© Boardworks Ltd 2006 1 of 35 © Boardworks Ltd 2006 1 of 35 A-Level Maths: Core 3 for Edexcel C3.3 Trigonometry 1 This icon indicates the slide contains

© Boardworks Ltd 20061 of 35 © Boardworks Ltd 20061 of 35

A-Level Maths: Core 3for Edexcel

C3.3 Trigonometry 1

This icon indicates the slide contains activities created in Flash. These activities are not editable.

For more detailed instructions, see the Getting Started presentation.

© Boardworks Ltd 20062 of 35

The inverse trigonometric functions

The reciprocal trigonometric functions

Trigonometric identities

Examination-style questionCo

nte

nts




The inverse of the sine function

Suppose we wish to find θ such that

In other words, we want to find the angle whose sine is x.

In this context, sin–1 x means the inverse of sin x.

sin θ = x

θ = sin–1 x or θ = arcsin x

This is not the same as (sin x)–1 which is the reciprocal of

sin x, .1sinx

Is y = sin–1 x a function?

This is written as



The inverse of this graph is not a function because it is one-to-many:

We can see from the graph of y = sin x between x = –2π and x = 2π that it is a many-to-one function:

y

x

y = sin x

y = sin–1 x

x

y



There is only one value of sin–1 x in this range, called the principal value.

However, remember that if we use a calculator to find sin–1 x (or arcsin x) the calculator will give a value between –90° and 90° (or between – ≤ x ≤ if working in radians).2

2

So, if we restrict the domain of f(x) = sin x to – ≤ x ≤ we have a one-to-one function:

2

2

y

2 2

1

–1

x

y = sin x


y = sin x

x

y

2

2

1–1

y = sin–1 x2

2

1–1

y = sin–1 x

The graph of y = sin–1 x

The graph of y = sin–1 x is the reflection of y = sin x in the line y = x:

The domain of sin–1 x is the same as the range of sin x :

The range of sin–1 x is the same as the restricted domain of sin x :

–1 ≤ x ≤ 1

2 2

1

–1

x

y

2

2– ≤ sin–1 x ≤

(Remember the scale used on the x- and y-axes must be the same.)

Therefore the inverse of f(x) = sin x, – ≤ x ≤ , is also a one-to-one function:

2

2

f –1(x) = sin–1 x


The inverse of cosine and tangent

We can restrict the domains of cos x and tan x in the same way as we did for sin x so that

if f(x) = cos x for 0 ≤ x ≤ π

– < x <And if f(x) = tan x for 2

2

The graphs cos–1 x and tan–1 x can be obtained by reflecting the graphs of cos x and tan x in the line y = x.

f –1(x) = cos–1 xthen for –1 ≤ x ≤ 1.

f –1(x) = tan–1 xthen for x .


y = cosx

x0

y

2

1–1

y = cos–1 x

The graph of y = cos–1 x

The domain of cos–1 x is the same as the range of cos x :

The range of cos–1 x is the same as the restricted domain of cos x :

0 ≤ cos–1 x ≤ π

–1 ≤ x ≤ 1

–1

x0

1

2

y

2

1–1

y = cos–1 x


y = tanx

x

y y = tan x

y = tan–1 x2

2

x2

2

y = tan–1 x2

2

The graph of y = tan–1 x

The domain of tan–1 x is the same as the range of tan x :

The range of tan–1 x is the same as the restricted domain of tan x :

y

x

– < tan–1 x <2

2


Find the exact value of sin–1 in radians.3

2

Problems involving inverse trig functions

To solve this, remember the angles whose trigonometric ratios can be written exactly:

10

01

10 12

1212

32

32

13

12

tan

cos

sin

90°60°45°30°0°degrees

0radians 6

4

3

2

3

From this tablesin–1 =3

2 3


Find the exact value of sin–1 in radians.


22

This is equivalent to solving the trigonometric equation

22cos θ = – for 0 ≤ θ ≤ π

this is the range of cos–1x

We know that cos =4 1

2

Sketching y = cos θ for 0 ≤ θ ≤ π :

–1

θ0

1

2 3

4

4

22

22

From the graph, cos =34 2

2–

22=

So, cos–1 = 342

2


Find the exact value of cos (sin–1 ) in radians.


74

Let sin–1 = θ74 so sin θ = 7

4

Using the following right-angled triangle:

θ

47

The length of the third side is 3 so

3

7 + a2 = 16

cos θ = 34

But sin–1 = θ so74

cos (sin–1 ) = 34

74

a = 3






nte

nts





The reciprocal trigonometric functions are cosecant, secant and cotangent.

They are related to the three main trigonometric ratios as follows:

This is short for cosecant.

cosec x =1

sin x

This is short for secant.

sec x =1

cos x

This is short for cotangent.

cot x =1

tan x

Notice that the first letter of sin, cos and tan happens to be the same as the third letter of the corresponding reciprocal functions cosec, sec and cot.


The graph of sec x


The graph of cosec x


The graph of cot x


Properties of the graph of sec x

The properties of the graphs of sec x, cosec x and cot x can be summarized in the following table:

f(x) = cot x

f(x) = cosec x

f(x) = sec x

odd or

evenperiod

f(x) = 0 when x =

asymptotes at x =

rangedomainfunction

odd180°

odd360°neverf(x) ≤ –1,

f(x) ≥ 1

even360°neverf(x) ≤ –1,

f(x) ≥ 1

90°+180n°,

n

180n°,

n

x

x ≠ 90°+180n° n

f(x)

x

x ≠ 180n°

n

90°+180n°,

n 180n°,

n

x

x ≠ 180n°

n


Transforming the graph of f(x) = sec x


Transforming the graph of f(x) = cosec x


Transforming the graph of f(x) = cot x


Problems involving reciprocal trig functions

Use a calculator to find, to 2 d.p., the value of:

a) sec 85° b) cosec 200° c) cot –70°

a) sec 85° =1

cos85 = 11.47 (to 2 d.p.)

b) cosec 220° =1

sin220 = –1.56 (to 2 d.p.)

c) cot –70° =1

tan 70 = –0.36 (to 2 d.p.)



Find the exact value of:

a) cosec b) cot c) sec –6 2

3 3

4

a) sin =6 1

2 so, cosec = 26

so, cot = –23 1

3= – 3

3

so, sec – = –34 2

b) tan = –tan (π – )23 2

3

= – tan3

= – 3

θ is in the 2nd quadrant so tan θ = tan (π – θ)

c) cos – =34 –cos(– + π)3

4

4= – cos

θ is in the 3rd quadrant so cos θ = –cos (θ + π)

= – 12


Given that x is an acute angle and tan x = find the exact values of cot x, sec x and cosec x.


34

Using the following right-angled triangle:

x

4

3

9 +16 = 5

The length of the hypotenuse is5

So

tan x = 34 cos x = 4

5 sin x = 35

Therefore

cot x = 43 sec x = 5

4 cosec x = 53


Prove that


tan sin sec cos .x x x x

LHS = tan sinx x

sin= sin

cos

xx

x

2sin=

cos

x

x

21 cos=

cos

x

x

Using sin2x + cos2x = 1

21 cos=

cos cos

x

x x

= sec cos = RHSx x



sec( + 20 ) = 2x

cos (x + 20°) =1

2

x + 20° = 60° or 300°

x = 40° or 280°

1= 2

cos( + 20 )x

Solve sec (x + 20°) = 2 for 0 ≤ x ≤ 360°.






nte

nts





Earlier in the course you met the following trigonometric identities:

We can write these identities in terms of sec θ, cosec θ and cot θ.

1Using

sintan (cos 0)

cos

1

22 2sin cos 1

1cot =

tan

sincos

1=

cos=

sin

So

coscot (sin 0)

sin



Dividing through by cos2θ gives2

22 2sin cos 1

2 2

2 2 2

sin cos 1+

cos cos cos

2 2tan +1 sec

Dividing through by sin2θ gives2

2 2

2 2 2

sin cos 1+

sin sin sin

2 21+ cot cosec



2

tanShow that cot .

1 sec

xx

x

2

tanLHS =

1 sec

x

x

2

tan=

1 tan 1

x

x

2

tan=

tan

x

x

1=

tan x

Using sec2x = tan2x + 1

= cot = RHSx



cosec x = 5

cosec2 x = 25

Using cosec2 x ≡ 1 + cot2 x,

1 + cot2 x = 25

cot2 x = 24

cot x = ±√24 = ±2√6

x is obtuse and so cot x is negative (since tan x is negative in the second quadrant). Therefore:

cot = 2 6x 1tan =

2 6x

Given that x is an obtuse angle and cosec x = 5, find the exact value of tan x.


Trigonometric equations

Solve 22sec = 2 + tan for 0 360

Using :2 2sec 1+ tan 22(1+ tan ) = 2 + tan

22 + 2tan 2 tan = 0 22tan tan = 0

tan (2tan 1) = 0

θ = 0°, 180°, 360° θ = 26.6°, 206.6° (to 1 d.p.)

The complete solution set is θ = 0°, 26.6°, 180°, 206.6°, 360°.

ortan = 0 tan = ½






nte

nts


Examination-style question


Examination-style question

a) Prove that sec θ ≡ cos θ + sin θ tan θ.

b) Hence solve the equation 2 cos θ = 3 cosec θ – 2 sin θ tan θ in the interval 0° < θ < 360°. Give all solutions in degrees to 1 decimal place.

a)sin

= cos + sincos

RHS

2 2cos sin= +

cos cos

1=

cos

= sec = LHS



b)

Using the result given in part a):

2sec = 3cosec

2cos = 3cosec 2sin tan

2cos + 2sin tan = 3cosec

2(cos + sin tan ) = 3cosec

2 3=

cos sin sin 3

=cos 2

32tan =

56.3 , 236.3 (to 1 d p.)= .

Documents

© Boardworks Ltd 2006 1 of 35 © Boardworks Ltd 2006 1 of 35 A-Level Maths: Core 3 for Edexcel C3.3 Trigonometry 1 This icon indicates the slide contains