1. AS-Level Maths: Core 2 for Edexcel
C2.4 Trigonometry 1
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2. The sine, cosine and tangent of any angle
The graphs of sin θ, cos θ and tan θ
Exact values of trigonometric functions
Trigonometric equations
Trigonometric identities
Examination-style questions
Contents
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© Boardworks Ltd 2005

3. The three trigonometric ratios
The three trigonometric ratios, sine, cosine and tangent, can be defined using the ratios of the sides of a right-angled triangle as follows:
Sin θ =
Opposite
Hypotenuse
S O H θ O P S I T E H Y N U
A D J A C E N T
Cos θ =
Adjacent
Hypotenuse
C A H
Tan θ =
Opposite
Adjacent
T O A
Remember:
S O H
C A H
T O A

4. The sine, cosine and tangent of any angle
These definitions are limited because a right-angled triangle cannot contain any angles greater than 90°.
To extend the three trigonometric ratios to include angles greater than 90° and less than 0° we consider the rotation of a straight line OP of fixed length r about the origin O of a coordinate grid. x y O
P(x, y) r
Angles are then measured anticlockwise from the positive x-axis. θ
Explain that a line can be dropped from any point P to the x-axis to form a right-angled triangle containing the associated acute angle α. α
For any angle θ there is an associated acute angle α between the line OP and the x-axis.

5. The sine, cosine and tangent of any angle
The three trigonometric ratios are then given by:
The x and y coordinates can be positive or negative, while r is always positive.
This means that the sign of the required ratio will depend on the sign of the x-coordinate and the y-coordinate of the point P.

6. The sine, cosine and tangent of any angle
If we take r to be 1 unit long then these ratios can be written as:
The relationship between θ measured from the positive x-axis and the associated acute angle α depends on the quadrant that θ falls into.
For example, if θ is between 90° and 180° it will fall into the second quadrant and α will be equal to (180 – θ)°.

7. The sine of any angle
If the point P is taken to revolve about a unit circle then sin θ is given by the y-coordinate of the point P.
Explain that if P is fixed at 1 unit from the origin on a coordinate grid, sin θ will be given by the y-coordinate of the point P. This is shown by the length of the red line.
Explain that moving the point P in an anticlockwise direction increases the angle between OP and the x-axis. Point out that in mathematics an anticlockwise rotation is a positive rotation.
Slowly drag the point P through 360° starting at 0° and ending at 360°. Observe the length of the line representing the y-coordinate of the point P as it is rotated. Point out that when the line is in the first and the second quadrant (that is, when θ is between 0° and 180°) it is above the x-axis and therefore positive. In other words, the sine of angles between 0° and 180° is positive. When the line representing sin θ is in the third and the fourth quadrant (that is, when θ is between 180° and 360°) it is below the x-axis and therefore negative. In other words, the sine of any angle between 180° and 360° is negative.
In this activity, P can be moved through any angle between –360° and 720°. Demonstrate the sine of these angles if required. Explain that P can be moved through any positive or negative angle in this way.
Focusing on angles between 0° and 180° (that is, angles in the first and second quadrants), demonstrate the relationships between pairs of angles that have the same sine. For example, show that sin 32° = sin 148°. Conclude that sin θ = sin (180° – θ).
Focus next on angles between 180° and 270° (that is, angles in the third quadrant). By looking at the sine of the associated acute angle show that sin θ = –sin (θ – 180°).
In the fourth quadrant, examine angles between 270° and 360° and between 0° and –90°. Conclude that sin θ = –sin (360° – θ) for angles between 270° and 360°, and sin –θ = –sin θ for angles between 0° and –90°.

8. The cosine of any angle
If the point P is taken to revolve about a unit circle then cos θ is given by the x-coordinate of the point P.
Explain that cos θ is given by the x-coordinate of the point P. This is shown by the length of the red line.
Slowly drag the point P through 360° starting at 0° and ending at 360°. Observe the length of the line representing the x-coordinate of the point P as it is rotated. Point out that when the line is in the first and the fourth quadrants (that is, when θ is between 0° and 90°/–90° or between 270° and 360°) it is to the right of the y-axis and therefore positive. When the line representing cos θ is in the second and third quadrants (that is, when θ is between 90° and 270°, or between –90° and –270°) it is to the left the y-axis and therefore negative. In other words, the cosine of an angle in the second and third quadrants is negative.
In this activity, P can be moved through any angle between –360° and 720°. Demonstrate the cosine of these angles if required. Explain that P can be moved through any positive or negative angle in this way.
Focusing on angles between 0° and 180° (that is, angles in the first and second quadrants), demonstrate the relationships between pairs of angles that have the same cosine, but are of opposite sign. For example, show that cos 148° = –cos 32°. Conclude that cos θ = –cos (180° – θ).
Focus next on angles between 180° and 270° (that is, angles in the third quadrant). Show that cos θ = –cos (θ – 180°).
In the fourth quadrant, examine angles between 270° and 360° and between 0° and –90°. Conclude that cos θ = cos (360° – θ) for angles between 270° and 360°, and cos θ = cos –θ for angles between 0° and –90°.

9. The tangent of any angle
Tan θ is given by the y-coordinate of the point P divided by the x-coordinate.
Remind students that tan θ is given by sin θ/cos θ. Tan θ is therefore given by the y-coordinate of the point P divided by the x-coordinate of the point P.
Slowly drag the point P through 360° starting at 0° and ending at 360°. Observe the change in the value of tan θ as the point P is rotated. Point out that when P is in the first and the third quadrants (that is, when θ is between 0° and 90° or between 180° and 270°) the sine and cosine of the required angle is of the same sign. The tangent is therefore positive in these quadrants. The tangent of 90° and 270° is undefined when cos θ = 0 (because we cannot divide by 0).
When the point P is in the second and fourth quadrants (that is, when θ is between 90° and 180° and between 270° and 360°) the sine and cosine of the required angle is of different sign. The tangent is therefore negative in these quadrants.
Focusing on angles between 0° and 180° (that is, angles in the first and second quadrants), demonstrate the relationships between pairs of angles that have the same tangent, but are of opposite sign. For example, show that tan 127° = –tan 53°. Conclude that tan θ = –tan (180° – θ).
Focus next on angles between 180° and 270° (that is, angles in the third quadrant). Show that tan θ = tan (θ – 180°).
In the fourth quadrant, examine angles between 270° and 360° and between 0° and –90°. Conclude that tan θ = –tan (360° – θ) for angles between 270° and 360°, and tan –θ = –tan θ for angles between 0° and –90°.

10. The tangent of any angle
Tan θ can also be given by the length of tangent from the point P to the x-axis.
This demonstration shows more clearly how the value of tan θ varies as θ varies. In particular, it can be shown that tan 90° and tan 270° are undefined. This is because the tangent is parallel to the x-axis at these points.
The circle has radius 1. Explain to students that the length of the tangent at P from P to the x-axis gives us the value of tan θ. Remind students that the tangent of a circle forms a right angle with the radius. We therefore have a right-angled triangle with opposite side of length tan θ and adjacent side of length 1. Opposite/adjacent = tan θ as required.
Tan can also be demonstrated by extending the line OP to meet the tangent to the circle at the point (1, 0).
Slowly move the point P through 0° < θ < 360° and observe how the value of tan θ varies. Draw students’ attention to the fact that when θ is close to 90° and 270° it gets very large very quickly. Establish that at these points the tangent at P is parallel to the x-axis. Since the tangent will never meet the x-axis at 90° or 270° (or any odd multiple of 90°) tan θ is undefined for these angles.
The slope of the tangent defines whether it is positive or negative. A positive gradient gives a negative value and a negative gradient gives a positive value.

11. Remember CAST
We can use CAST to remember in which quadrant each of the three ratios are positive.
2nd quadrant
1st quadrant
Sine is positive S
All are positive A
3rd quadrant
4th quadrant
Introduce students to CAST to help them remember in which quadrant each of the three ratios are positive.
The disadvantage of this mnemonic is that we have to remember that the first letter refers to the fourth quadrant, the second letter to the first quadrant and so on. Students may find it easier to make up their own mnemonic using the letters in the order ASTC.
Tangent is positive T
Cosine is positive C

12. The sine, cosine and tangent of any angle
The sin, cos and tan of angles in the first quadrant are positive.
In the second quadrant:
sin θ = sin α
cos θ = –cos α
tan θ = –tan α
In the third quadrant:
sin θ = –sin α
cos θ = –cos α
tan θ = tan α
where α is the associated acute angle.
In the fourth quadrant:
sin θ = –sin α
cos θ = cos α
tan θ = –tan α

13. The sine, cosine and tangent of any angle
The value of the associated acute angle α can be found using a sketch of the four quadrants.
For angles between 0° and 360° it is worth remembering that:
when 0° < θ < 90°, α = θ
when 90° < θ < 180°, α = 180° – θ
when 180° < θ < 270°, α = θ – 180°
when 270° < θ < 360°, α = 360° – θ
For example, if θ = 230° we have:
α = 230° – 180° =
50°
230° is in the third quadrant where only tan is positive and so:
sin 230° = –sin 50°
cos 230° = –cos 50°
tan 230° = tan 50°

14. The graphs of sin θ, cos θ and tan θ
The sine, cosine and tangent of any angle
The graphs of sin θ, cos θ and tan θ
Exact values of trigonometric functions
Trigonometric equations
Trigonometric identities
Examination-style questions
Contents
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15. The graph of y = sin θ
Trace out the shape of the sine curve and note its properties.
The curve repeats itself every 360°. Also, –1 < sin θ < 1.

16. The graph of y = sin θ
The graph of sine is said to be periodic since it repeats itself every 360°.
We can say that the period of the graph y = sin θ is 360°.
Other important features of the graph y = sin θ include the fact that:
It passes through the origin, since sin 0° = 0.
The maximum value is 1 and the minimum value is –1. Therefore, the amplitude of y = sin θ is 1.
The definition of an odd function is a function for which f(x) = –f(–x), or f(–x) = –f(x), for all values of x. Odd functions are symmetrical about the origin.
It has rotational symmetry about the origin. In other words, it is an odd function and so sin (–θ) = –sin θ.

17. The graph of y = cos θ
Trace out the shape of the cosine curve and note it properties.
The curve repeats itself every 360°. Also, –1 < cos θ < 1.
The cosine curve is symmetrical about the vertical axis.

18. The graph of y = cos θ
Like the graph of y = sin θ the graph of y = cos θ is periodic since it repeats itself every 360°.
We can say that the period of the graph y = cos θ is 360°.
Other important features of the graph y = cos θ include the fact that:
It passes through the point (0, 1), since cos 0° = 1.
The maximum value is 1 and the minimum value is –1. Therefore, the amplitude of y = cos θ is 1.
The definition of an even function is a function for which f(x) = f(–x) for all values of x. Even functions are symmetrical about the y-axis.
It is symmetrical about the y-axis. In other words, it is an even function and so cos (–θ) = cos θ.
It the same as the graph of y = sin θ translated left 90°. In other words, cos θ = sin (90° – θ).

19. The graph of y = tan θ
Trace out the shape of the tangent curve and note it properties.
The curve repeats itself every 180°.
Tan θ is undefined at 90° (and 180n + 90° for integer values of n).

20. The graph of y = tan θ
You have seen that the graph of y = tan θ has a different shape to the graphs of y = sin θ and y = cos θ.
Important features of the graph y = tan θ include the fact that:
It is periodic with a period of 180°.
It passes through the point (0, 0), since tan 0° = 0.
Its amplitude is not defined, since it ranges from +∞ to –∞.
It is symmetrical about the origin. In other words, it is an odd function and so tan (–θ) = –tan θ.
tan θ is not defined for θ = ±90°, ±270°, ±450°, … that is, for odd multiples of 90°. The graph y = tan θ therefore contains asymptotes at these points.

21. Exact values of trigonometric functions
The sine, cosine and tangent of any angle
The graphs of sin θ, cos θ and tan θ
Exact values of trigonometric functions
Trigonometric equations
Trigonometric identities
Examination-style questions
Contents
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© Boardworks Ltd 2005

22. Sin, cos and tan of 45°
A right-angled isosceles triangle has two acute angels of 45°.
Suppose the equal sides are of unit length.
45° 1
Using Pythagoras’ theorem:
The hypotenuse
45° 1
Ensure that students can see that these ratios would be the same for any right-angled isosceles triangle.
If the equal sides were of a different length (for example, 3) the hypotenuse would be of length 32. In each ratio the 3’s would cancel and so simplify to those shown above.
If required, verify using a scientific calculator that sin and cos of 45° = … = 1/2
Stress that 1/2 is an exact answer. Sine and cosine of 45° cannot be written exactly as a decimal.
We can use this triangle to write exact values for sin, cos and tan 45°:
sin 45° =
cos 45° =
tan 45° = 1

23. Sin, cos and tan of 30°
Suppose we have an equilateral triangle of side length 2. 2
60°
30° 1 2
60°
If we cut the triangle in half then we have a right-angled triangle with acute angles of 30° and 60°.
Using Pythagoras’ theorem:
The height of the triangle
We can use this triangle to write exact values for sin, cos and tan 30°:
sin 30° =
cos 30° =
tan 30° =

24. Sin, cos and tan of 60°
Suppose we have an equilateral triangle of side length 2. 2
60°
30° 1 2
60°
If we cut the triangle in half then we have a right-angled triangle with acute angles of 30° and 60°.
Using Pythagoras’ theorem:
The height of the triangle
We can also use this triangle to write exact values for sin, cos and tan 60°:
sin 60° =
cos 60° =
tan 60° =

25. Sin, cos and tan of 30°, 45° and 60°
The exact values of the sine, cosine and tangent of 30°, 45° and 60° can be summarized as follows:
30°
45°
60°
sin
cos
tan
Establish that 135° is in the second quadrant where cos is negative. The associated acute angle is 45°. So cos 135 = –cos 45°.
Ask students to give the exact values of sin 135° and tan 135° and for sin, cos and tan of other angles associated with 45° (for example, 225°, 315°, 405°, – 45° or –135°).
Use CAST to remind students in which quadrant each ratio is positive.
Use this table to write the exact value of cos 135°
cos 135° = –cos 45° =

26. Sin, cos and tan of 30°, 45° and 60°
Write the following ratios exactly:
1) cos 300° =
2) tan 315° = -1
3) tan 240° =
4) sin –330° =
5) cos –30° =
6) tan –135° = 1
Ask students to complete this exercise individually before revealing the answers.
Suggest to students that they write down which quadrant each angle falls into before writing down their solutions.
7) sin 210° =
8) cos 315° =

27. Trigonometric equations
The sine, cosine and tangent of any angle
The graphs of sin θ, cos θ and tan θ
Exact values of trigonometric functions
Trigonometric equations
Trigonometric identities
Examination-style questions
Contents
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© Boardworks Ltd 2005

28. Equations of the form sin θ = k
Equations of the form sin θ = k, where –1 ≤ k ≤ 1, have an infinite number of solutions.
If we use a calculator to find arcsin k (or sin–1 k) the calculator will give a value for θ between –90° and 90°.
There is one and only one solution is this range.
This is called the principal solution of sin θ = k.
Other solutions in a given range can be found using the graph of y = sin θ or by considering the unit circle.
Note that sin–1 θ means ‘inverse sin θ’. It does not mean 1/sin θ. To avoid confusion use arcsin.
For example:
Solve sin θ = 0.7 for –360° < θ < 360°
arcsin 0.7 = 44.4° (to 1 d.p.)

29. Equations of the form sin θ = k
Using the graph of y = sin θ between –360° and 360° and the line y = 0.7 we can locate the other solutions in this range.
44.4°
y = 0.7
y = sin θ
–315.6°
–224.4°
135.6°
Use the graph to show that the other solutions in the range are:
θ = –360° ° = –315.6°
θ = –180° – 44.4° = –224.4°
θ = 180° – 44.4° = 135.6°.
So the solutions to sin θ = 0.7 for –360° < θ < 360° are:
θ = –315.6°, –224.4°, 44.4°, 135.6° (to 1 d.p)

30. Equations of the form sin θ = k
Demonstrate how the graph of y = sin θ can be used to solve equations of the form sin θ = k in the range –360° < θ < 360°. Change the value of k by moving the point on the y-axis up and down.
Tell students that they can find the principal solution using a calculator and a sketch of the graph to find the other solutions.
An alternative would be to consider angles in a unit circle as shown on the previous slide.

31. Equations of the form sin θ = k
We could also solve sin θ = 0.7 for –360° < θ < 360° by considering angles in the first and second quadrants of a unit circle where the sine ratio is positive.
Start by sketching the principal solution 44.4° in the first quadrant.
Next, sketch the associated acute angle in the second quadrant.
44.4°
–224.4°
–315.6°
135.6°
Moving anticlockwise from the x-axis gives the second solution:
180° – 44.4° = 135.6°
Moving clockwise from the x-axis gives the third and fourth solutions:
–(180° °) = –224.4°
–(360° – 44.4°) = –315.6°

32. Equations of the form cos θ = k and tan θ = k
Equations of the form cos θ = k, where –1 ≤ k ≤ 1, and tan θ = k, where k is any real number, also have infinitely many solutions. For example:
Solve tan θ = –1.5 for –360° < θ < 360°
Using a calculator, the principal solution is θ = –56.3° (to 1 d.p.)
–56.3°
123.7°
Now look at angles in the second and fourth quadrants of a unit circle where the tangent ratio is negative.
–236.3°
303.7°
This gives us four solutions in the range –360° < θ < 360°:
θ = –236.3°, –56.3°, 123.7°, 303.7°

33. Equations of the form cos θ = k
Demonstrate how the graph of y = cos θ can be used to solve equations of the form cos θ = k in the range –360° < θ < 360°. Change the value of k by moving the point on the y-axis up and down.
Tell students that they can find the principal solution using a calculator and a sketch of the graph to find the other solutions.
This can be used as an alternative to using the unit circle.

34. Equations of the form tan θ = k
Demonstrate how the graph of y = tan θ can be used to solve equations of the form tan θ = k in the range –360° < θ < 360°.

35. Equations involving multiple or compound angles
Multiples angles are angles of the form aθ where a is a given constant.
Compound angles are angles of the form (θ + b) where b is a given constant.
When solving trigonometric equations involving these types of angles, care should be taken to avoid ‘losing’ solutions.
Solve cos 2θ = 0.4 for –180° < θ < 180°
Start by changing the range to match the multiple angle:
–180° < θ < 180°
–360° < 2θ < 360°
Next, let x = 2θ and solve the equation cos x = 0.4 in the range –360° ≤ x ≤ 360°.

36. Equations involving multiple or compound angles
Now, using a calculator: x = 66.4° (to 1 d.p.)
Using the unit circle to find the values of x in the range –360° ≤ x ≤ 360° gives:
–293.6°
66.4°
293.6°
x = 66.4°,
293.6°,
–66.4°,
–293.6°
–66.4°
But x = 2θ so:
θ = 33.2°, 146.8°, –33.2°, –146.8°
This is the complete solution set in the range –180° < θ < 180°.

37. Equations involving multiple or compound angles
Solve tan(θ + 25°)= 0.8 for 0° < θ < 360°
Start by changing the range to match the compound angle:
0° < θ < 360°
25° < θ + 25° < 385°
Next, let x = θ + 25° and solve the equation tan x = 0.8 in the range 25° < x < 385°.
Using a calculator: x = 38.7° (to 1 d.p.)
38.7°
Using the unit circle to find the values of x in the range 25° ≤ x ≤ 385° gives:
218.7°
x = 38.7°, 218.7° (to 1 d.p.)
But x = θ + 25° so:
θ = 13.7°, ° (to 1 d.p.)

38. Trigonometric equations involving powers
Sometimes trigonometric equations involve powers of sin θ, cos θ and tan θ. For example:
Solve 4sin2θ – 1= 0 for –180° ≤ θ ≤ 180°
Notice that (sin θ)2 is usually written as sin2θ.
To find the second solution when sin θ = 1/2 we can imagine the associated acute angle, 30°, in the second quadrant. This gives the second solution as 180° – 30° = 150°. With practice this can be done without drawing.
When sin θ = ,
θ = 30°, 150°
When sin θ = – ,
θ = –30°, –150°
So the full solution set is θ = –150°, –30°, 30°, 150°

39. Trigonometric equations involving powers
Solve 3cos2θ – cos θ = 2 for 0° ≤ θ ≤ 360°
Treat this as a quadratic equation in cos θ.
3cos2θ – cos θ = 2
3cos2θ – cos θ – 2 = 0
Factorizing:
(cos θ – 1)(3cos θ + 2) = 0
cos θ = 1 or cos θ = –
Some students may find it easier to let c = cosθ so that the equation can be written as 3c2 – c = 2.
To find the second solution when cos θ = –2/3 we can imagine the associated acute angle, 180° –131.8° = 48.2°, in the third quadrant. This gives the second solution as 180° ° = 228.2°. Again, with practice this can be done without drawing.
When cos θ = 1,
θ = 0°, 360°
When cos θ = – ,
θ = 131.8°, °
So the full solution set is θ = 0°, 131.8°, 228.2°, 360°

40. Trigonometric identities
The sine, cosine and tangent of any angle
The graphs of sin θ, cos θ and tan θ
Exact values of trigonometric functions
Trigonometric equations
Trigonometric identities
Examination-style questions
Contents
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© Boardworks Ltd 2005

41. Trigonometric identities
Two important identities that must be learnt are:
sin2θ + cos2θ ≡ 1
The symbol ≡ means “is identically equal to” although an equals sign can also be used.
An identity, unlike an equation, is true for every value of the given variable so, for example:
sin24° + cos24° ≡ 1,
sin267° + cos267° ≡ 1,
sin2π + cos2π ≡ 1, etc.

42. Trigonometric identities
We can prove these identities by considering a right-angled triangle: x y r θ
Also:
But by Pythagoras’ theorem x2 + y2 = r2 so:

43. Trigonometric identities
One use of these identities is to simplify trigonometric equations. For example:
Solve sin θ = 3 cos θ for 0° ≤ θ ≤ 360°
Dividing through by cos θ:
Using a calculator, the principal solution is θ = 71.6° (to 1 d.p.)
71.6°
So the solutions in the given range are:
251.6°
θ = 71.6°, 251.6° (to 1 d.p.)

44. Trigonometric identities
Solve 2cos2θ – sin θ = 1 for 0 ≤ θ ≤ 360°
We can use the identity cos2θ + sin2 θ = 1 to rewrite this equation in terms of sin θ.
2(1 – sin2 θ) – sin θ = 1
2 – 2sin2θ – sin θ = 1
2sin2θ + sin θ – 1 = 0
(2sin θ – 1)(sin θ + 1) = 0
Note that when sin θ = 1 or –1 there will only be one solution in the range 0° ≤ θ ≤ 360°.
So: sin θ = 0.5 or sin θ = –1
If sin θ = 0.5, θ = 30°, 150°
If sin θ = –1, θ = 270°

45. Examination-style questions
The sine, cosine and tangent of any angle
The graphs of sin θ, cos θ and tan θ
Exact values of trigonometric functions
Trigonometric equations
Trigonometric identities
Examination-style questions
Contents
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© Boardworks Ltd 2005

46. Examination-style question
Solve the equation
3 sin θ + tan θ = 0
for θ for 0° ≤ θ ≤ 360°
Rewriting the equation using
So sin θ = or cos θ + 1 = 0
3 cos θ = –1
cos θ = –

47. Examination-style question
In the range 0° ≤ θ ≤ 360°,
when sin θ = 0, θ = 0°, 180°, 360°.
when cos θ = – :
cos–1 – = 109.5° (to 1 d.p.)
cos θ is negative in the 2nd and 3rd quadrants so the second solution in the range is:
109.5°
θ = ° (to 1 d.p.)
So the complete solution set is:
θ = 0°, 180°, 360°, 109.5°, °
250.5°