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AS-Level Maths: Core 2for Edexcel

C2.4 Trigonometry 1

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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



θ

OPPOSITE

HY

PO

TE

NU

SE

A D J A C E N T

The three trigonometric ratios

Sin θ =Opposite

Hypotenuse S O HS O H

Cos θ =Adjacent

Hypotenuse C A HC A H

Tan θ =OppositeAdjacent T O AT O A

Remember: S O HS O H C A HC A H T O AT O A

The three trigonometric ratios, sine, cosine and tangent, can be defined using the ratios of the sides of a right-angled triangle as follows:


x

y

O

P(x, y)

r


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.

Angles are then measured anticlockwise from the positive x-axis.

For any angle θ there is an associated acute angle α between the line OP and the x-axis.

α

θ



The three trigonometric ratios are then given by:

sin =y

r

cos =x

r

tan =y

x

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.



The relationship between θ measured from the positive x-axis and the associated acute angle α depends on the quadrant that θ falls into.

If we take r to be 1 unit long then these ratios can be written as:

sin = =1

yy

cos = =1

xx

tan =y

x sin

tan =cos

For example, if θ is between 90° and 180° it will fall into the second quadrant and α will be equal to (180 – θ)°.


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.


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.


The tangent of any angle

Tan θ is given by the y-coordinate of the point P divided by the x-coordinate.


The tangent of any angle

Tan θ can also be given by the length of the tangent from the point P to the x-axis.


3rd quadrant

2nd quadrant 1st quadrant

4th quadrant

Tangent is positiveTT

Sine is positiveSS All are positiveAA

Remember CAST

We can use CAST to remember in which quadrant each of the three ratios is positive.

Cosine is positiveCC



The sin, cos and tan of angles in the first quadrant are positive.The sin, cos and tan of angles in the first quadrant are positive.

In the second quadrant: sin θ = sin αcos θ = –cos αtan θ = –tan α

sin θ = sin αcos θ = –cos αtan θ = –tan α

In the third quadrant: sin θ = –sin αcos θ = –cos αtan θ = tan α

sin θ = –sin αcos θ = –cos αtan θ = tan α

In the fourth quadrant: sin θ = –sin αcos θ = cos αtan θ = –tan α

sin θ = –sin αcos θ = cos αtan θ = –tan α

where α is the associated acute 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:

230° is in the third quadrant where only tan is positive and so:

α = 230° – 180° = 50°

sin 230° = –sin 50°

cos 230° = –cos 50°

tan 230° = tan 50°


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The graph of y = sin θ


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.

It has rotational symmetry about the origin. In other words, it is an odd function and so sin (–θ) = –sin θ.


The graph of y = cos θ


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.

It is symmetrical about the y-axis. In other words, it is an even function and so cos (–θ) = cos θ.

It is the same as the graph of y = sin θ translated left 90°. In other words, cos θ = sin (90° – θ).


The graph of y = tan θ


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 passes through the point (0, 0), since tan 0° = 0.

It is symmetrical about the origin. In other words, it is an odd function and so tan (–θ) = –tan θ.

It is periodic with a period of 180°.

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.

Its amplitude is not defined, since it ranges from +∞ to –∞.


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Sin, cos and tan of 45°

A right-angled isosceles triangle has two acute angles of 45°.

45°

45°Suppose the equal sides are of unit length.

1

1

Using Pythagoras’ theorem:

We can use this triangle to write exact values for sin, cos and tan 45°:

cos 45° = tan 45° = 1

2

2

The hypotenuse 2 21 1

1

2sin 45° =

1

2


2 2

2

60° 60°

60°

2

60°

30°

1

3


Suppose we have an equilateral triangle of side length 2.

We can use this triangle to write exact values for sin, cos and tan 30°:

If we cut the triangle in half then we have a right-angled triangle with acute angles of 30° and 60°.


The height of the triangle 2 22 1

3

sin 30° =1

2cos 30° =

3

2tan 30° =

1

3



Suppose we have an equilateral triangle of side length 2.

We can also use this triangle to write exact values for sin, cos and tan 60°:

3

sin 60° =3

2cos 60° =

1

2tan 60° = 3

2 2

2

60° 60°

60°

2

60°

30°

1

3

If we cut the triangle in half then we have a right-angled triangle with acute angles of 30° and 60°.


The height of the triangle 2 22 1


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°

sin

cos

tan

45° 60°

Use this table to write the exact value of cos 135°.

cos 135° = –cos 45° =

12

1

21

21

3

12

32

32

31

1

2


Sin, cos and tan of 30°, 45° and 60°

Write the following ratios exactly:

1) cos 300° =

3) tan 240° =

5) cos –30° =

7) sin 210° =

2) tan 315° =

4) sin –330° =

6) tan –135° =

8) cos 315° =

12

12

12

32

1

2

3

–1

1


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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°.

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.

For example:

There is one and only one solution in this range.

Solve sin θ = 0.7 for –360° < θ < 360°.

arcsin 0.7 = 44.4° (to 1 d.p.)


44.4°

y = 0.7y = sin θ


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.

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.)

135.6°–224.4°–315.6°


44.4°


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.

135.6°–224.4° –315.6°

Moving anticlockwise from thex-axis gives the second solution:

Moving clockwise from thex-axis gives the third and fourth solutions:

180° – 44.4° = 135.6°

–(180° + 44.4°) = –224.4°–(360° – 44.4°) = –315.6°


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°

–236.3°123.7°

303.7°

Now look at angles in the second and fourth quadrants of a unit circle where the tangent ratio is negative.

This gives us four solutions in the range –360° < θ < 360°:

θ = –236.3°, –56.3°, 123.7°, 303.7°


Equations of the form cos θ = k


Equations of the form tan θ = k


Equations involving multiple or compound angles

Multiple 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:

Next, let x = 2θ and solve the equation cos x = 0.4 in the range –360° ≤ x ≤ 360°.

–180° < θ < 180°

–360° < 2θ < 360°



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:

66.4°

293.6°–66.4°

–293.6°

x = 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°.

293.6°, –66.4°, –293.6°



Solve tan(θ + 25°)= 0.8 for 0° < θ < 360°.

Start by changing the range to match the compound angle:

Next, let x = θ + 25° and solve the equation tan x = 0.8 in the range 25° < x < 385°.

0° < θ < 360°

25° < θ + 25° < 385°

Using the unit circle to find the values of x in the range 25° ≤ x ≤ 385° gives:

x = 38.7°, 218.7° (to 1 d.p.)

But x = θ + 25° so:

θ = 13.7°, 193.7° (to 1 d.p.)

Using a calculator: x = 38.7° (to 1 d.p.)38.7°

218.7°


Trigonometric equations involving powers

Sometimes trigonometric equations involve powers of sin θ, cos θ and tan θ. For example:

Notice that (sin θ)2 is usually written as sin2θ.

Solve 4sin2θ – 1= 0 for –180° ≤ θ ≤ 180°.

24sin 1= 0 24sin =12 1

4sin =12sin =

12When sin θ = ,

12When sin θ = – ,

θ = 30°, 150°

θ = –30°, –150°

So the full solution set is θ = –150°, –30°, 30°, 150°


Trigonometric equations involving powers

Treat this as a quadratic equation in cos θ.

Solve 3cos2θ – cos θ = 2 for 0° ≤ θ ≤ 360°.

When cos θ = 1, θ = 0°, 360°

θ = 131.8°, 228.2°

So the full solution set is θ = 0°, 131.8°, 228.2°, 360°

3cos2θ – cos θ = 2

3cos2θ – cos θ – 2 = 0Factorizing:

(cos θ – 1)(3cos θ + 2) = 0

cos θ = 1 or cos θ = – 23

When cos θ = – ,23


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Two important identities that must be learnt are:

sintan (cos 0)

cos

sin2θ + cos2θ ≡ 1

An identity, unlike an equation, is true for every value of the given variable so, for example:

The symbol ≡ means “is identically equal to” although an equals sign can also be used.

sin24° + cos24° ≡ 1, sin267° + cos267° ≡ 1, sin2π + cos2π ≡ 1, etc.



We can prove these identities by considering a right-angled triangle:

x

yr

θ

y

rsin =

x

rand cos =

sin=

cos

y

r

xr

=y

x= tan as required.

Also:y x

r r

2 22 2sin + cos = +

2 2

2

+=

x y

r

But by Pythagoras’ theorem x2 + y2 = r2 so:2

2 22

sin + cos = =1 as required.r

r


71.6°


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:sin 3cos

=cos cos

tan = 3Using a calculator, the principal solution is θ = 71.6° (to 1 d.p.)

251.6°

So the solutions in the given range are:

θ = 71.6°, 251.6° (to 1 d.p.)



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

So: sin θ = 0.5 or sin θ = –1

If sin θ = 0.5, θ = 30°, 150°

If sin θ = –1, θ = 270°


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Examination-style question

Solve the equation

3 sin θ + tan θ = 0

for θ for 0° ≤ θ ≤ 360°.

Rewriting the equation usingsin

tan :cos

sin3sin + = 0

cos

3sin cos + sin = 0 sin (3cos +1) = 0

So sin θ = 0 or 3 cos θ + 1 = 0

3 cos θ = –1

cos θ = – 13


Examination-style question

In the range 0° ≤ θ ≤ 360°,

when sin θ = 0, θ = 0°, 180°, 360°.

13when cos θ = – :

cos–1 – = 109.5° (to 1 d.p.)13

cos θ is negative in the 2nd and 3rd quadrants so the second solution in the range is:

250.5°

109.5°θ = 250.5° (to 1 d.p.)

So the complete solution set is:

θ = 0°, 180°, 360°, 109.5°, 250 .5°

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