# TOPIC 7: TRIGONOMETRY (I) ~ ADV MATHEMATICS FORM 5

## 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

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