1

Chapter 7: Trigonometry

In this chapter, we will study:

7.1 Trigonometric functions in a right triangle. Values of trigonometric functions for 30 , 45 and 60o o o . The

special cases of 0 and 90o o .

7.2 Positive and negative angles. Trigonometric functions of any angle. Signs of trigonometric functions.

The reference angle theorem.

7.3 The graphs of sin( ) and cos( ) for 0,360o . The graph of tan( ) for 090 ,90o . Extensions of

these graphs by periodicity. Properties of the graphs.

7.4 Solving trigonometric equations. Principal values of the inverse functions and derived solutions.

7.5 Radian measure. The length of an arc of a circle. The area of a sector of a circle.

7.6 Transformations of graphs.

7.1.A Trigonometric functions in a right triangle. Values of trigonometric functions for 30 , 45 and 60o o o .

i) Definitions of trigonometric functions of an acute angle:

Remember that in a right triangle ABC, as shown in Figure 1:

DEFINE :

(1)

oppositesin

hypothenuse

adjacentcos

hypothenuse

opposite sintan

adjacent cos

y

r

x

r

y

x

Figure1: Definitions of trigonometric functions of acute angles

By using the Pythagoras theorem in triangle ABC, we see that

(3) 2 2

2 2 2 2 21 sin cos 1x y

x y rr r

in our notation.

(3) is the fundamental trigonometric identity, which we will see that holds in general.

r=hypotenuse y=opposite side

x=adjacent side

θ A

B

C

2

ii) Particular cases : 30 ,45 and 60o o o

Using definitions (1), we obtain the values of the 3 trigonometric functions of 30 ,45 and 60o o o , listed as in the

following table:

Table 1: Values of trigonometric

functions of acute angles

Proof (optional) :

i) Calculate the trigonometric functions for 30 and 60o o by considering an equilateral triangle ABC of side a and

By drawing the height AD (which is also median and bisector) from vertex A ( D BC ).

Apply the Pythagoras’ theorem in the right triangle ADB to find: 3

2

aAD , then in the right triangle ADB:

332sin 60

2

o

a

a and

12cos 602

o

a

a , therefore

sin 60tan 60 3

cos 60

o

o

o also:

12sin 302

o

a

a and

332cos 30

2

o

a

a and therefore

sin 30 1 3tan 30

3cos 30 3

o

o

o

ii) To calculate the trigonometric functions of 45o , consider a right isosceles triangle of sides a (and therefore of

hypotenuse 2a to obtain :

sin 452sin 45 cos 45 tan 45 1

2 cos 45

o

o o o

o .

Note: Memorize the values of the trigonometric functions listed in Table 1 above, as we will use them often.

sin cos tan

030 1

2 3

2

3

3

045 2

2

2

2

1

060 3

2

1

2 3

3

Also, memorize the Pythagorean identity:

(3) 2 2sin cos 1 and the definition (31 )

sintan

cos

,

as we will use these often as well.

Note that (3) and (31 ) imply that:

(4)

2

2

11 tan

cos

and also that (5) 1 sin 1 and that 1 cos 1

The identities (3) , (31 ) , (4) and (5) are very important and need to be known.

7.1. B (Optional) Trigonometric functions for the special angles of 0 and 90o o

For the special angles of 0 and 90o o , we consider a triangle as shown in Figure 2 below, where 0o

(consider X to be gradually closer to 0o)

Figure 2: A right triangle with 0oX . Note that in this case:

and 0x adjacent a r hypotenuse a y opposite

Using our usual definitions (1) we obtain:

(6)

sin 0 0; cos 0 1 and tan 0 0 and also that

sin 90 1; cos 90 0 and tan 90

o o o

o o o undefined

Exercises:

From Exercise Set 7A do problems 1, 4, 5 and 6 , from Exercise Set 7B do problems 1, 5, 8, 9, 10 and 11.

From the Practice Book (Chapter 7: Trigonometry):

From Exercise Set 7.1 (page 141): do problems 1 , 2 , 3, 4 ii)

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7.2 Positive and negative angles. Trigonometric functions of any angle.

Signs of trigonometric functions. The reference angle theorem.

In this section we generalize definitions (1) (valid only for acute angles) to any angle.

Figure 3: A point on the unit circle.

For this reason, we consider a general angle as the angle formed by the radius OP (where P is on the unit circle:

2 2 1x y ) and the x axis.

Note that:

The origin of the angle is the x axis.

0 if is measured counter-clockwise, and 0 if measured clock-wise.

0 ,90 when P Q I, 90 ,180 when P Q II,

180 ,270 when P Q III, 270 ,360 when P Q IV

o o o o

o o o o

Exercise:

1. Draw the following angles:

0 045 ,90 ,300 ,230 , 30 , 300 and 440o o o o o .

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Using Figure 3 above, define:

(7)

sin1

cos1

sintan

cos

yy

xx

y

x

.

Note that definitions (7) are natural generalizations of (1) (as they are identical with (1) for acute angles ( P Q I ), but

that these definitions apply now for any angle .

The line OP is called the terminal line of the angle.

In order to most easily calculate the trigonometric functions of any angle as in (7), it is best to define the reference angle

of .

Definition 1: The reference angle ' of a generic angle is the smallest positive angle formed between the terminal

line of the angle and the x axis.

Note that for a generic angle , its reference angle ' is always acute.

Exercise 2:

Calculate the reference angles of the following angles:

0 045 ,90 ,300 ,230 , 30 , 300 and 440o o o o o .

Note: The applet shown here https://www.mathopenref.com/reference-angle.html indicates clearly reference angles

for any angle.

Theorem 1: (Reference Angle Theorem):

Consider a generic angle and its reference angle ' . Then:

(8)

sin sin '

cos cos '

tan tan '

Where the correct sign of each trigonometric function above is given by applying definition (7).

An easy way to remember the correct sign of each trigonometric function by quadrant is to use the mnemonic All

Students Take Calculus (read counter-clockwise ) shown below:

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Figure 4: Signs of trigonometric functions.

Figure 4 indicates which functions are positive in each quadrant (by using definition (7)) (in quadrants II,

III and IV the indicated function is positive and the remaining 2 are negative).

Exercise 3:

Using (8) and Figure 4, calculate:

sin 135 , cos 135 , tan 135

sin 240 , cos 240 , tan 240

sin 300 , cos 300 , tan 300

sin 60 , cos 60 , tan 60

sin 420 , cos 420 , tan 420

sin 90

o o o

o o o

o o o

o o o

o o o

o

, cos 90 , tan 90 o o

Note that identities (3), ( 31 ) , (4) and (5) still hold for general trigonometric functions defined as in (7) (since

2 2 1x y now). We will use these identities often.

Example:

Do Examples 7.3 and 7.4 in the textbook.

From Exercise Set 7.2 (Practice Book : page 144) do problems 1, 2 and 3 . Note that for trigonometric identities we often

make use of the following algebraic identities:

2 2 2

2 2 2

2 2

2

2

a b a ab b

a b a ab b

a b a b a b

.

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7.3 The graphs of sin( ) and cos( ) for 0,360o

The graph of tan( ) for 090 ,90o

Extensions of these graphs by periodicity. Properties of the graphs.

Our main goal for this section is to graph sin for 0,360 and cos for 0,360o of g .

We also want to obtain general properties of the functions sin and g cos f .

The graph of sin f :

To do this, we start by obtaining the values of sinf for main angle values between 0 and 360o o.

Using the reference angle theorem learned in section 7.2 and the values from Table 1 (Section 7.1 A) , we obtain the

following values:

0o 30o

60o

90o

120o

150o

180o

210o

240o

270o

300o

330o

360o

sin 0

10.5

2

30.866

2

1 3

2

1

2 0

1

2

3

2

1

3

2

1

2 0

Table 2: The values of the sine function for 0 ,360o o

These values are best represented in the full graph of the sine function shown below:

Figure 4: The sine function for 0 ,360o o

8

Remember the shape of the sine graph above, since we will use it often.

Note also that the angle 390o gives the same point

1P on the unit circle as the angle 30o, the angle 420o

gives the same

point 2P and so on. Therefore, you can see that for 360 ,720o o the sine function will repeat itself as shown in

Figure 5 below.

Figure 5: The sine function for 0 ,720o o

The function ( ) sinf is called a periodic function of period 360o ,

since it repeats itself after every whole multiple (positive or negative) of 360o.

The graph of cos f :

Similarly, for the cosine function for between 0 and 360o o, using the reference angle theorem learnt in section 7.2

and the values from Table 1 (Section 7.1 A) , we obtain the following values:

0o

30o

60o

90o

120o

150o

180o

210o

240o

270o

300o

330o

360o

cos

1 3

0.8662

10.5

2

0 1

2

3

2

1

3

2

1

2 0

1

2

3

2 1

Table 2: The values of the cosine function for 0 ,360o o

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These values are represented in the full graph of the cosine function shown below:

Figure 6: The cosine function for 0 ,360o o

Since the points for 360 ,720o o overlap with the points for 0 ,360o o , therefore for angles in

360 ,720o o the cosine function will repeat itself. The cosine function ( ) cosf is also periodic of period

360o . A graph of cos outside of 0 ,360o o is shown in the graph below:

Figure 7: The cosine function

You can also note that the shape of the graph of cosine is identical to the shape of the graph of sine.

Looking carefully at the two graphs of sine and cosine (Figures 4 and 6), you can see that sin 90 coso , that

is the graph of cosine can be thought as a left shift by 90o of the graph of sine.

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The graph of tan f :

Remember that sin

tancos

y

x

.

We obtain therefore the following values:

90o 60o 30o 0o 30o

60o 90o

cos

Undefin

ed

( 0x )

3 1.73

1 30.577

33

0

1 30.577

33

3 1.73

Undefin

ed

( 0x )

Table 2: The values of the tangent function for 90 ,90o o

Note:

sin 210 sin 30

tan 210 tan 30cos 210 cos 30

o o

o o

o o

and so on for all angles 180 ,360o o

This shows that the tangent function is periodic of period 180o .

These values and properties of tan f are represented in the full graph of the tangent function shown below:

Figure 8: The tangent function

As a conclusion of this section, review and remember the graphs of sine and cosine for 0 ,360o o shown in Figures

4 and 6, and the graph of tangent shown in Figure 8. Also remember that sine and cosine are periodic of period 360o, as

Figures 5 and 7 show, and that tangent is periodic of 180o, as Figure 8 shows.

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7.4 Solving trigonometric equations

Principal values of the inverse functions and derived solutions

Goals:

Solve trigonometric equations of the form:

sin for 1,1 and 0 ,360

cos for 1,1 and 0 ,360

tan for and 90 ,90

o o

o o

o o

a a

a a

a a R

Extend the solutions above to larger intervals for ;

Understand what the principal values of 1 1 1sin ,cos and tana a a are, and how to use these.

Definition 2: A trigonometric equation is an equation which contains a trigonometric function it contains at least

one of sin ,cos , tan ) or a combination of these.

Notes:

1. Our main tool for solving the trigonometric equations in this section will be the knowledge of the graphs of

sin ,cos and tan which we have covered in 7.3 .

Therefore, make sure that you can sketch these graphs before starting.

2. From the graph of sin , note that the equation sin a has no solution if 1 or if 1a a

(look also at the property (5) in 7.1A).

Similarly, the equation cos a has no solution if 1 or if 1a a .

Therefore, we consider only sin and cos with 1,1a a a .

Looking at the graph of tan in 7.3, we see that the equation tan a has a solution for any a R .

This is one major difference between the 3 trigonometric equations above, which needs to be remembered.

3. From the graph of sin for 0 ,360o o note that the equation sin with 1,1a a has two

solutions (there are two values of 0 ,360o o which satisfy sin a with 1,1a ) except for 1a

(for which there is only one solution).

4. Similarly, from the graph of cos for 0 ,360o o note that the equation cos with 1,1a a has

two solutions (there are two values of 0 ,360o o which satisfy cos a with 1,1a ) except for

1a (for which there is only one solution).

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5. From the graph of tan for 90 ,90o o , we see that the equation tan for a a R has only one

solution (there is only one 90 ,90o o such that tan for a a R ).

These are the solutions that we will find in the examples which follow.

Example:

1. Let us solve 1

sin for 0 ,3602

o o

Figure 9: 1

sin2

Remembering the values of trigonometric functions learned in 7.1, we realize that the first solution is: 0 30o .

Using the reference angle theorem learned in 7.2 and the graph above, we realize that the second solution 1 is:

1 0180 180 30 150o o o o .

The method outlined above works for any equation of the form sin for 1,1 and 0 ,360o oa a , when we

know 0 based on the values learned in 7.1.

2. Let us solve sin 0.4 for 0 ,360o o

The graph of this equation is very similar with the graph shown in Figure 9. This time, however, we cannot find

0 based on the values learnt in 7.1. The value 0 is found now using 1

0 sin 0.4 23.6o to 1 d.p. (using a

calculator. Make sure that your calculator is in degree mode when using this function).

We will learn more about 1sin ( )a

soon. As in Example 1 above, 1 0180 180 23.6 156.4o o o .

Exercise:

Following the method shown above:

1. Solve cos 0.5 for 0 ,360o o

Solution: Your values of should be: 1cos 0.5 60o

o and 1 360 60 240o o o .

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2. Solve cos 0.4 for 0 ,360o o

Solution: Your values of should be: 1cos 0.4 66.4o

o to 1 d.p. (using a calculator)

and 1 360 66.4 293.6o o o .

3. Solve tan 1 for 90 ,90o o .

Using the values learn in sections 7.1 and 7.2 and the graph of tan shown in Figure 8 above, note that the

only solution is 0 45o

4. Solve tan 2 for 90 ,90o o .

Using the the graph of tan shown in Figure 8 above, note that the only solution is 1

0 tan 2 63.4 to 1

d.p. (using a calculator).

Note: About 1 1 1sin ,cos ( ), tan ( )a a a and the principal values of these functions.

6. The trigonometric function ( ) sinf is NOT one to one for 0,360o since it does not pass the

horizontal line test. Therefore, the function ( ) sinf is not invertible (it does not have an inverse) for

0,360o . To make the function ( ) sinf invertible, we need to restrict the domain of the

function. Looking at the graph of sin , we see that many such restrictions are possible,

but we choose the (most natural ) restriction: 90 ,90o o for which the function ( ) sinf is

now one to one and therefore invertible.

Therefore, the function : 90 ,90 1,1 , sino of f is invertible (it has an inverse) and its

inverse is: 1sin : 1,1 90 ,90o o . Note that 1sin ( )a

applies only to values in 1,1 and produces

an angle in 090 ,90o such that sin a .

This choice of the inverse function for ( ) sinf is called the principal value of 1sin ( )a

(Figures of

these principal values are shown on page 230 of the textbook).

Similarly, cos : 0 ,360 1,1o o is not invertible, but the restriction cos : 0 ,180 1,1o o is

invertible. Its inverse: 1cos : 1,1 0 ,180o o is called the principal value of 1cos ( )a

. Note that

1cos ( )a produces the angle in

00 ,180o such that cos a .

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1tan : 90 ,90o oR is the principal value of 1tan ( )a

. 1tan ( )a

produces the angle in 90 ,90o o

such that tan a .

It is very important to remember the domains and the ranges of the principal values of the trigonometric functions

defined above.

Example:

Do examples 7.7 and 7.8 in the textbook.

Note:

7. To extend the solutions of trigonometric equations to larger intervals for , remember and use the fact that

these functions are periodic, that is:

sin 360 sin ; cos 360 cos and tan 180 tano o o .

Therefore, we first solve the equation in the basic interval of the function, and the other solutions are found by

adding or subtracting (as needed) 360o (for the functions sine and cosine) or 180o

for tangent.

Example:

1. Solve tan 1 for 0 ,360o ox x .

Consider first 90 ,90o ox : the solution of tan 1 for 90 ,90o ox x is 0 45ox .

To find the other solutions, add / subtract 180o until the solutions will fall out of the required interval

0 ,360o o .

In our case: 1 45 180 225o o ox is the only other solution in 0 ,360o o .

Therefore, the solutions of tan 1 for 0 ,360o ox x are: 145 and 225o o

ox x .

Exercise:

From Exercise Set 7C in the textbook (page 233) do problems 1, 3 ii), iii), iv) and v) 4 i), ii), vii, viii and ix, 5, 8 i),

iv), v, vi, vii, viii, ix) and x, 10, 11, 12 and 13.

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7.5 Radian measure. The length of an arc of a circle. The area of a sector of a circle. Other formulas.

In this section:

We define the notion of a radian , convert between radian and degrees and graph trigonometric functions

for angles measured in radians;

We give the formula for the length of an arc of a circle and use it to find arc lengths;

We give the formula for the area of a sector of a circle and use it;

We recall other trigonometric formulas learnt last year, useful for calculating angles, sides or areas.

7.5. A The notion of a radian:

A radian is a specific amount of degrees. More precisely:

Definition 3:

A radian is the measure of the central angle in a circle with radius r which subtends an arc of length r , as shown in

Figure 10 below:

Figure 10: A radian

Since there are 360o in a full rotation of the radius, these will correspond to 2 radians (remember that 1 radian

is the central angle which covers 1 r on the circumference).

Therefore:

(9) 360 2 radians

180 radians

o

o

1801 radian 57.3

oo

(to 1 d.p.).

Using formula (9), we can convert any amount of degrees to radians.

To convert from degrees to radians , multiply by 180o

. To convert from radians to degrees, multiply by

180o

.

A few examples are shown below:

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The trigonometric functions sin( ) and cos( ) can also be graphed on the (radians domain) of 0,2 instead of

0,360o .

These graphs are shown below:

Figure 11: Graphs of sin and cos for 0,2

Similarly, the graph of tan for , is shown in Figure 12 below:

Figure 11: Graph of tan for ,

Exercises:

From Exercise Set 7D, do problems 1 (choose 5 angles), 2 (choose 5 angles) , 3 (choose 5 functions) , 4 and 5.

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7.5.B Arc length:

Remember that the circumference of a circle with radius R is: 2C R .

Therefore, the length of a semi-circle is: L R , and, more generally, the length of an arc of a circle of radius R and

central angle is:

(10) L R , where the central angle is measured in radians.

7.5.C The area of a sector of a circle:

Remember that the area of a circle with radius R is: 2Area R . Therefore, the area of a semi-circle is

2

2

R , and,

more generally, the area of a sector of a circle of radius R and central angle is:

(11)

2

Area(sector)2

R , where the central angle is measured in radians.

Example: Do Example 7.12 in the textbook.

7.5.D. Other formulas:

Last year you have learned other formulas which are useful to measure sides or angles in a generic triangle ABC , as that

shown in Figure 12 , which are:

Law of sines:

(12) sin sin sin

or sin sin sin

a b c A B C

A B C a b c

Law of cosines:

(13) 2 2 2 2 cosa b c bc A or

2 2 2

cos2

b c aA

bc

and similarly for sides and and angles B and Cb c

Figure 12: A generic triangle ABC and The area of the triangle ABC:

(14) sin( ) sin( ) sin( )

2 2 2

ab C ac B bc AArea

Memorize formulas (12)-(14) together with the other formulas in this Chapter, as we will use them often.

Exercise:

Do Example 7.12 from the textbook, and from Exercise set 7E do Exercises 1, 2, 3, 6, 8 and 10.

From Exercise Set 7.4 of the the Practice Book (page 151) do Exercises 1, 2,3, 4 iv and vi and from Exercise Set 7.5 do

Exercises 1, 2 and 7.

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7.6 Transformations of graphs.

7.6. A General transformations

Consider a generic function ( )f x , whose graph is known.

As an example, consider 2( ) 1f x x , shown in Figure 13 below.

Figure 13: The graph of 2( ) 1f x x

General transformations:

The graph of ( )f x a is a vertical translation (or vertical shift) of the graph of ( )f x : ( )f xG by:

a units upwards ( if 0a ) and downwards ( if 0a ) .

The graph of ( )f x a is a horizontal translation (or horizontal shift) of ( )f xG by:

a units to the left ( if 0a ) and a units to the right ( if 0a ) . Pay close attention to the two cases here, as

they may appear counter-intuitive.

The graph of ( )f x is a reflection of ( )f xG with respect to the x axis ;

The graph of ( )a f x for 0a is a vertical stretch a times of ( )f xG if 1a and a vertical shrink

(contraction) a times if 0 1a ;

The graph of ( )f a x is a horizontal stretch a times if 0 1a and a vertical shrink (contraction) a times if

1a . Again, pay close attention at the two cases here, as they may appear counter-intuitive.

Illustrations of these transformations are shown below:

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Figure 14: Vertical translations of a graph :

2

2

2

1

(1 ) 2

(1 ) 2

red x

orrange x

purple x

Figure 15: Horizontal translations of a graph :

2

2

2

1

1 1 (right shift)

1 1 (left shift)

red x

orrange x

purple x

Figure 16: Vertical stretch and shrink of a graph :

2

2

2

1

2 1 (vertical stretch)

11 (vertical shrink)

2

red x

orrange x

purple x

Figure 17: Horizontal stretch and shrink of a graph :

2

2

2

1

1 (horizontal stretch) !! 2

1 2 (horizontal shrink) !!

red x

xorrange

purple x

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7.6.B Transformations of the graphs of trigonometric functions.

We will use the knowledge of the transformations learned in 7.6.A to transform the graphs of trigonometric functions.

Examples:

1. Graph ( ) sin( ) and g( ) 2 sin( ) for 0 360of x x x x x on the same axes (Example 7.3 in the textbook)

2. Graph ( ) sin( ) and g( ) sin( 45 ) for 0 360o of x x x x x on the same axes (Example 7.4 in the textbook)

3. Graph ( ) sin( ) and g( ) sin( ) for 0 360of x x x x x on the same axes (Example 7.5 in the textbook)

4. Graph ( ) sin( ) and g( ) 2 sin( ) for 0 360of x x x x x on the same axes (Example 7.6 in the textbook)

5. Graph ( ) sin( ) and g( ) sin(2 ) for 0 360of x x x x x on the same axes (Example 7.6 in the textbook)

The following properties of trigonometric functions will be very useful when graphing:

The function ( ) sin( )f x x has amplitude 1. The amplitude of a trigonometric function is defined as

1

Max-min2

A , where Max and min are the maximum and the minimum of the function ;

The function ( ) sin( )f x a x has amplitude (for 0)a a . This follows from performing a vertical stretch (or

shrink) and then using the definition above ;

The function ( ) sin( )f x x has period 2 (if working in radians) or 360 (if working in degrees)o . The

period is the smallest number which can be added to x without changing the value of ( )f x ; The function

( ) cos( )f x x has period 2 and the function ( ) tan( )f x x has period .

The function ( ) sin( )f x b x has period 2

(for 0)bb

. This follows from performing a horizontal stretch

or shrink and using the definition above; Similarly for ( ) cos( )f x b x

The function ( ) sin( )f x a b x has amplitude (for 0)a a and period 2

(for 0)bb

.

The function ( ) sin( )f x a b x c has amplitude (for 0)a a , period 2

(for 0)bb

and vertical shift of

c .

Example:

Do Example 7.14 from the textbook and from Exercise set 7F: problems 6, 7, 8, 9 and 10.

As a summary of this chapter, read the following page:

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