?

Move forward one step at a time. Change the values and observe what happens.

One turn, three languagesxyr = 3 u
Drawing description

A circle of radius 3 u is centred at the origin; the axes have the same scale.

Step 1 of 6

Start at the center

Change r and compare distance to center in several directions.

The origin is the axes' intersection. Every point on the green circle is distance r from it; increasing r enlarges the circle.

Radius
3 u

What if you change this?

1 u4 u
0 °360 °

Before moving on

Choose r=2 and θ=150°. Which quadrant, and what is each coordinate's sign?

Compare with the explanation

Second quadrant: x=−√3≈−1,73 and y=1. The squares sum to 4.

You decide when to mark this lesson as studied.

One turn, three languages

Slide the drawing to see the details.

Explore the full explanationThe foundations, principles, worked examples and all the exercises, at your own pace.

What you will understand

  • Relate coordinates, distances and geometric equations.
  • Use the trigonometric circle to build periodic functions.
  • Solve problems using domains, equivalence and geometric verification.

Before you start

You do not need to know everything by heart. The explanation revisits the necessary ideas and shows how they connect. If a foundation is still unclear, you can return to these lessons:

Review the three languages

A point can be given by coordinates; a distance, by a segment; a geometric condition, by an equation. Coordinate geometry makes these descriptions comparable. For two points A=(a,b) and B=(c,d), the horizontal and vertical differences form the legs of a right triangle. Pythagoras gives AB=√[(c−a)²+(d−b)²]. The result is a nonnegative distance and has the same unit as the coordinates.

A function adds a correspondence: each permitted input has a unique output. An entire circle is not the graph of a single function y=f(x), since many horizontal coordinates correspond to two vertical coordinates. It can, however, be divided into branches or described by a parameter. Choosing the appropriate representation avoids trying to apply function rules to relations that are not functions in that form.

Equations of geometric loci

Points at distance r>0 from C=(h,k) satisfy (x−h)²+(y−k)²=r². This equation follows directly from the distance formula, squared with both sides nonnegative. The centre is (h,k), not (−h,−k). Completing squares lets us recognise the same form when the expression is expanded.

For example, x²+y²−6x+4y−12=0 becomes (x−3)²+(y+2)²=25. We added 9 and 4 to complete the squares and compensated for the change on the other side. Therefore the centre is (3,−2) and the radius is 5. If the right side were negative, there would be no real points; if zero, there would be one point. Algebraic manipulation must produce a coherent interpretation, not just a familiar expression.

The circle explains the identities

On a circle of radius r, a point at angle θ can be written as x=r cos θ and y=r sin θ. Substituting into x²+y²=r² and dividing by r² gives cos²θ+sin²θ=1. Thus, the fundamental identity is Pythagoras in angular form. The signs of sine and cosine come from the coordinates: in the second quadrant, x<0 and y>0, so cosine is negative and sine positive.

A radian measures arc length divided by radius: θ=s/r. One turn corresponds to 2π rad. Using radians is essential in differentiation formulas such as (sin x)′=cos x, because the angular scale is part of the rate. Tangent is sin θ/cos θ, defined only when cos θ≠0. Dividing an equation by cos θ without checking its zeros can erase legitimate solutions or introduce meaningless expressions.

Parameterise circular motion

With a parameter t, we can describe the circle by x(t)=h+r cos(ωt+φ) and y(t)=k+r sin(ωt+φ). Here r is the radius, ω the angular rate and φ the initial phase. If t is time in seconds, ω has unit rad/s; the phase and argument are angular quantities. The period for ω≠0 is T=2π/|ω|. The sign of ω determines the direction of travel.

The vertical projection is a sinusoidal function with amplitude r and midline k. An observer seeing only y(t) loses the horizontal coordinate but retains periodicity. Differentiating, x′=−rω sin(ωt+φ) and y′=rω cos(ωt+φ). The speed is r|ω|, since sin²+cos²=1. This example shows geometry, trigonometry and derivatives confirming one description, without confusing a circular path with a sinusoidal graph.

Intersections and additional solutions

To find the intersection of a line and a circle, solve the system of their two equations. Substituting y=mx+b into the circle generally produces a quadratic equation in x. A positive discriminant indicates two distinct intersections, zero a tangency and negative no real intersection. Each x root needs its corresponding y coordinate.

Squaring requires care with equivalence. Solving √(x+2)=x starts with x≥0, since the left side is nonnegative. Squaring gives x+2=x², with roots 2 and −1; the second fails in the original equation. Transformations may enlarge the candidate set. Returning to the original condition is compulsory, especially with roots, denominators or inverse functions with restricted domains. A graph is a good visual check, but does not replace domain analysis.

Build a model and question it

For a periodic height, the form H(t)=c+A sin(ωt+φ) separates four decisions: c is the midline, |A| the amplitude, 2π/|ω| the period and φ fixes the initial position. The maximum and minimum values are c+|A| and c−|A|. One initial condition alone may not determine all parameters: the direction of change or another instant must also be known.

The synthesis should end with consistency questions. Do the units allow the terms to be added? Is the input in the domain? Does the result respect the geometry? How many solutions are expected? A negative distance reveals an error; two positions with the same height may be perfectly natural. Instead of collecting separate formulas, identify the object, choose coordinates or a parameter, solve and translate the result back into the original situation. Connections help both discovery and correction.

Worked examples, without skipping steps

Follow the reasoning first. Then cover the solution and try to reconstruct it in your own words.

Example 1

Line and circle

Find the intersections of x²+y²=25 and y=3.

  1. Substitute the fixed vertical coordinate: x²+9=25, so x²=16.
  2. There are two real roots, x=−4 and x=4. Each must be combined with y=3.
  3. Obtain the points (−4,3) and (4,3); both are distance 5 from the origin.
Result

The intersections are (−4,3) and (4,3).

Check the result

In both cases, 16+9=25. Symmetry about the vertical axis explains the pair of points.

Example 2

A wheel and its projection

A point moves according to x=2 cos(πt/3), y=2 sin(πt/3), with coordinates in metres and t in seconds. Determine the period, position at t=1 and speed.

  1. The radius is 2 m and the angular rate is π/3 rad/s. Thus T=2π/(π/3)=6 s.
  2. At t=1, the angle is π/3: x=1 m and y=√3 m.
  3. The velocity components are −(2π/3)sin(πt/3) and (2π/3)cos(πt/3). The magnitude is 2π/3 m/s.
Result

T=6 s; position (1,√3) m; speed approximately 2,094 m/s.

Check the result

x²+y²=1+3=4 confirms the radius. A turn of length 4π m in 6 s produces the same speed.

Example 3

Height with an initial phase

A height oscillates between 2 and 8 m, has period 12 s and starts at maximum height. Construct a model.

  1. The midline is (8+2)/2=5 m and the amplitude is (8−2)/2=3 m.
  2. The angular rate is 2π/12=π/6 rad/s. Cosine starts at value 1.
  3. Choose H(t)=5+3 cos(πt/6), with t in seconds.
Result

H(t)=5+3 cos(πt/6) m, over the observation interval in which the oscillation is regular.

Check the result

H(0)=8; H(6)=2; H(12)=8. The three tests verify extrema and repetition.

Now try it yourself

There is no timer. Write what you know, choose a relationship and justify each step. Hints and solutions are available whenever you need them.

Exercise 1

Identify the centre and radius of (x+2)²+(y−1)²=9.
A hint
Compare with the formula for distance from the centre.
See the worked solution
The centre is (−2,1) and the radius is 3. The sign change comes from x−(−2), and the radius is the positive square root of 9.

Exercise 2

Solve sin θ=1/2 on [0,2π).
A hint
Use quadrants with positive sine.
See the worked solution
The reference angle is π/6. In the first quadrant θ=π/6 and in the second θ=5π/6. The remaining quadrants have negative sine, and the interval endpoints add no solutions.

Exercise 3

Solve √(x+6)=x.
A hint
Require x≥0 before squaring.
See the worked solution
Squaring gives x²−x−6=0, so x=3 or x=−2. The restriction removes −2. Substituting 3 gives √9=3; therefore the only solution is 3.

Exercise 4

A circle has radius 4 cm. What arc corresponds to 1,5 rad?
A hint
The radian measure is arc length divided by radius.
See the worked solution
s=rθ=4·1,5=6 cm. The ratio 6 cm/4 cm is 1,5, dimensionless; the unit rad identifies the angular quantity.

Exercise 5

Does the projection y=7−2 sin(4t) have negative amplitude?
A hint
Distinguish the coefficient from the distance to the midline.
See the worked solution
The amplitude is |−2|=2, the midline is 7 and the period is 2π/4=π/2. The negative sign reverses the phase, not the amplitude's sign. The range is [5,9].

Mistakes worth avoiding

An entire circle is always y=f(x).
It may require two branches or a parameterisation to represent all its positions.
Squaring always preserves exactly the solutions.
It may add candidates; check all of them in the original relation.
Phase does not affect the initial condition.
Phase shifts the oscillation and determines the position at the initial instant.

The essentials to take with you

  • Use Pythagoras to connect distance and equation.
  • Radians relate arc length and radius.
  • Parameterisation represents two coordinates with one input.
  • Verify the solution in algebra and geometry.

If you still cannot explain one of these ideas, return to the corresponding section and work through an example again. Recognising the difficulty is part of studying.

References for further study

Text and examples prepared for this learning path. These sources help check the principles and continue studying:

The organisation by school year is a study proposal. Pedagogical review by teachers remains a stage of the project.