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
But = tanθ
SPECIAL ANGLES
These are the angles which we can find their trigonometric ratios without mathematical tables or scientific calculators.
The angles are 00, 300, 450, 600, 900, 1800, 2700, 3600.
Finding the trigonometric ratios for special angles.
Case 1: Consider 300 and 600
Here use an equilateral triangle with unit sides
That is
From AMB (right angled)
Then from the fig above
Sin 300 = =
Case 2 Consider 450
Here use are square with unit sides (1 unit)
That is
From ABC (right angled)
= +
= 1² + 1² = 2
=
Then sin 450 = =
Cos 450 = =
Tan 450 = = 1
Trigonometric ratios for 00, 900, 1800 and 2700 and 3600.
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
From a unit circle we use
X= cosine value of an angle
Y= sine value of an angle
Hence consider angles 00, 900, 1800, 2700, 3600 and their corresponding coordinates in a unit circle.
00 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<900
The all trig ratios are positive and are obtained directly from four figure (mathematical figure
2nd Quadrant angles
The range of the angles is 900 < θ <1800
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 00 and 3600 inclusive.
Solution:
1 + – =0
But from Pythagoras identity
cosθ = 0,cos θ =-1
case of cosθ = 0
θ=cos–(0)
θ=900
θ=900,2700
case of cosθ = 0
θ=cos–(0)
θ=900
θ=900,2700
Example 2.
Solve for the values of x between 00 and 3600 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
m2 -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) = 450
X = 1800 + 450 = 2250
Tan x =-1
X= tan -1(-1)
X =180 450 =1350
X = 3600 -450=3150
Case 2: m3
Tan2x = 3, tanx=
Tan x =
X = tan-1( =600
X =1800 + 600 =2400
tan x =-
x = tan -1(-
= 1800 -600=1200
X=3600 -600=3000
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)
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
=1 + tan2θ -(1 – sin2θ)
=1 + tan2θ -1 + sin2θ
=tan2θ+ sin2θ
tan2θ+ sin2θ L.H.S proved
tan2θ+ sin2θ L.H.S proved
ii) cot4θ+ cot²θ= cosec4θ – cosec2θ
solution.
Dealing with L.H.S
Dealing with L.H.S
Proof
=Cot4θ + cot2θ
then
then
=Cot2θ(cot2θ + 1)
But Cot2θ+ 1 =cosec2θ
Cot2θ =cosec2θ -1
(cosec2θ -1) cosec2θ
Cosec4θ – cosec2θ R.H.S
Cot4θ + cot2θ= cosec4θ – 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
(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 by both sides
+ =
1+ =
But
Then 1 + =
1 + =
1 + =
iii) X = 1 +
= x – 1 ……….. (i)
= y
Refer, + = 1
Dividing by both sides
+ =
ÆŸ + 1 =
+ 1=
+ 1 =
= 1
Solution (iv)
x = – ………….(a)
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
From OTR
=
But TR = TS + SR
=
= + , but TS = PQ
= +
Multiplying by and by
But from the figure above
= =
= , =
Then substituting into
= +
From (1) if B=–B
But =
But =
=â»
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=â» 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
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) =
a) =
= –
=
=
=
= 1
=
= =
=
=
= –
=
.
If = , find the tangent of x in terms of and then find tan x when = 45° and = 60° (leaving your answer in surd form)
If = , find the tangent of x in terms of and then find tan x when = 45° and = 60° (leaving your answer in surd form)
: = cos
+ = +
– =cos x cos – sin
=
=
= =
=
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 – –
= 1 – –
Or
= –
= –
= 1 –
=
= – 1 +
= 2 – 1
TRIPLE ANGLE FORMULAE
i) Consider
sin(2θ+θ) =sin2θcosθ +
= 2
= –
= 2
= 2 + –
3 –
But θ = 1 –
= –
= 3θ –
=3 – 4
ii) Consider =
= –
But = –
= 2
= –
-2
=cos3θ
But =1 –
– 3
+3
