Mathematics · Simulation and virtual lab Free
Trigonometry: unit circle, graphs and a clinometer
Turn the angle on the unit circle and watch the sine and cosine graphs being drawn. Then use trigonometry outside: measure the height of a tree.
- You will learn to
- Link the unit circle to the graphs of sin θ, cos θ and tan θ.
- Use the signs of the ratios in each quadrant (All, Sin, Tan, Cos).
- Calculate heights with the angle of elevation and describe transformations of y = sin x.
- You will measure
- sin, cos and tan of angles; angle of elevation and the height of a tree (m); amplitude and period.
- Time needed
- About 30–40 min
- Equipment
- Nothing extra. Real version: clinometer (protractor, straw, string and a small weight), tape measure, graph paper.
- Price
- Free: no payment, no account.
Virtual lab
Follow the steps with the simulation above (go back to the simulation ↑). Each time you tap Record, a row is added to the table and a point to the graph.
Aim
To link the unit circle to the graphs of sin θ, cos θ and tan θ, to find the signs of the ratios in each quadrant, to calculate heights using the angle of elevation, and to describe transformations of y = sin x.
What to do
- Choose Unit circle and graphs. Set θ = 0°, 30°, 60°, … up to 360° and tap Record each time.
- Look at the graph below the table: your points build the curves y = sin θ and y = cos θ. Where are they positive? Where negative?
- Turn θ slowly and read the quadrant sign rule (All, Sin, Tan, Cos). Check it with θ = 150°, 210° and 330°.
- Show the tan graph. What happens near 90° and 270°? Why?
- Choose Clinometer. Read the angle of elevation at d = 10 m, 20 m and 30 m and record. Calculate the height h = d tan θ + 1.5 each time.
- Choose Transformations. Change a, then b, then c, then d (one at a time) and record. Describe the effect of each.
Results
Questions
Using the unit circle, explain why sin² θ + cos² θ = 1.
The point P(cos θ, sin θ) is on a circle of radius 1. By Pythagoras, cos² θ + sin² θ = 1² = 1.
Find the exact values of sin 30°, cos 60° and tan 45°.
sin 30° = 1/2, cos 60° = 1/2, tan 45° = 1.
In which quadrant is sin θ positive and cos θ negative? Give an example.
The second quadrant (90° < θ < 180°), for example θ = 150°: sin 150° = 0.5 and cos 150° = −0.866.
Why is tan 90° undefined?
tan θ = sin θ ÷ cos θ, and cos 90° = 0. Division by zero is not possible, so the graph has an asymptote at 90°.
From 25 m away, the angle of elevation of the top of a tree is 32°. The eye is 1.5 m above the ground. Find the height of the tree.
h = 25 × tan 32° + 1.5 = 25 × 0.625 + 1.5 = 15.6 + 1.5 = 17.1 m.
State the amplitude and the period of y = 3 sin 2x.
Amplitude 3; period 360° ÷ 2 = 180°.
For teachers: expected results and how the model works
Unit circle: values are exact to 3 decimal places; the quadrant shading and the ASTC rule follow θ. Clinometer: the tree height is random between 9 and 25 m (new tree button); the angle is shown to the nearest 0.5°, as on a real clinometer, so the calculated height is usually within about 0.3 m of the true height (the “Check with a tape measure” button reveals it). Transformations: the period bracket shows one full cycle; c shifts the graph by −c ÷ b degrees.
Exam practice Exam set
National-exam style questions on this simulation (15 marks). Try each one on paper first, then open the marking scheme.
Solve sin x = 0.5 for 0° ≤ x ≤ 360°. (3 marks)
Marking scheme: x = 30° (1); sin is also positive in the second quadrant, so x = 180° − 30° = 150° (1); x = 30° or 150° (1).
From the top of a building 20 m high, the angle of depression of a car on the road is 35°. Calculate the distance from the foot of the building to the car. (3 marks)
Marking scheme: tan 35° = 20 ÷ d (1); d = 20 ÷ tan 35° = 20 ÷ 0.700 (1) = 28.6 m (1).
For y = 2 sin(x + 30°) + 1: (a) state the amplitude, the period and the range; (b) find the maximum value and the first value of x where it occurs. (5 marks)
Marking scheme: (a) Amplitude 2 (1), period 360° (1), range −1 ≤ y ≤ 3 (1). (b) Maximum y = 3 (1) when x + 30° = 90°, so x = 60° (1).
Given that cos θ = −0.6 and θ is in the third quadrant, find sin θ and tan θ without a calculator. (4 marks)
Marking scheme: sin² θ = 1 − 0.36 = 0.64, so sin θ = ±0.8 (1); in the third quadrant sin θ is negative: sin θ = −0.8 (1); tan θ = sin θ ÷ cos θ = −0.8 ÷ −0.6 (1) = 4/3 (1).
Donat Sciences and MathsMathematics · Virtual lab worksheet
School: ........................................ Name: ................................ Class: ........ Date: ............
Trigonometry: unit circle, graphs and a clinometer
Aim: To link the unit circle to the graphs of sin θ, cos θ and tan θ, to find the signs of the ratios in each quadrant, to calculate heights using the angle of elevation, and to describe transformations of y = sin x.
Apparatus (real laboratory): clinometer (protractor, straw, string and a small weight), tape measure, graph paper.
What to do (with the simulation on www.donatstem.co.rw)
- Choose Unit circle and graphs. Set θ = 0°, 30°, 60°, … up to 360° and tap Record each time.
- Look at the graph below the table: your points build the curves y = sin θ and y = cos θ. Where are they positive? Where negative?
- Turn θ slowly and read the quadrant sign rule (All, Sin, Tan, Cos). Check it with θ = 150°, 210° and 330°.
- Show the tan graph. What happens near 90° and 270°? Why?
- Choose Clinometer. Read the angle of elevation at d = 10 m, 20 m and 30 m and record. Calculate the height h = d tan θ + 1.5 each time.
- Choose Transformations. Change a, then b, then c, then d (one at a time) and record. Describe the effect of each.
Results
| # | Activity | θ (°) | sin θ | cos θ | tan θ | Result |
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Graph
On the same axes plot sin θ and cos θ for θ = 0°, 30°, 60°, … 360° (horizontal axis θ from 0° to 360°, vertical axis from −1 to 1). Join each set of points with a smooth curve.
Questions
- Using the unit circle, explain why sin² θ + cos² θ = 1. [2]
- Find the exact values of sin 30°, cos 60° and tan 45°. [2]
- In which quadrant is sin θ positive and cos θ negative? Give an example. [2]
- Why is tan 90° undefined? [2]
- From 25 m away, the angle of elevation of the top of a tree is 32°. The eye is 1.5 m above the ground. Find the height of the tree. [2]
- State the amplitude and the period of y = 3 sin 2x. [2]
- Solve sin x = 0.5 for 0° ≤ x ≤ 360°. [3]
- From the top of a building 20 m high, the angle of depression of a car on the road is 35°. Calculate the distance from the foot of the building to the car. [3]
- For y = 2 sin(x + 30°) + 1: (a) state the amplitude, the period and the range; (b) find the maximum value and the first value of x where it occurs. [5]
- Given that cos θ = −0.6 and θ is in the third quadrant, find sin θ and tan θ without a calculator. [4]
Donat Sciences and MathsMathematics · Virtual lab worksheet
School: ........................................ Name: ................................ Class: ........ Date: ............
Marking scheme: Trigonometry: unit circle, graphs and a clinometer
Expected results
Unit circle: values are exact to 3 decimal places; the quadrant shading and the ASTC rule follow θ. Clinometer: the tree height is random between 9 and 25 m (new tree button); the angle is shown to the nearest 0.5°, as on a real clinometer, so the calculated height is usually within about 0.3 m of the true height (the “Check with a tape measure” button reveals it). Transformations: the period bracket shows one full cycle; c shifts the graph by −c ÷ b degrees.
Answers
- Using the unit circle, explain why sin² θ + cos² θ = 1. [2]
The point P(cos θ, sin θ) is on a circle of radius 1. By Pythagoras, cos² θ + sin² θ = 1² = 1.
- Find the exact values of sin 30°, cos 60° and tan 45°. [2]
sin 30° = 1/2, cos 60° = 1/2, tan 45° = 1.
- In which quadrant is sin θ positive and cos θ negative? Give an example. [2]
The second quadrant (90° < θ < 180°), for example θ = 150°: sin 150° = 0.5 and cos 150° = −0.866.
- Why is tan 90° undefined? [2]
tan θ = sin θ ÷ cos θ, and cos 90° = 0. Division by zero is not possible, so the graph has an asymptote at 90°.
- From 25 m away, the angle of elevation of the top of a tree is 32°. The eye is 1.5 m above the ground. Find the height of the tree. [2]
h = 25 × tan 32° + 1.5 = 25 × 0.625 + 1.5 = 15.6 + 1.5 = 17.1 m.
- State the amplitude and the period of y = 3 sin 2x. [2]
Amplitude 3; period 360° ÷ 2 = 180°.
- Solve sin x = 0.5 for 0° ≤ x ≤ 360°. [3]
x = 30° (1); sin is also positive in the second quadrant, so x = 180° − 30° = 150° (1); x = 30° or 150° (1).
- From the top of a building 20 m high, the angle of depression of a car on the road is 35°. Calculate the distance from the foot of the building to the car. [3]
tan 35° = 20 ÷ d (1); d = 20 ÷ tan 35° = 20 ÷ 0.700 (1) = 28.6 m (1).
- For y = 2 sin(x + 30°) + 1: (a) state the amplitude, the period and the range; (b) find the maximum value and the first value of x where it occurs. [5]
(a) Amplitude 2 (1), period 360° (1), range −1 ≤ y ≤ 3 (1). (b) Maximum y = 3 (1) when x + 30° = 90°, so x = 60° (1).
- Given that cos θ = −0.6 and θ is in the third quadrant, find sin θ and tan θ without a calculator. [4]
sin² θ = 1 − 0.36 = 0.64, so sin θ = ±0.8 (1); in the third quadrant sin θ is negative: sin θ = −0.8 (1); tan θ = sin θ ÷ cos θ = −0.8 ÷ −0.6 (1) = 4/3 (1).