ADVANCED MATHEMATICS FORM 5 – TRIGONOMETRY

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

             
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)

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 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)

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 =

=⁻

 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 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 –
–

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

iii)             Consider

=

But =

=

=

Alternative: Using =

Dividing by cos³θ numerator and denominator