ADVANCED MATHEMATICS FORM 5 – TRIGONOMETRY

Applications of the double and triple formulae

A.   Proving Identities

Examples: Prove the following identities

(i)                           +

(ii)                        =

(iii)                     

Solution(i)

      I.            Proof

Dealing with L.H.S

=

=cos²A+cos²A-sin²A
=2cos²A-sin²A

–

=2 –

= R.H.S

   II.            Solution(ii)

Dealing with L.H.S

But

A = R.H.S

III.       Solution(iii)

=

=

=

Work on the following problems prove the identities

i)                  =

ii)                =

iii)            

iv)              =

v)                + =2

vi)              =

vii)           =

viii)         =

ix)             = 2

x)                =

xi)             + =

Warm up with:

i)                  Find  tan without calculate mathematical tables

ii)                

HALF ANGLES FORMULAE

From = –

Then

=

=

= = 1 –

=

= – 1 +

=

Again from = –

But = 1 –

=1 – –

= 1 -2

2 = 1 –

For =

= =

Similarly the formulae can be expressed as

EQUATION OF THE FORM

a = c

where a, b and c are real constant.

The task here is to solve the equation. The are two ways to solve.

       i.            Using t –fomulae

     ii.            Using R – fomula (or transforming a function a + b= c as a single function)

I.                 USING t- FORMULAE

Consider

Concept of t formulae From =

=

=

=

But =

=

Again = 1 +

=

Let = y

from Pythagoras theorem

+ =

= –

= (1+y²) – ²

=

=

=

=

Then =

Cos2ÆŸ = …………………… (ii)

= =

= ………………………..(iii)

From equations (i) (ii) and (iii) it follows that

=

=

=

Let t =, then we get

Equation (1), (2) and (3) are called t-substitution formulae

Solving the equation

+ b = c

Let t =

= ,

+ b = c

=c

a – at² + 2bt = c(1 + t²)

a – at² + 2bt = c + ct²

at² + ct² – 2bt + c – a =o

(a + c)t² – 2bt + c –a =o

Quadratic equation

Solve for it

t=

=

=

=

t =

=

t =

but t =

tan =

=

Example:

Solve for values of θ between 0° and 180° if 2cos θ+ sin θ= 2.5

Solution: let t = tan

2 + 3 =2.5

Then =

Sin θ=

2 + 3 2.5

+ 3 = 2.5

2       -2t² + 6t =2.5

2– 2t² + 6t = 2.5 + 2.5t²

2

4– 4t² + 12t = 5 + 5t²

9t² – 12t + 1 = 0

t=

=

= =

=

=

t = = 1.244 or t =

t = 0.00893

case 1:
t =1.244, t= tan tan = 1.244

= tan

=

=51.2° = θ = 51.2 x 2

= 102.4°

case 2:
t = 0.0893

= 0.0893

=

=

 =5.1°, θ = 10.20

θ=

Example 2: solve the equation

5 – 2=2 for

     for -180⁰
x

Using t formula, let t =

5cosx – 2sin x=2

5=2

=2

5 – 5t² -4t = 2

5 – 5t² – 4t = 2 + 2t²

7t² + 4t -3 =0

7t² + 7t – 3t -3 =0

7t (t + 1) -3(t + 1) =0

(7t – 3) (t + 1)=0

7t – 3 = 0 or t + 1=0

7t =3 t= -1

t =

Case 1.
t= = 0.42857

= 0.42857

=

= 23.2° = 23.2°x2=46.4°

Case2,
t=^–1, tan = ⁻1

= =

II.            SOLVING THE EQUATION

acosθ = C

R-formula or simply transforming a function acosÆŸ bsinÆŸ as a single function.

From acosθ bsinθ = c

Consider acosθ + bsinθ – this can be expressed transformed into form

here R >O

R is the maximum value of a function (or Amplitude)

is a phase angle and it is an acute angle

Then from acosθ + bsinθ =C

acosθ + bsinθ = Rcos(θ – )

acosθ + bsinθ= R

Square equation (i) and (ii) then sum

(Rcos + = a² + b²

R²cos²+ R²sin² = a² + b²

R² = a² + b²

But + =1

R².1 = a² + b²

R² =a² + b²

Then from

acosÆŸ + bsinÆŸ =c = Rcos (ÆŸ –

Rcos(

=

– =

=

Example

 Rcos cosx = 3cosx

Rcos = 3 —- (i)

-4sinx = Rsinxsin

Sin = 4 —– (ii)
Dividing (ii) by (i), then we get

=

(i) and (ii) then sum

+

= 9 + 16

R² + R = 25

R²1 =25

R= 25, R= R=5

But

5

C = 1.5 , = 53.12°

5 = 1.5

=

Cos =0.3

X + 53,12°=

X + 53.12° = 72.54°

X = 72.54° – 53.12°

x = 19.42°

Example 2: solve for between 0° and 180° if

2= 2.5

Solution

2= 2.5

R=23

R

R =2

R =2 —(i) and

R

R = 3 ………. (ii)

Dividing (ii) by (i)

=

,

= 56.3°

Squaring (i) and (ii) then add

+ = 2² + 3²

R² + R = 4 + 9

R²

R² = 13, R=

Then 

θ- 56.3°=

θ=

= + 56.3°

= 46.1° + 56.4°= 102.4°

θ= 313.9° + 56.3°= 370.2°

= 370.2° – 360°=10.2°

θ=10.2°,102.4°

Example: 3
solve for x iƒ5 – 2sinx =R=2

5 – 2= R

5 = R

R = 5 ……………………. (i)

2 = R

R = 2 ……………………..(ii)

Dividing (ii) by (i)

= , = =

= = =21.8°

Squaring equations (i) and (ii) the add

2² + 5²

R² = 29

+ = 1

R²x1 =29, R²=29, R =

From R = 2

= 2

=

X + 21.8 =

X + 21.8° = 68.2° , -68.2°

X= 68.2° – 21.8° = 46.40°

Also x + 21.8° = ⁻68.2°

X = ⁻68.2° -21.80 =-90

x =

NB: The R- formula ( Transformation) can also be done using an auxiliary angle approach; where we substitute constants a and b as functions of sine or cosine.

Thus considering the same problem solving  5  – 2 =2

Imagine a triangle

Using Pythagoras theorem

= + ²

= 5² + 2² = 25 + 4 = 29

=

From the figure above, it follows that

= , 2 = cos

Then from 5cos x – 2sin x = 2

– = 2

= 2

=2

=

– x =

-x = 21.8°

So, the principle angle = 21.8°

Using the general solution of sin

– x = 21.8°, thus 180°n + ⁿ

= 68.2°

X = – 21.8°

X = –

X= 68.2° –

n=

find x values according to the limits given in the question

OR imagine a triangle

Then sin, 2= sin

cos = , 5= cos

from 5cosx – 2

– = 2

= 2

=2

=

+ x ==68.2°

Using the general solution of cosine

+ x =360°n 68.2°

X = –

= 68.2°

X= – 21.8

n =