sin x = 0.5: two answers
Calculator: sin⁻¹ 0.5 = 30°
The hump is mirrored in x = 90°: the other is 180° − 30° = 150°
x = 30° or x = 150°
Graphs of sin, cos and tan Why does sin x = 0.5 have two answers between 0° and 360°, when the calculator gives one?
Higher · AQA · 45 minutes
Students move a level line across the sine wave and count where it meets it, then build sin and cos from heights on a circle, find each graph's key points and mirror lines, and use them to find the second answer the calculator never gives. Slides to present, a vote that shows who's stuck, practice and homework.
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This is y = sin x from 0° to 360°. The level line is y = k ÷ 2. Move k: how many times can a level line meet the wave? Can it meet it only once?
They cross at (30, 0.5) and (150, 0.5)
The idea in one minute
Turn a point round a circle of radius 1. Its height is sin x and its distance across is cos x. So sin x starts at 0, rises to 1 at 90°, is 0 again at 180°, −1 at 270° and 0 at 360°; cos x is the same wave started at its top, 1 at 0°. tan x = sin x ÷ cos x has no value where cos x is 0, at 90° and 270°, and repeats every 180°. For most heights k between −1 and 1, a level line y = k meets the sine wave twice between 0° and 360°, at two angles that are mirror images: the calculator gives one, and the graph's symmetry gives the other.
The lesson goes picture first. Students move a level line across the sine wave and count how often it meets it before anyone explains. Then they build sin and cos from heights on a circle, find each graph's key points and mirror lines, and use them to solve sin x = k and cos x = k for both answers, including a negative k. It is Higher.
By the end, students can
- sketch y = sin x, y = cos x and y = tan x from 0° to 360°, with their key points
- explain the shapes of sin and cos from heights and distances on a circle
- use each graph's mirror line to find a second solution
- solve sin x = k and cos x = k between 0° and 360°, for k positive or negative
Key words
- sine
- cosine
- tangent
- period
- symmetry
- asymptote
- solution
Before this lesson
Students should use sin, cos, tan and sin⁻¹ on a calculator in degrees, know that sin 30° = 0.5, and describe a graph's translation.
Taught in Transforming graphs.
The lesson, step by step
60 minutes in total. Each step says what changes for this length.
5 min 1 · Hook · students think
How many answers?
Students count how often a level line meets the sine wave before anyone explains.
Say“This is y = sin x from 0° to 360°. Your calculator says sin⁻¹ 0.5 = 30°. Is 30° the only angle between 0° and 360° with sin x = 0.5? Write your answer.”
Ask- Where does the line y = 0.5 meet the wave?
- Where must the line be to meet the wave only once? Never?
Watch forSaying one answer, because the calculator gives one: keep it. The next step shows where the second comes from.
For the teacher
Give a minute of silent thinking, then vote. Many will say one: the calculator gives one.
Move k slowly. Two crossings for most heights, one at the top and the bottom, three on the axis (0°, 180°, 360°), none above 1 or below −1. Don't explain yet: the question is where the second answer comes from.
Take a vote with Ask the class: how many angles between 0° and 360° have sin x = 0.5, one or two? Then move k from −3 to 3 and count the crossings each time.
Shorter: Vote only; move k in the next step (−2 min).
12 min 2 · Explain · you explain
From a circle to three graphs
Heights on a circle give sin; then cos and tan, their key points and mirror lines; then two answers to sin x = k.
Say“Sin x is the height of a point turning round a circle. Every feature of the graph comes from that.”
Ask- Where on the circle is the point at 90°? So what is sin 90°?
- Which angle between 90° and 180° has the same height as 30°?
- Why can't tan x have a value at 90°?
Watch forPutting the top of the sine wave at 180°: the top of the circle is at 90°.
cos x = 0.5: a different mirror
Calculator: cos⁻¹ 0.5 = 60°
The cosine wave is mirrored in x = 180°: the other is 360° − 60° = 300°
x = 60° or x = 300°
cos is sin moved 90° left
cos x = sin(x + 90°)
So the cosine wave is the sine wave translated by the vector (−90, 0)
For the teacher
First slide: turn a point round a circle of radius 1, starting on the right. Its height is sin x: 0 at 0°, 1 at the top (90°), 0 at 180°, −1 at the bottom (270°), 0 again at 360°. At 30° it is halfway up: sin 30° = 0.5. Its distance across is cos x.
Second and third slides: y = sin x and y = cos x. cos x starts at the top, 1 at 0°, and is the sine wave moved 90° to the left: cos x = sin(x + 90°), a translation from the transforming graphs lesson. The sine wave's first hump is mirrored in x = 90°; the cosine wave between 0° and 360° is mirrored in x = 180°. Both repeat every 360° and stay between −1 and 1. Fourth slide: tan x = sin x ÷ cos x has no value where cos x = 0, at 90° and 270°: asymptotes, never joined. It repeats every 180°.
Fifth slide: sin x = 0.5. The line y = 0.5 meets the wave twice. The calculator gives 30°, which is 60° before the mirror line x = 90°; the other answer is 60° after it, 150° (180° − 30°). Sixth slide: for cos the mirror is x = 180°: 60° after 0° matches 60° before 360°, so cos x = 0.5 gives 60° and 300° (360° − x). For a negative k, the dip below the axis is the hump upside down, mirrored in x = 270°. Read the second answer from the graph; the rules come from it. Ask each time: is the second answer above or below the axis, as k is?
Vote before each graph (Ask the class): where is the top of the sine wave? What is cos 0°? Then, on the last slide: where is the second answer to sin x = 0.5?
Shorter: Show tan on its slide without the vote; skip the cos example (−4 min).
8 min 3 · Your turn · students write
Worked, then yours
Each worked example beside a twin that students solve themselves, to 1 decimal place.
Say“Watch me do one. Sketch first, then the calculator. Then you do its twin.”
Ask- Is k above or below the axis? So where are the answers?
- Sin or cos? So which mirror line?
Watch forLeaving −30° from the calculator as an answer: it is not between 0° and 360°.
sin x = k: two answers
sin x = 0.8
Calculator: 53.1°
Mirror in 90°: 180° − 53.1° = 126.9°
Then theirs
sin x = 0.3, to 1 decimal place
A negative k
sin x = −0.5: the calculator gives −30°, not in range
Below the axis: the dip between 180° and 360°, mirrored in x = 270°
It is the hump upside down: 30° after 180° and 30° before 360°, so 210° and 330°
Then theirs
sin x = −0.8, to 1 decimal place
cos x = k
cos x = 0.3
Calculator: 72.5°
Mirror in 180°: 360° − 72.5° = 287.5°
Then theirs
cos x = −0.4, to 1 decimal place
For the teacher
Work the first example aloud: sketch, draw y = k, count the crossings, then use the calculator. Students solve its twin while you watch who finds the second answer from the mirror.
Do the same for each pair. The second pair catches students who keep a negative calculator answer; the third, students who use 180° − x for cos.
Open the answers slide once most have finished their twins.
Shorter: Do the first and third pairs only (−3 min).
25 min 4 · Who's stuck? · students vote
Vote, then practise
One question shows who's stuck; then practice in levels, and a stretch.
For this length: The stretch, then: sketch y = sin(x − 90°) and say which graph it is
Say“Choose an answer on your own. Nobody talks until everyone has chosen.”
Ask- Sketch it: where does y = 0.6 meet the wave?
- Put your second answer into the calculator: is its sine 0.6?
Watch forAnswers not checked: every angle given should go back into the calculator and give the k in the question.
Sketch the graph and the line before using the calculator. Give angles in degrees, to 1 decimal place unless they are exact. The stretch is optional.
The question that shows who's stuck
- sin⁻¹ 0.6 = 36.9°. Solve sin x = 0.6 for 0° ≤ x ≤ 360°.
In your Classroom, as they answer (an example) - A9
- B8
- C5
- D3
Each wrong answer is a different mistake. While they answer, your Classroom shows who chose which, so you know in seconds whom to help with what.
- sin 90° = ___, cos 90° = ___ and cos 180° = ___
- Which graph goes through the point (0°, 1)?
- y = tan x has no value at two angles between 0° and 360°: the smaller is ___ degrees and the larger is ___ degrees
- cos 30° = √3/2. Solve cos x = √3/2 for 0° ≤ x ≤ 360°: the smaller x = ___ and the larger x = ___
- How many solutions does sin x = 1.5 have between 0° and 360°?
- sin 40° = 0.643 (to 3 decimal places). Which other angle between 0° and 360° has the same sine?
- sin 60° = √3/2. Solve sin x = −√3/2 for 0° ≤ x ≤ 360°: the smaller x = ___ and the larger x = ___
- What is the greatest value of y = sin x + 2?
- Stretch: solve tan x = 1 for 0° ≤ x ≤ 360°: the smaller x = ___ and the larger x = ___
Support · For question 4: sketch y = cos x and the line y = √3/2, about 0.87. How many times do they meet? The cosine wave is mirrored in x = 180°: where is the second meeting?
For the teacher
Everyone chooses before anyone explains. Then ask every student: how many times does y = 0.6 meet the wave? Is your second angle above or below the axis? Put it into the calculator: is its sine 0.6?
Questions 1–2 build fluency; 3–4 change the look; 5–6 ask for reasons; 7–8 are exam-style, 8 a translation from the transforming graphs lesson. The stretch uses tan.
Students show their working on paper or the board; the quiz keeps their final answers.
Whatever the vote, before they practise: sketch the wave with y = 0.6 on the board, ask how many times they meet and on which side of the mirror line x = 90°, and have everyone put each angle they chose into the calculator.
Shorter: After the vote, questions 1–4 only (−6 min).
Longer: The stretch, then: sketch y = sin(x − 90°) and say which graph it is (+10 min).
10 min 5 · Exit · students show
Solve it, show it
One equation with a negative k shows who finds both answers from the graph.
For this length: Check both answers on the calculator, and explain any that don't work
Say“Solve cos x = −0.5 for 0° ≤ x ≤ 360°. Sketch y = cos x and the line y = −0.5, and mark both answers.”
Watch forOne answer only, or 180° − 120° = 60°, which is above the axis.
A sketch with both answers marked.
- Solve cos x = −0.5 for 0° ≤ x ≤ 360°. Sketch y = cos x and the line y = −0.5, and mark both answers on your sketch.
For the teacher
Solve cos x = −0.5 for 0° ≤ x ≤ 360°.
Sketch y = cos x with the line y = −0.5, and mark both answers on it.
The sketches show who reads both answers from the graph: start the next lesson with the hook's line at a negative height if few do. Set the homework.
Longer: Check both answers on the calculator, and explain any that don't work (+5 min).
One to one: for every equation, have the student sketch the graph and the line y = k first and count the crossings, then use the calculator, and check both answers by putting them back in.
Homework: sketches and two answers
Sketch the graph and the line y = k before using the calculator. Give angles to 1 decimal place unless they are exact.
- Sketch y = sin x and y = cos x for 0° ≤ x ≤ 360° on one pair of axes. Label where each crosses the x-axis, and its highest and lowest points.
- Use the graph of y = cos x to find the angle between 0° and 360°, other than 70°, whose cosine is cos 70°.
- Solve sin x = 0.4 for 0° ≤ x ≤ 360°.
- Solve cos x = −0.7 for 0° ≤ x ≤ 360°.
- Sketch y = tan x for 0° ≤ x ≤ 360°, with its asymptotes. Then solve tan x = −1.
- Explain why sin x = 2 has no solutions, and why sin x = 1 has only one between 0° and 360°.
Print the worksheet with working space and homework. Teacher answers come with your Library copy.
Teaching graphs of sin, cos and tan: common questions
Why does sin x = k have two solutions between 0° and 360°?
Because sin x is a height on a circle, and a point turning round the circle is at each height twice in one turn: once on the way up or down on the right, once on the left. On the graph, a level line y = k meets the wave twice. The two angles are mirror images: in 90° for a positive k, in 270° for a negative k. Since sin(180° − x) = sin x, one answer always gives the other.
How do students remember the shapes of sin, cos and tan?
Build them rather than memorise them. The height of a point on a circle gives sin: 0, up to 1 at 90°, 0, down to −1 at 270°, 0. The distance across gives cos, which starts at 1. tan is sin ÷ cos, so it has no value where cos is 0, at 90° and 270°. A short table of calculator values every 30° confirms each shape.
Should students find the second answer from the graph or from a rule?
From the graph first. A sketch of the wave with the line y = k shows how many answers there are and roughly where, and the mirror line gives the second one. Rules such as 180° − x for sin and 360° − x for cos come from that picture; learnt alone, they get swapped, or used when the calculator's answer is negative and out of range.
Is this trig graphs lesson Higher only?
Yes. Recognising and sketching the graphs of sin, cos and tan, and using them to solve equations, is Higher content. Students need right-angled trigonometry on a calculator first; the translation questions use the transforming graphs lesson.
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Yes. Read the lesson, print the worksheet and open its slides without signing in; nothing you draw there is kept. A free teacher account keeps the whole lesson in your Library to change and teach, with the quizzes, the answers and the answer slides.
Curriculum reference: AQA GCSE Mathematics 8300: A12 (the graphs of y = sin x, y = cos x and y = tan x, Higher) and A13 (translations of a given function, with trigonometric graphs, Higher).
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