# Investigating the Idea of Sin

Purpose

In this unit we explore the sine and related functions to find out specific values, relations between them, and applications.

Achievement Objectives
GM5-10: Apply trigonometric ratios and Pythagoras' theorem in two dimensions.
Specific Learning Outcomes
• use sin to solve problems involving right-angled triangles
• solve equations of the form sin(θ) = a, for θ between –180º and 360º
• state the value of sin(θ) in special cases
• graph y = sin(θ)
• describe some of the ways in which the sine, cosine and tangent functions are related
Description of Mathematics

This unit takes an investigative approach to sine by encouraging students to start with a definition and develop the standard results themselves based on this. It is best if this unit is completed after the similar units on tan and cos: Investigating the Idea of Tan and Investigating the Idea of Cos.

Required Resource Materials

Strips of paper or light cardboard to make rulers calibrated to non-standard unit lengths.

Activity

#### Session 1

Here we revise the basic values of sin, cos, and tan, and where these values come from.

1. Begin this session with a light-hearted test (with a serious purpose). This is revision of material learned in two previous units. Present, in random order, each of the values that have tans and cosines that are easy to name: 0º, 30º, 60º, 120º, 180º, 45º, 135º, 225º, -45º, -135º, -120º, 90º, -90º, 270º, and so on. Get the students to write down, based on their understanding of the definition (and their memorisation of √3 and its reciprocal, √3/2, 1/2, and 1/√2), the tan and cos of each of these angles. Then give them questions like this:
If you know that an angle has a cos of -1/√2, name two values that angle could have?
These light-hearted tests are very important because they consolidate the ideas presented above. Mostly the answers depend on little more than a sound understanding of what tan and cos mean. The only exceptions are the cosines of multiples of 45º, and the surd values, but these involve memorising simple facts.

2. In this session students are introduced to the notion of sin of an angle. ‘Sin’ is short for the fuller word ‘sine’. Sine is, for reasons which will become clear, like tan and cos, what is called a circular function.

3. Get each student to draw a set of axes and a unit circle (say 5cm in radius). Thus 5cm is equivalent to one unit in length.

4. Next get the students to make a ruler calibrated in units and tenth of units using a strip of paper (or light cardboard) as the ruler. This ruler will be used to make measurements of sin of an angle for a unit circle of radius 5cm without the students needing to continually convert from a standard ruler. This special ruler will have 1 unit equivalent to 5cm and 0.1 of a unit equivalent to 0.5cm.

5. Show them the definition (the idea) of sin of an angle. They did see this briefly when they did the unit on cos. See diagram below.

1. Explain that sin of an angle is the y-coordinate for the point P on the circle. Remind them that the x-coordinate is called cos of the angle. Get the students to carefully draw an angle of 50º and to measure (with their newly made ruler), the sin of 50º. They should get a value of about 0.77 units. Ensure that every student understands this definition of sin(50). Now do the same for sin(65) and sin(40). Their answers should be approximately 0.9 units and 0.64 units respectively.

2. Once you are confident that the students understand what the definition of the sin of angles between 0º and 90º is, move on to discuss the definition for angles between 0º and -90º. Explain that the definition is the same as above, and that clearly, in this case, the values of sin will be negative.

3. Ask the students to use the unit circle (and the paper ruler) to find sin(-35º) and sin(-80º). The answers should be about -0.57 and -0.98 respectively.

4. Now move on to explain how the definition is to be understood for angles in the second and third quadrants (angles between 90º and 180º, and angles between 180º and 270º, respectively).

5. Get the students to draw and measure sin(260º). Their answers should be approximately -0.98. Now try sin(115º). Their answers should be approximately 0.91.

6. It is crucially important that every student fully understands the definition of sin and can find sin of the sorts of angles mentioned above.

7. At this point you can check to see if they have understood what you have shown them by asking them to find the angle which has a sin of 0.7 units. Those who understand should mark the point on the unit circle which has a y-coordinate of 0.7 and draw a line between it and the origin and measure the angle. The answer is approximately 44º. Now ask them if this is the only answer. Hopefully the students will realise, if they haven’t already done so, that there is another angle that also has a sin of 0.7. It is in the second quadrant and is approximately 136º. The fact that there are at least two answers to these sorts of questions is a very important point and worth stressing.

8. Finish this session by getting the students to make a poster that shows what the meaning of sin is.

#### Session 2

In this session students are asked to fully investigate the nature of sin.

1. Begin this session with another light-hearted test (with a serious purpose). Present, in random order, each of the values that have tans and cosines that are easy to name: 0º, 30º, 60º, 120º, 180º, 45º, 135º, 225º, -45º, -135º, -120º, 90º, -90º, 270º, and so on. Get the students to write down, based on their understanding of the definition (and their memorisation of the special surd expressions), the tans and cosines of each of these angles. Then give them questions like this:
If you know that an angle has a tan of 1/√3, name two values that angle could have?

2. Now revise the definition of sin introduced in Session 1.

3. Investigation Instructions:Using pencil and paper find out as many interesting facts about sin as you can. You should draw a unit circle of size 5cm and use the special ruler calibrated for this circle.

4. If a student has fully understood the definition of sin they will be able to participate in this investigation. The investigation is mainly allowing the students to pinpoint and articulate the sorts of results that they will know by intuition based on the work done in Session 1. In this investigation they will systematically explore the nature of sin and write down what they find.

5. The following investigation is similar to one they did in the cos unit. So it is assumed that it will be done more quickly than the cos investigation.

6. This is a broad domain of investigation and many students will be able to work out for themselves many important results. Some may need some guidance but do not give this too readily because this will detract from the fact that this activity is an investigation of an idea. You need to trust that the work you did in Session 1 is sufficient for the students to do this investigation. However, if guidance is needed, here are some suggestions:

• Why is it easy to name the value of sin(0º)? Which other values are similarly easy to name? List them all.
• Why is the value of sin(90º) easy to name? Which other angles are similarly easy to name? List them all.
• Describe in your own words the behaviour of the value of sin as the angle moves from 0º round to 360º.
• If you know that the sin of an angle is 0.4, use a scale drawing to find one value of that angle. There is another angle that has the same sin. What is it?
• Explain why it is that if you know the sin of 10º you can easily name sin(–10º), sin(170º) and sin(190º).
• If sin of an angle is negative, what can you say about the angle?
• Graph the value of sin, between 0º and 360º, by carefully drawing the angles between 0º and 90º in steps of 10 and measuring the lengths of sin.
• If sin(θ) = cos(θ) can you say what θ is?
• Does sin(θ/2) = sin(θ)/2? What other questions could you ask of this sort?
• Does sin(θ + 2) = sin(θ) + 2? Does sin(-θ) = -sin(θ)? Can you explain this geometrically? What other questions could you ask of this sort?
• Does sin(90º - θ) = cos(θ)? Can you explain this geometrically? What other questions could you ask of this sort? (Note: when you ask the students to explain the last of these geometrically you are asking them to use a diagram such as the diagram below. sin(90º - θ) = AN. But this is the same as OM which is cos(θ).)

• What does cos(θ) x cos(θ) + sin(θ) x sin(θ) equal? Can this be shown geometrically? (Note: this result follows from the application of Pythagoras’ Theorem to the triangle OAB in the diagram below.

1. Students should make a poster that explains what they have discovered about sin.

#### Session 3

In this session we continue the exploration of sin to include the inverse sin.

1. Begin the session with another light-hearted test. Name angles such as 90º, 180º, 270º, 360º, -90º, -180º, and 450º, in random order and ask them to write down the sin. Next name values such as 0, 1 and -1. In each case ask the students to name two angles with a sin equal to that value.

2. When the students studied the unit on cos they found surd  and other expressions for cos of multiples of 30º and 45º degrees. In a similar way surd and other simple expressions can be found for sin of these same angles.

3. First find expressions for sin of 30º and 60º. Use the following diagram and Pythagoras to find BC

4. Most students will wonder why it is important to know the surd expression for sin 60o. They will want to just use a decimal approximation, for instance, and not one of these strange surd expressions. Point out that the square root of a prime number is irrational. And so is half the square root of an irrational number. To put it in plain language, the value can never be written down as a decimal, no matter how may significant figures you use. And it cannot be written as a recurring decimal either. So the surd expression is exact. It also turns out that many properties of sin that the students will meet later are simpler to find if the surd (and not the decimal approximation) is used.

5. Spend some time memorising the surd expression (that is, the expression itself, not its decimal approximation): √3/2. Spend time memorising the value of sin(60º).Spend some more time becoming familiar with the method of producing these values using an equilateral triangle and Pythagoras. Make a poster showing this method.

6. Now find a surd expression for sin(45º). A similar method can be used, but this time a different starting triangle is used. See diagram below.

Use Pythagoras to show that AC = √2.  So AB = BC = 1, and sin(45º) = 1/√2.

1. Spend some time memorising the surd expression (that is, the expression itself, not its decimal approximation): 1/√2. That is, spend time memorising the value of sin(45º). Spend some more time becoming familiar with the method of producing this value using an isosceles right angled triangle and Pythagoras. Make a poster showing this method.

2. Introduce the students to the ‘sin’ and ‘inv’ ‘sin’ buttons on the calculator. Find two values, between 0º and 360º, for θ if sin(θ) is 0.34. Do other similar examples.
Important: Note again that there are at least 2 angles that have a sin of 0.34, but the calculatoronly give one of these. But because the students have been introduced to sin as they have been, it is easy for them to find the values that the calculator cannot give them.

#### Session 4

In this session we move on to using sin values to find the unknown sides in a right angles triangle.

1. Begin this session with yet another light-hearted test. Present, in random order, each of the values that have cosines that are easy to name: 0º, 30º, 60º, 120º, 180º, 45º, 135º, 225º, -45º, -135º, -120º, 90º, -90º, 270º, and so on. Get the students to write down, based on their understanding of the definition (and their memorisation of √3/2, 1/2 and 1/√2), the sin of each of these angles. Then give them questions like this:
If you know that an angle has a sin of -√3/2, name two values that angle could have?
These light-hearted tests are very important because they consolidate the ideas presented above. Mostly the answers depend on little more than a sound understanding of what sin means. The only exceptions are the surd values and the value 1/2, but these involve memorising relatively simple facts. It would be wise to continue these little light-hearted tests once the unit is completed. For instance, continue them daily for the first week following the unit and then weekly after that.

2. Now, all the above has been based on the unit circle.
But what happens if we do not have a unit circle?
What if the circle were of size 2, for instance?

All that happens is that the previous situation has been enlarged by scale factor 2.

Thus the point P on the unit circle is now (2cos(θ), 2sin(θ)).

In general, if the circle has size r, the coordinates are (rcos(θ), rsin(θ)).

1. The session is now finished off by applying the above to the solving of some particular right-angled triangles. This now becomes a straightforward application of the above ideas. Go through the following examples.

2. Example 1. Find the unknown, a, in the diagram below.

The important step is this: the student must decide which size of circle to use. In this case the circle must be of size 8. Thus the problem can be shown as on diagram below.

It is clear that the coordinates of P are (8cos(34), 8sin(34)). Clearly the unknown, a, is just the y-value of P. So a = 8sin(34). That is, a = 4.47 2DP.

1. Example 2. Find the unknown angle, θ, in the diagram below.

Again the important step is deciding the size of circle to use. In this case a circle of size 4 is the appropriate one to use.

It is clear that 4sin(θ) = 3. This can be done using the sin-1 button on the calculator. Thus θ = 48.6º to 1DP.

1. Give the students a range of similar problems to practise on. Mix in a few examples from the tan and cos units so that the students learn to distinguish sin from cos and tan cases.

#### Session 5

Here we look at the squared relation involving sin and cos; look at the relation between slopes of lines and tan; and simplify some sin, cos and tan expressions.

1. No! Not another light-hearted test!

2. Problem: Use Pythagoras to show that for any angle θ, between 0º and 90º, sin2(θ) + cos2(θ) = 1. Clearly this rather strange notation will need to be explained first. Check out this result for a range of values, such as: 20º, 40º, 70º. Check it out for values larger than 90º.
Do you think the identity still holds?

3. Consider the diagram below.

Why does the point T have the coordinates (1, tan(θ))?

1. Exercise: Consider the slope of the line OP. This line has the same slope as the line OT.
Why?
Get them to explain first why tan(θ) = sin(θ)/cos(θ) and then why sin(θ)/cos(θ) has no value when θ is 90º or 270º. (Hint: you cannot divide by 0.)

2. Find simpler expression for

(1)   sin(90º – θ)

(2)   cos(90º – θ)

(3)   tan(90º – θ) [Hint: convert from tan to sin and cos using the identity tan(θ) = sin(θ)/cos(θ)]

(4)   sin(90º + θ)

(5)   cos(90º + θ)

(6)   tan(90º + θ)

(7)   sin(180º – θ)

(8)   sin(180º + θ)

(9)   tan(180º – θ)

(10) -sin(θ)

(11) -cos(θ)

(12) -tan(θ)
In each case ask them to show geometrically why the result holds.

1. Explain why the line y = mx + c cuts the x-axis at an angle tan-1(m) [Hint: m is the slope of the line. Imagine the line cuts the x-axis at an angle θ. Now tan(θ) = sin(θ)/cos(θ) and is also a measure of the slope of the line. So tan(θ) = m. Using the reverse flowchart method, θ = tan-1(m)].