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MATHEMATICS FORM FIVE TOPIC 7: TRIGONOMETRY I

TRIGONOMETRY

Trigonometry is the study of angle measurement and functions that depends on angle.
The fundamental trigonometric ratios are
Sine (sin)
Cosine (Cos)
Tangent (Tan)
Others are cosecant (cosec)
Secant (sec)
Cotangent (cot )
Let θ be the angle in a right angled triangle; then we say
Sin θ
COS θ
Tan θ
And= Cosecant θ = Cosec θ
= secant = sec θ
= Cotangent θ = cot θ
Consider a right angled triangle below
Sin θ ==………..(i)
cos θ ==……….(ii)
tan θ==…………(iii)
= cosec θ==(iv)
= sec θ ==(v)
= cot θ ==(vi)


But= tanθ
SPECIAL ANGLES
These are the angles which we can find their trigonometric ratios without mathematical tables or scientific calculators.
The angles are 0, 30, 45, 60, 90, 180, 270, 360.
Finding the trigonometric ratios for special angles.
Case 1: Consider 30 and 60
Here use an equilateral triangle with unit sides
That is
FromAMB (right angled)

Then from the fig above
Sin 30 ==

Case 2 Consider 45
Here use are square with unit sides (1 unit)
That is
FromABC (right angled)
=+
= 1² + 1² = 2
=
Then sin 45 ==
Cos 45 ==
Tan 45== 1
Trigonometric ratios for 0, 90, 180and 270 and 360.
Here use a unit circle ‘Discussed also in O level’
A unit circle is a circle with radius (1 unit)
Suppose p(x,y) is a point in a unit circle

Generally in a unit circle
X = cosine value of an angle
Y= sine value of an angle
= Tangent of an angle

Angle measurement can be in two ways.
Clockwise direction (-ve angles)
Anticlockwise direction (+ve angles)
From a unit circle we use
X= cosine value of an angle
Y= sine value of an angle
Hence consider angles 0, 90, 180, 270, 360 and their corresponding coordinates in a unit circle.
means




360°means
Summary:-
The concept of picture and negative angles.

 

But sine function and tangent function are odd functions
Cosine function is an even function

Fig above
From Sin θ =
Sin ( -θ) = –= -sinθ
cos θ =
cos(-θ) = Cos θ
THE IDEA OF QUADRANTS
The idea is discussed in O’Level form IV Basic Mathematics, but let us recall the idea.
1st Quadrant Angles
The range of the angles is 0°< θ<90
The all trig ratios are positive and are obtained directly from four figure (mathematical figure
2nd Quadrant angles
The range of the angles is 90 < θ <180
3rd Quadrant
Ranges from 180°< θ< 270°
4th Quadrant
Ranges from 270°<θ<360°




Eg: Sin 315° = -Sin (360° -315°)
=-
= -tan (360° – 330°
= -tan 30°
=
=
==
PYTHAGORAS THEOREM (IDENTITY)
Consider a right angled
From Pythagoras theorem
+ b² = c²
Dividing by C²
+=
+ (= 1————–
Substitute equations (i) and (ii) into (*)
Then we get
Is the Pythagoras Identity.
Dividing equation (1) by

 

dividing equation (i) by Sin2θ

APPLICATIONS OF PYTHAGORAS IDENTITY
I. SOLVING TRIG EQUATIONS
Example 1.
Solve the equation 1 += 0 for the values of the values (θ) between 0 and 360 inclusive.
Solution:
1 +=0
But from Pythagoras identity
cosθ = 0,cos θ =-1
case of cosθ = 0
θ=cos(0)
θ=900
θ=90,2700

Example 2.
Solve for the values of x between 0 and 360 inclusive of
(i) Tan 4x + 7 = 4sec2x
(ii) -6sm2x – cosx + 5 =0
Solution
Tan4x + 7 =4sec2x
But tan2x + 1 =sec2x
Tan4x + 7=4(tan2x + 1)
Tan4x + 7 =4tan2x + 4
Tan4x +7-4tan2x -4 =0
Tan4x -4tan2x + 3 =0
Let tan2x =m
Then m2 – 4m +3 =0
m-3m –m + 3 =0
m(m -3)-1(m-3)=0
(m – 1)(m-3) =0
m – 1 =0, m- 3=0
m= 1, m=3
Case 1 m =1 =tan2x
Tan x =
Tan x = 1
X = tan-1(1) = 45
X = 180 + 45 = 225
Tan x =-1
X= tan -1(-1)
X =180 45=135
X = 360 -45=315
Case 2: m3
Tan2x = 3, tanx=
Tan x =
X = tan-1(=60
X =180 + 60 =240
tan x =-
x = tan -1(-
= 180 -60=120
X=360 -60=300
x=work on (ii)
II PROVING IDENTITIES
Examples: prove the following identify
i) Tan2θ + sin2θ =(secθ + cosθ) (secθ – cosθ)
ii) Cot4θ + cot2θ =cosec4θ – cosec2θ
iii)= cosecθ – cotθ
iv)
v) cosecθ –sinθ = cotθ
Solution: (i)
tan2θ + sin2θ = (secθ+ cosθ) (secθ –cosθ)
Delaying with R.H.s
Proof = (secθ + cosθ)(secθ – cosθ)
Then
=sec2θ – cos2θ
But sec2θ = 1+ tan2θ and
Cos2θ = 1 –sin2θ
=1 + tan2θ -(1 – sin2θ)
=1 + tan2θ -1 + sin2θ
=tan2θ+ sin2θ
tan2θ+ sin2θ L.H.S proved

ii) cot4θ+ cot²θ= cosec4θ – cosec2θ
solution.
Dealing with L.H.S
Proof
=Cot4θ + cot2θ
then
=Cot2θ(cot2θ + 1)
But Cot2θ+ 1 =cosec2θ
Cot2θ =cosec2θ -1
(cosec2θ -1) cosec2θ
Cosec4θ – cosec2θ R.H.S
Cot4θ + cot2θ= cosec4θ – cosec2θ
iv) sin θtanθ + cosθ=secθ
solution.
Proof
Dealing with L.H.S
Sinθtanθ+ cosθ
But tanθ =
Then
Sinθ+ cosθ
== secθ
sin²θ + cos²θ =1 (Pythagoras identity)
sin


III) ELIMINATION PROBLEMS
Examples:
Eliminate ÆŸ from the following equations
i) Cosθ + 1 =x and sinθ =y
ii) X= a sinθ and y= btan θ
iii) X= 1 + tanθ and y = cos θ
iv) X= sinθ – cosθ
Y= cotθm+ tanθ
Solution.
(i) Cosθ + 1 =x
Cosθ=x – 1 ……… (i)
sinθ = y…………..(ii)
squaring equations (i) and (ii) the sum
cos²θ+ sin²θ= (x -1)² + y²
but sin²θ + cos²θ =1
then 1= (x – 1)² + y²
1 = x² – 2x + 1 + y²
x² + y2 -2x + 1 – 1 =0
x² +y²- 2x =0
ii) from x = a sinθ, sinθ=
and from y=btanθ, tanθ=
refer+=1
dividing byboth sides
+=
1+=
But
Then 1 +=
1 +=
1 +=
iii) X = 1 +
= x – 1 ……….. (i)
= y
Refer,+= 1
Dividing byboth sides
+=
ÆŸ + 1 =
+ 1=
+ 1 =
= 1
Solution (iv)
x =………….(a)
Y =+……….(b)
From (b)
=+


Y==
Y =
Squaring
x² =
x² =-2+
=+-2
x² = 1- 2
then
x² = 1 – 2
but=
x² = 1 – 2
x² =1 –
x² +-1 =0
NB: In elimination problems concept is to eliminate the trig function in the equation, then try the possibilities of eliminating it by connecting it to the pythageras theorem (identity)
COMPLEMENTARY ANGLES
Consider the triangle below
=(i)=(iv)
=(ii)=(v)
=(iii)=.(vi)
Thus
Is the condition for complementary angles
Definition: Complementary angles are angles whose sum is 90°
E.g: A + B = 90°
30° + 60° = 90°
30° and 60° are complementary angles.
NB: Supplementary angles are angles whose sum is 180°
Eg: A + B = 180°
Then A and B are supplementary angles
COMPOUND ANGLES FORMULA
Consider two angles say A and B then the angles A + B are called compound angles.
The concept here is to obtain
Sin (A ±B), Cos (A ±B), Tan (A ± B)
However it is easier to say that
Sin(A + B) = sin A + sin B
Testing if it is true
Let A= 60 and B= 30°
Sin(A + B) = sin(60° + 30°) = sin 90° = 1
Sin A + sin B = sin 60°+ sin 30°
Consider the figure below
FromOTR
=
But TR = TS + SR
=
=+, but TS = PQ
=+
Multiplyingbyandby
But from the figure above
==
=,=
Then substituting into
=+
From (1) if B=B

But=
=⁻
Again from the figure above
=
But OT =

For tan
Refer=
=

Dividing numeration and denomination by


=
From above equation
If B = -B, then
Tan( A+=
But tan=⁻ tanB
=
Or, shown by
=
Use procedure (5) obtain (6)
APPLICATION OF THE COMPOUND FORMULAE
I. PROVING OF IDENTITIES
Examples:
Prove the following trig identities
i)=+
ii)=
iii)=
Proof(i)=
Dealing with L.H.S
II. COS(A+B)COS(A-B) =
Proof dealing with L.H.S
B –
=1-and
= 1 –then
-(sin2A-cos2Asin2B)
cos2A-cos2Asin2B-sin2A+cos2Asin2B
R.H.S
=
III.=
Proof
Dealing with L.H.S
=
=1
=
But=
=+ 1
1 –
=
=
=
IV. FINDING VALUES OF TRIG RATIOS
Examples: Evaluate
a)b)c)d)
Solution:
a)=
=
=
=
=
= 1
=
==
=
=
=
=
.
If=, find the tangent of x in terms ofandthen find tan x when= 45° and= 60° (leaving your answer in surd form)
:= cos
+=+
=cos x cossin
=
=
==
=
Given=45°,= 60
=


=
DOUBLE ANGLE FORMULAE
Recall (a)=
If B = A
=
=2
b)
If B = A
=
c)
If B = A
=………………….. (iii)

Also from
=
But= 1 –
=(1 –)-
1 –
Or
=
= 1 –
=
=– 1 +
= 2– 1

TRIPLE ANGLE FORMULAE
i) Consider
sin(2θ+θ) =sin2θcosθ +
= 2
=
= 2
= 2+
3

Butθ = 1 –
=
= 3θ –
=3– 4
ii) Consider=
=
But=
= 2
=
-2
=cos3θ
But=1 –
– 3
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Most mathematical activity involves the use of pure reason to discover or prove the properties of abstract objects, which consist of either abstractions from nature or—in modern mathematics—entities that are stipulated with certain properties, called axioms. A mathematical proof consists of a succession of applications of some deductive rules to already known results, including previously proved theorems, axioms and (in case of abstraction from nature) some basic properties that are considered as true starting points of the theory under consideration.

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