Home ADVANCED LEVEL TOPIC 7: TRIGONOMETRY(II) ~ ADV MATHEMATICS FORM 5

TOPIC 7: TRIGONOMETRY(II) ~ ADV MATHEMATICS FORM 5

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iii) Consider
=
But =
=
=
Alternative: Using =
Dividing by cos3θ 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
=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 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
= 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
+ 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
+ R = 4 + 9
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
= 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 =
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
=
= tan⁻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 7cosθ+ 24 sinθ in the form of Rcos(10 –
5. Solve for θ
3cosθ + 4sinθ =2
6. 5cos2θ– sin 2θ=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 = 4sin sinsin
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)
=
=
But
=
= =
=
=
Solution(ii)
,
=
=



Solution (iii)
=
= R.H.S
=
Solution(iv)
= 4
+3A = 2
=2
=2cos2Acos
=2
+=2
=2
=2
==
=2
Then
=2 + 2
=2
=2
=2
=2
=2
=4 R.H.S
Solution(V).
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, 20
2x= 2=1
2x=
X=
X=
X=
X=
x=
Questions
1. Solve for the value of x between 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) θ
iv) 2
9. Show without using tables or calculators
i)
ii) 2

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