03 Trigonometry

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3 Trigonometry

Trigonometric functions are usually defined in terms of ratios of sides of a right-

angled triangle.

A B

C

AC is the hypotenuse. For the angle , BC is

the side opposite and AB is the side adjacent.

Then cos =ADJ

HYP

sin =OPP

HYP

tan =OPP

ADJ=

sin

cos .

We can define trigonometric functions as functions of real numbers and for this

pupose, we consider a circle of unit radius (i.e. r = 1); x2 + y2 = 1.

-x

6y

1

x

y

P(x, y)For each value of there is a unique point

on the unit circle. As changes the posi-

tion ofP varies and so the values ofx and

y depend on .

From the triangle, we have cos =x

1and sin =

y

1

x = cos

y = sin

Note that x and y are parametric functions of .

Now, as 1 cos 1, then 1 x 1and as 1 sin 1, then 1 y 1

We now introduce a third quantity;

tan =y

x(where x = 0)

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We also define the reciprocal functions

sec =1

cos , cos = 0

cosec = 1sin

, sin = 0

cot =1

tan , tan = 0

Since the equation of our circle is x2 + y2 = 1, then

cos2 + sin2 = 1

This is an important identity for the trigonometric functions.

Dividing this identity by cos2 , we get

1 + tan2 = sec2

or dividing by sin2 , we get

cot2 + 1 = cosec2

These identities are also important. As an example of how these identities can be

used, consider the following.

Example: Simplify (cos + sin )2 + (cos

sin )2.

(cos + sin )2 + (cos sin )2= cos2 + 2 sin cos + sin2 + cos2 2sin cos + sin2 = 2cos2 + 2 sin2

= 2(cos2 + sin2 )

= 2 1 = 2.

EXERCISES 30:

1. Find the five remaining trigonometric ratios in each of the following cases:

(a) cos = 3/5, 0 < < /2

(b) sec = 3/2, /2 < < (c) cot = 3, < < 3/2

2. Prove the following trigonometric identities:

(a) tan x + cot x = sec x csc x (b)1 + cos x

sin x=

sin x

1 cos x3. Show that sec2 x can be written as 1 + sin2 x + sin4 x + sin6 x + .

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Radians and Degrees:

The size of an angle can be measured in degrees or radians. The angle of a full circle

is 360 degrees (360) or 2 radians (2c ). Thus

360 = 2c

One half of this is 180 or c . Note that this does not mean that and 180 are in

any way the same. It is only when we put in the units that we get 180 = c . Note

that we also have

1 = (/180)c

or 1c = (180/)

Note that, whenever an angle is measured in degrees it is important to include the

units (the degree sign) explicitly. Often though, if the measurement is in radians,

the units can be omitted. Thus if the units are not specified, it is understood that

the angle is in radians. Therefore, sin x will always mean the sine of x radians.

Example 1: Convert the following angles into radians.

30 = 30(/180)c = (/6)c

45 = 45(/180)c = (/4)c

90 = 90(/180)c = (/2)c

Example 2: Convert the following angles into degrees.

3c

2=

3

2 180

= 270

4c

5=

4

5 180

= 144

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Returning to the unit circle, x = cos and y = sin .

-x

6y

A

B

C

D

(1, 0)

At the point A = (1, 0), = 0.

cos 0 = x = 1, sin 0 = y = 0.

i.e.cos 0 = 1

sin0 = 0

At the point B = (0, 1), =

2.

cos

2= 0

sin

2= 1

At C = (1, 0), = .

cos = 1sin = 0

At D = (0, 1), = 32

.

cos

3

2 = 0

sin3

2= 1

IfP() is the point on a circle at angle , then the points P(), P( +2), P( +4),

etc. all coincide. Therefore, in general

cos = cos( + 2k)

sin = sin( + 2k)

for any integer value of k.

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The coordinate axes divide the unit circle into four quadrants.

-x

6y

(1, 0)12

3 4

We can use symmetry properties to relate the values of trigonometric functions when

> /2 to values of these functions in the first quadrant (0 /2).Second Quadrant: /2 < < .

-x

6y

1 1y y

x x

P( ) P()

Consider the point P() in quadrant 1

and the point P( ) in quadrant 2.

By symmetry, the y coordinates are equal and the x coordinates are equal in mag-

nitude but opposite in sign.

Hence, cos( ) = x cos( ) = cos()

sin( ) = y sin( ) = sin()

tan( ) = yx = y

x

tan( ) = tan()

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Example 3: Rewrite the following trigonometric expressions with angles in the first

quadrant.

(1) sin

2

3 = sin

3

= sin

3

(2) cos3

4= cos

4

= cos

4

Third Quadrant: < < 3/2.

-x

6y

1

1

x xy

y

P( + )

P() Consider the point P() in quadrant

1 and the point P( + ) in quad-

rant 3. By symmetry, their coordinatesare equal in magnitude but opposite in

sign.

Hence, cos( + ) = x cos( + ) = cos()

sin( + ) = y sin( + ) = sin()

tan( + ) =yx =

y

x

tan( + ) = tan()

Fourth Quadrant: 3/2 < < 2.

-x

6y

1

1

x

y

y

P()

P(2 )

Consider the point P() in the quad-rant 1 and the point P(2) in quad-rant 4. By symmetry, their x coordi-

nates are equal and their y coordinates

are equal in magnitude but opposite in

sign.

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Hence, cos(2 ) = x

cos(2 ) = cos()sin(2 ) = y

sin(2 ) = sin()

tan(2 ) = yx

= yx

tan(2 ) = tan()

Example 4: Rewrite the following trigonometric expressions with angles in the first

quadrant.

(1) sin5

4= sin

+

4

= sin

4

(2) cos7

4= cos

2

4

= cos

4

These identities can be summarized by remembering that any angle between 0c and

2c can be related to an angle, , in the first quadrant by writing it as

(2nd

quadrant) + (3rd quadrant)

2 (4th quadrant).The sign of the function can be deduced using the following diagram.

C

AS

T

All functions are positive in the first quadrant. Sin is positive in the second quad-

rant. Tan is positive in the third quadrant. Cos is positive in the fourth quadrant.

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There are particular angles for which we should know values of the trigonometric

functions. These values can be obtained from the following triangles.

1

1/4

2

4

cos 4

= 12

sin

4=

12

tan

4= 1

1 1

2 2

/3

3

6

cos

3 =1

2 cos

6 =

3

2

sin

3=

3

2sin

6=

1

2

tan

3=

3 tan

6=

13

Examples : cos2

3

= cos

3 =

cos

3

=

1

2sin

3

4= sin

4

= sin

4=

12

tan7

4= tan

2

4

= tan

4= 1

Complementary Angles:

Complementary angles add to 90.

2

sin

2

= cos

cos

2

= sin

tan

2

= cot

In general if the angle for the trigonometric function is of the form

2 or 3

2 ,

the trigonometric function changes

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i.e. sin coscos sintan

cot

Example 1: sin

2+

4

= + cos

4=

12

.

Example 2: tan

3

2+

6

= cot

6=

3.

EXERCISES 31: Calculate sine, cosine and tangent for the following angles:

1. 34

2. 43

3. 76

4. 114

5. 6. 74

7. 56

Addition Formulae:

The remaining trigonometric identities are obtained from the addition formulae.

sin(x + y) = sin x cos y + cos x sin y

cos(x + y) = cos x cos y sin x sin y

From these we can obtain the formulae for sin(x y) and cos(x y)sin(x y) = sin[x + (y)]

= sin x cos(y) + cos x sin(y)

sin(x y) = sin x cos y cos x sin y

cos(x

y) = cos[x + (

y)]

= cos x cos(y) sin x sin(y)

cos(x y) = cos x cos y + sin x sin y

tan(x + y) =sin(x + y)

cos(x + y)

=sin x cos y + cos x sin y

cos x cos y sin x sin y

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Dividing top and bottom by cos x cos y, we get

tan(x + y) =

sin x cos y

cos x cos y+

cos x sin y

cos x cos y

cos x cos ycos x cos y

sin x sin ycos x cos y

i.e. tan(x + y) =tan x + tan y

1 tan x tan yReplacing y with y, we have

tan(x y) = tan x tan y1 + tan x tan y

Example 1: Expand sin34 +

.

sin

3

4+

= sin

3

4cos + cos

3

4sin

= sin

4cos cos

4sin

=1

2cos 1

2sin .

Example

2:

If cos A =

1

2 ,

3

2 < A < 2

and sin B =3

5,

2< B <

Find (a) sin(A + B), and

(b) cos(A B) without using a calculator.

First, recall the diagram:C

AS

T

The expansions of sin(A + B) and cos(AB) require values for sin A and cos B. Weare given that 3/2 < A < 2, which indicates that A lies in quadrant 4, so thatsin A < 0. We have also been given that /2 < B < , which indicates that B lies

in quadrant 2, so that cos B < 0.

Using cos A =1

2, find sin A:

y

1

2

AUsing Pythagoras theorem we have 12 + (y)2 = 22 ory2 = 22 12, so y = 3.

sin A =

3

2

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Using sin B =3

5, find cos B:

3

x5

B

Using Pythagoras theorem we have 32 + (x)2 = 52 orx2 = 52 32, so x = 4.

cos B = 45

(a) sin(A + B) = sin A cos B + cos A sin B

=3

2 4

5+

1

2 3

5

=4

3 + 3

10

.

(b) cos(A B) = cos A cos B + sin A sin B

=1

2 4

5+

32

35

=4 33

10.

EXERCISES 32:

1. Use addition formulae to simplify the following:

(a) sin 70 cos20 + cos 70 sin 20

(b) cos65 cos25 sin 65 sin 25(c) sin 2x cos x + cos 2x sin x

(d) cos2 cos + sin 2 sin

(e)tan75 tan45

1 + tan 75 tan45

(f)tan15 + tan30

1 tan 15 tan30(g)

tan x tan2y1 + tan x tan2y

2. If tan A = 3 and tan B = 2, find the values of:

(a) tan(A + B) (b) tan(B A)3. If cos = 3/5, where 0 < < /2 and cos = 4/5, where /2 < < ,

find

(a) sin , tan

(b) sin , tan

(c) sin( + )

(d) cos( )(e) tan( )(f) tan(+ /4)

4. Prove that tan 75 = 2 +

3 using an appropriate addition formula.

5. Find tan 15 in the simplest surd form.

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6. Simplify without expanding each term:

cos( + )cos( ) sin( + ) sin( )7. Use the identities

sin( + ) = sin cos + cos sin , sin( ) = sin cos cos sin ,cos( + ) = cos cos sin sin , cos( ) = cos cos + sin sin .

to find expressions for sin cos , cos cos and sin sin .

Double Angle Formulae:

If we let y = x in the addition formula

sin(x + y) = sin x cos y + cos x sin y

we get

sin(x + x) = sin x cos x + cos x sin x

i.e. sin 2x = 2 sin x cos x

Similarly, cos(x + y) = cos x cos y sin x sin ybecomes cos(x + x) = cos x cos x

sin x sin x

i.e. cos 2x = cos2 x sin2 x

Also, tan(x + y) =tan x + tan y

1 tan x tan y

becomes tan 2x =2tan x

1 tan2 x

By using the identity cos2 x + sin2 x = 1, we obtain the alternative forms of the

double angle formula for cos 2x

i.e.cos2x = 1 2sin2 xcos2x = 2 cos2 x 1

Example 1: If sin =4

5,

2< < , find (a) sin 2, (b) cos2, and (c) tan2.

First, recall the diagram:C

AS

T

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The expansion of sin 2, cos 2, and tan 2 will require values for cos . We are given

that /2 < < , which indicates that lies in quadrant 2, so that cos < 0.

4

x

5

Using Pythagoras theorem, we find x = 3.

cos = 35

(a) sin 2 = 2 sin cos

= 2 45

35

= 2425

.

(b) cos 2 = cos2

sin2

=9

25 16

25= 7

25.

(c) tan 2 =sin2

cos2

=24/257/25 =

24

7.

Example 2: Simplify 2 sin 2 cos2.

2sin2 cos2 = sin 2(2)

= sin 4.

Example 3: Simplify cos2 15 sin2 15.

cos2 15 sin2 15 = cos 2(15)

= cos30 =

3

2.

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Example 4: If cos2 = 79

and is an acute angle (i.e. less than 90), find (i)

cos , (ii) sin , (iii) tan .

(i) cos 2 = 2 cos2

1 cos2 =

cos2 + 1

2

cos =

cos2 + 1

2

Take the positive square root, as we are in quadrant 1.

cos = +

cos2 + 1

2

=

7/9 + 1

2

=

2/9

2=

1

9=

1

3.

(ii) As cos 2 = 1 2sin2 sin2 =

1 cos22

sin = +

1 cos22

(Quadrant 1)

=

1 + 7/9

2=

16/9

2

=

8

9=

2

2

3.

Note that as you have cos from part (i) you could also have made use of the

identity sin2 + cos2 = 1.

sin2 = 1 cos2 = 1

19

2

= 89

sin = +

8

9=

2

2

3. (Quadrant 1)

Another approach would involve using cos from part (i), then drawing up a triangle

to get a value for sin as follows

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1

x3

Using Pythagoras theorem we find x =

8,

sin =

83 = 2

23 .

(iii) tan =sin

cos

=2

2/3

1/3= 2

2.

EXERCISES 33:

1. Express the following in terms of 2:

(a) 2sin cos

(b) 4cos sin

(c) 2cos2 1(d) cos2 sin2

(e) sin2 cos2

(f)2tan

1 tan2 2. If cos = 1/3 and < < 3/2, find the values of:

(a) sin 2 (b) cos2 (c) tan2

3. If cos2 = 7/9 and is acute, find the values of:(a) cos (b) sin (c) tan

4. Prove the following trigonometric identities:

(a) cos4 x sin4 x = cos 2x(b) sin 3x = 3 sin x 4sin3 x

(c) cos3x = 4 cos3 x 3cos x

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Example 1: Sketch the graph of f() = 3 cos 2, 0 2.

Here a = 3 and n = 2. Hence the amplitude is |a| = 3 and the period is T = 2n

=

2

2 = . That is, the curve completes one cycle in a distance of . For 0 2,the curve completes two cycles.

-

6f()

3

3

22

4

7

4

3

2

5

4

3

4

Example 2: Sketch the graph of f() = sin 2

, 0 2.

Here a = 1 and n = 12

. Hence the amplitude is |a| = 1 and the period isT =

2

n=

2

1/2= 4. That is, the curve completes one cycle in a distance of 4.

For 0 2, we have only half a cycle. Note the effect of the negative number

a is to reflect the graph of f() = sin 2

in the x-axis.

-

6f()

2

1

1 f() = sin 2

Graphs of a sin n( ) and a cos n( )We now have replaced by . This has the effect of translating the graph units to the right along the -axis.

is called the phase.

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Example 3: Sketch the graph of f() = cos

4

, 0 2.

Here a = 1 and n = 1. Hence the amplitude is |a| = 1 and the period is T = 2n

=

21

= 2. The phase is = 4

, which has the effect of translating the basic cosine

curve

4units to the right.

-

6f()

4

2

3

4

5

4

3

2

7

4

2

1

1

-

f() = cos

f() = cos

4

Example 4: Sketch the graph of f() = 3 cos 2

+

4

, 0 2.

Here a = 3 and n = 2. Hence the amplitude is |a| = 3 and the period is T =2

n=

2

3. The phase is =

4, which has the effect of translating the curve of

f() = 3 cos 2,

4 units to the left.

-

6f()

4

2

3

4

5

4

3

2

7

4

2

3

3

f() = 3 cos 2

f() = 3 cos 2

+

4

Graphs of a sin n( ) + b and a cos n( ) + b.The effect of the constant b is obviously to translate the curve b units up or down

the y-axis according to whether b is positive or negative. It does not alter the period

or the amplitude of the curve.

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Example 5: Sketch the graph of f() = sin 2

+ 2, 0 2.

The constant 2 raises the graph of f() = sin 2

, two units up the y-axis. The

range is 1 f() 2, the amplitude is |a| = 1, and the period is 4. Note thatonly half of the cycle needs to be drawn as 0 2.

-

6f()

2

1

1

2 f() = sin 2

+ 2

f() = sin 2

The Tangent Function:

To sketch the graph of f() = tan we first locate the zeros of the function then

the vertical asymptotes. Since tan =sin

cos , the graph will cross the axis where

sin = 0 (i.e. at = n). The asymptotes will occur at cos = 0 or =

2+ n. It

follows that the graph looks like

-

6f()

32

2

0

2

3

2

Note that the period of the graph is T = .

In general the period, T, of the tangent function with angular frequency n is given

by T =

n. i.e., the function repeats itself every

nunits.

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Example 1: Sketch the graph of f() = tan 2, 2

2.

Here n = 2, so the period T =

n=

2.

-

6f()

2

4

4

2

f() = tan 2

EXERCISES 34:

1. State (i) the amplitude, (ii) angular frequency, (iii) horizontal translation, and(iv) vertical translation for

(a) cos

t

2

3 (b) 2 sin(4t ) + 1

2. Write down the trigonometric function described by:

(a) the cosine function with amplitude1

2, angular frequency 3, horizontal

translation2

3units to the right and vertical translation 2 units down-

wards

(b) the sine function with amplitude 10, angular frequency 5 and horizontal

translation

10units to the left.

3. A mass attached to the end of a spring oscillates so that its displacement scm from a central position is given by: s = 6 sin(3t), where t is the time inseconds.

(a) What is the maximum displacement from the central position?

(b) Where is the object 5 seconds into the motion?

(c) Where is the object 17.3 seconds into the motion?

4. Graph the functions:

(a) y = cos

x 3

(b) y = tan

x2

(c) y = sin

x +

4

+ 2

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Solving Trigonometric Equations:

Generally, trigonometric equations have an infinite number of solutions unless the

domain is restricted in some way.

Solving sin = a:

Example 1: Solve 2 sin x 3 = 0.Note that no domain is specified. Hence it is assumed we are solving for x R.Also, x is measured in radians.

Rewrite the equation as sin x =

3

2.

C

AS

T

Sine is positive in quadrants 1 and 2. From the triangle below

1

2

/3

3

6 sin

3=

3

2

x =

3(quadrant 1)

or x = 3

=2

3(quadrant 2)

Sine has period 2 and so has an infinite number of solutions.

x =

2n +

3

2n +2

3

(n is an integer)

The above solutions can be observed graphically by considering the values of x at

the points of intersection of the curves y = sin x and y =

3

2.

-x

6y

3

2

3

2 3

1

1

..

...

...

....

....

..

...

...

....

.... y =

3

2

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Solving cos = a:

Example 2: Solve

2cos x 1 = 0.

Note that no domain is specified, so we solve for all x R.Rewrite the equations as cos x =

12 C

AS

T

Cosine is positive in quadrants 1 and 4. From the triangle below

1

1/4

2

4 cos

4=

12

x =

4

(quadrant 1)

or x = 4

or 2

4

(quadrant 4)

But since cosine has period 2 we have an infinite number of solutions.

x =

2n +

4

2n 4

(n is an integer)

Solving tan = a:

Example 3: Solve 3 tanx 5 = 0.

tan x =5

3 C

AS

T

The tangent function is positive in quadrants 1 and 3. This ratio doesnt correspond

to on of the standard triangles, so we must use a calculator to find x (in radians)

x = tan1

53

1.030c (quadrant 1)and x + 1.030c (quadrant 3)

The general solution is therefore

x =

1.030 + 2n

+ 1.030 + 2n(n is an integer)

This result could also be written as x = 1.030 + n.

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Restricted Domain:

Consider the example for cos = a again.

Example 4: Solve

2cos x

1 = 0, 0 x 2.

This time our domain is restricted. We only want solutions that lie in the range

0 x 2. From before: x = 2n 4

.

If n = 0 x =

4(inside range)

or x = 4

(outside range)

n = 1 x = 2 +

4=

9

4(outside range)

or x = 2 4

= 74

(inside range)

For the domain specified we have only two solutions, namely

x =

4or x =

7

4.

EXERCISES 35:

1. Solve for x:

(a) 2cos x 1 = 0, x [0, 4](b) sin x = 0.9962, x [0, 360](c) tan x + 1 = 0, x [0, 2](d) sin

2x

6

= 1, x [0, 2)

(e) tan(3x + 80) = 0.8390, x [0, 180)2. Solve for where [0, 2]:

(a) sin + cos = 1

(b) cos2 + sin + 1 = 0

(c) sin + cos =1

2

(d) sin2 + cos2 = 1

(e) tan = cot(2)

(f) sin(2) = sin(7)

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03 Trigonometry