TRIGONOMETRY(II)

iii) Consider

=

But=

=

=

Alternative: Using=

Dividing by cos3Î¸ numerator and denominator

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

=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 tanwithout 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=

Cos2Ã†Å¸ =…………………… (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 Î¸ between 0° and 180° if 2cos Î¸+ sin Î¸= 2.5

Solution: let t = tan

2+ 3=2.5

Then=

Sin Î¸=

2+ 32.5

+ 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= tantan= 1.244

t =1.244, t= tantan= 1.244

= tan

=

=51.2° = Î¸ = 51.2 x 2

= 102.4°

case 2:

t = 0.0893

t = 0.0893

= 0.0893

=

=

=5.1°, Î¸ = 10.20

Î¸=

Example 2: solve the equation

5– 2=2 for

for -180x

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

t== 0.42857

= 0.42857

=

= 23.2° = 23.2°x2=46.4°

Case2,

t=–1, tan= Ã¢»1

t=–1, tan= Ã¢»1

==

II. SOLVING THE EQUATION

acosÎ¸= C

R-formula or simply transforming a function acosÃ†Å¸bsinÃ†Å¸ as a single function.

From acosÎ¸bsinÎ¸ = c

Consider acosÎ¸ + bsinÎ¸ – 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 acosÎ¸ + bsinÎ¸ =C

acosÎ¸ + bsinÎ¸ = Rcos(Î¸ –)

acosÎ¸ + bsinÎ¸= 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

acosÃ†Å¸ + bsinÃ†Å¸ =c = Rcos (Ã†Å¸ –

Rcos(

=

–=

=

Example

Rcoscosx = 3cosx

Rcos= 3 —- (i)

-4sinx = Rsinxsin

Sin= 4 —– (ii)

Dividing (ii) by (i), then we get

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 forbetween 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

Î¸- 56.3°=

Î¸=

=+ 56.3°

= 46.1° + 56.4°= 102.4°

Î¸= 313.9° + 56.3°= 370.2°

= 370.2° – 360°=10.2°

Î¸=10.2°,102.4°

Example: 3

solve for x iÆ’5– 2sinx =R=2

solve for x iÆ’5– 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° = Ã¢»68.2°

X = Ã¢»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°n68.2°

X =–

= 68.2°

X=– 21.8

n =

OTHER KIND OF QUESTIONS USING THE TRANSFORMING INTO A SINGLE FUNCTION CONCEPT

Example:1 Express

i) 4cosx – 5sinx in the form of Rcos(x +

ii) 2sinx + 5cosx in the form of Rsin(x +

Solution(i)

4cos x-5sinx =Rcos(x +

4cosx = Rcoscosx

Rcos= 4 ……… (i)

5sinx = Rsinsinx

Rsin=5 …………..(ii)

Dividing (ii)by (i)

=== tan

=

= tanÃ¢»¹

=

Squaring equations (i) and (ii) then add

+= 4² + 5²

R²cos+ R²= 16 + 25

R= 41

R=41, R=

4cos x -5 sin x =cos(x+ )

Rcossinx = 2sinx

Rcos=2 …………(i) and

Rcosxsin= 5cosx

Rsin= 5 ………….(ii)

Dividing (ii) by (i)

==

Tan= ,=

Squaring equations (i) and (ii) then add

+= 2² + 5²

R²cos²+ R² sin²= 4 + 25

R=29

But cos²

R²(1)=29

Example. Find the maximum value of 24sinx -7cosx and the smallest positive value of x that gives this maximum value.

Solution. 24sin x -7cosx = Rsin(x –

24sinx = Rcossinx

Rcos=24, 7cosx = Rsincosx

Rsin=7 ………(ii)

=

==

== 16.26°

Squaring equation (i) and (ii) then add

+=+

R=625

R=625

R²=625, R=

R =25

24– 7cosx = Rsin

=

=25sin

24sinx – 7cosx = 25sin

f(x)= 25sin(x – 16.26°)

Max value of sine function is when

Sin

X – 16.26°=90°

X = 90° + 16.26°

X= 106.26°

Hence max value f=y=25 sin 90°

=25

The maximum value is 25 obtained when x = 106.26°

Note. The maximum values of

Problems to work on

Using t formula and R –formula solve the following.

3. 6sinx + 8cosx =6

4. Express 7cosÎ¸+ 24 sinÎ¸ in the form of Rcos(10 –

5. Solve for Î¸

3cosÎ¸ + 4sinÎ¸ =2

6. 5cos2Î¸– sin 2Î¸=2

Note: If the question has no limits/boundaries write the answer using the general solution

FACTOR FORMULAE (SUM AND DIFFERENCE FORMULAE)

The concept here is to express the sum or difference of sine and cosine functions as product and vice versa

Refer

Sin(A +B) = sin AcosB + cosAsin B ……….(i)

Sin(A –B) = sinAcosB –cosAsinB ………….(ii)

Cos(A + B) =cosAcosB – sinA sinB …………(iii)

Cos(A+ B) =cosAcosB + sinAsinB ……………(iv)

Add (i) and (ii)

Sin(A + B) + sin(A +B) =2sin AcosB

Let f = A + B ………(i)

Q =A-B …….(ii)

(a) +(b) 2A = P+Q, A=

(a) –(b) 2B =P-Q, B=

Therefore sin(A+B)+sin(A-B)=2sinAcosBbecome

SinP + sinQ= 2sincos…(1)

Substract(i) –(ii)

Sin(A+B) –sin(A-B) = 2cosA sinB

But P=A+B, Q=A-B

Add (iii) and (iv)

Cos(A+B)+cos(A-B) = 2cosAcosB

CosP + cosQ = 2coscos

Substract (iii) – (iv)

Cos(A + B) –cos(A-B) = -2sinAsin B

Expressions (1) (2) (3) and ( 4) are called factor formulae

APPLICATIONS OF THE FACTOR FORMULAE

a) Proving problems

Examples

i)= cot 2x

ii)= cot

iii)= tan

v) If A, B and C are angles of a triangle prove that

cosA +cosB + cosC -1 = 4sinsinsin

vi) If A, B and C are angles of a triangle prove that

cos2A + cos2B + cos2C + 1 = 4cosAcosBcosC

vii)=tan A

viii)=

Solution (i)

(L.H.S)

(L.H.S)

=

=

But–

=

==

=

=

Solution(ii)

,

,

=

=

Solution (iii)

=

=R.H.S

=

Solution(iv)

= 4

= 4

+3A = 2

=2

=2cos2Acos

=2

+=2

=2

=2

==

=2

Then

=2+ 2

=2

=2

=2

=2

=2

=4R.H.S

Solution(V).

A, B, C are angles of a

A, B, C are angles of a

+

L.H.S

CosA + cosB + cosC – 1

2

=2-2………….(i)

But A + B + C= 180°

(Degree angle in

A + B = 180°-C

=

90 –=

Apply cos

cos= cos

Cos=

2

But

=1 ––

= 1 – 2

Substitute (ii) into (i)

=2cos-2sin

= 2-2

=2

=2

But=

Using factor formula

2

2

2

2

But

2

2

=

=4

solution(VI).

= 4

From factor fomulae

=

=2

=2

But A + B +C = 180° ()

A +B = 180° -C

Cos

=+

= –+ 0

= –

Substitute into (i)

=-2++ 1

=

=

=

=2

= -2

= -2

= -2+2

=2

But= –

2

= -2

= -2

= -2

= -2

=

= -4

=

=

=

=

Solution (vi)

L.H.S changing the products into sin or difference

Numerator:

From sinP +sinQ=2

=

=

Similarly=

Denominator

=

=

=

=

=

=

=RHS

Examples (i) solve for x if

+=for 0°

ii)

For

iii)

For

Solution (i)

+=

Writing using factor formulae

=2

=2

=2

=2

2

2

=0

=0, 2

2=1

=0=

3x ==0°, 180, 360°

X=540°

=0°,60°,120°, 180°

== 60°,300°

X=

X=30°, 150°

x=

iv)=

2=

2

2=

2

=0, 2

2x=2=1

2x=

X=

X=

X=

X=

x=

Questions

1. Solve for the value of x between 0° and 360° in the question

i)–=

ii)+=0

2. Prove that

i)+°=0

ii)=

3. Simplify

4. Evaluate

5. Prove that

2=

If+a and

+=b show that

7. Prove that

8. Express as a sum or difference

i) 2

ii)

iii)Î¸

iv) 2

9. Show without using tables or calculators

i)

ii) 2

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