Week 10 – SS2 First Term Further Mathematics Notes

 WEEK 10
TOPIC: Trigonometric Identities and graphs of inverse trigonometric ratios
Pythagoras Theorem

                          P
    
 
 
 
                            x

 
 
 
 
 
                 Fig. 10 a

 
 
 Fig 10a shows a unit circle ∆OPN is a right – angled triangle with OP = 1, ON = x and PN = y, PON – ѳ. Find the definition of trigonometric ratios.
    x = cosѳ                …(1)
    y = sinѳ                …(2)
From (1)    x² = cos²ѳ            …(3)

 From (2)    y² = sin²ѳ            …(4)

 Adding (3) and (4)
    x² + y² = cosѳ² + sinѳ²        …(5)

 Since ∆OPN is a right – angled triangle
    ON² + NP² = OP²
    x² + y² = 1                …(6)
Equating (5) and (6)
cos²ѳ+ sin²ѳ = 1                …(7)

 Dividing both sides of (7) by cos²ѳ
cos²ѳ     + sin²ѳ    =     1
cos²ѳ     cos²ѳ    cos²ѳ
: 1 + tan²ѳ = sec²ѳ
Dividing (7) through by sin²ѳ
Cos²ѳ + 1 = cosec²ѳ
: 1 + cot²ѳ = cosec²ѳ

 Example
Show that (3 – sin²ѳ) cosec²ѳ = 2cosec²ѳ + cot²ѳ
Solution
LHS = 3cose²ѳ – sin²ѳ cosec²ѳ
    = 3cosec²ѳ – sin²ѳ
            Sin²ѳ
    = 3cosec²ѳ – 1
    = 3cosec²ѳ – (cosec²ѳ- cot²ѳ)
    = 3cosec²ѳ – cosec²ѳ + cot²ѳ
    = 2cosec²ѳ + cot²ѳ
    = RHS

 Example
Prove that 2cosec²ѳ

 Solution
LHS =
=     2
1 – cos²ѳ

 LHS =     2
sins²ѳ
    = 2cosec²ѳ
    = RHs
Example
Show that = 4cot²ѳ + 2

 Solution
LHS

 = (1 + cosѳ)2 + (1 – cosѳ)²
(1 – cosѳ)(1 + cosѳ)
= 1 + 2cosѳ + cos²ѳ + 1 – 2cosѳ + cos²ѳ
        1 – cos²ѳ
= 2cos²ѳ + 2
Sin²ѳ
= 2(cos²ѳ + 1)
Sin²ѳ
= 2(cot²ѳ + cosec²ѳ)
= 2(cot²ѳ + 1 + cot²ѳ)
= 2(2cot²ѳ + 1)
= 4cot²ѳ + 2
= RHS

 Example
If asinѳ + bcosѳ = p
andacosѳ – bsinѳ = q
show that a² + b² = p² + q²

 Solution
asinѳ + bcosѳ = p                    …(1)
acosѳ – bsinѳ = q                    …(2)
Squaring both sides of (1)
    (asinѳ + bcos)² = p²
: a² sin²ѳ + 2absinѳ cosѳ + b²cos²ѳ = p²        …(3)
Squaring both sides of (2)
(acosѳ – bsinѳ)² = q²
a²cos²ѳ – 2absinѳ cosѳ + b²cos²ѳ = q²        …(4)
Adding (3) and (4)
a²sin²ѳ + 2absinѳ cosѳ + b²cos²ѳ + a²cosѳ – 2absinѳ cosѳ + b²sin²ѳ = p²q²
: a²sin²ѳ + a²cos²ѳ + b²cos²ѳ + b²sin²ѳ = p² + q²
a²(sin²ѳ + cos²ѳ) + b²(cos²ѳ + sin²ѳ) = p² + q²
But sin²ѳ + cos²ѳ = 1
: a² + b² = p² + q²

 Trigonometric Equations
A trigonometric equation is an equation containing trigonometric ratios.
Since trigonometric functions are periodic functions, there are infinite numbers of solutions for trigonometric equations. It is therefore the tradition to find some solutions within a given range.
If                asinѳ + = 0(a ≠ 0)
then,                asinѳ = -b
                
Ѳ = sin¹
Sin¹means the angles whose sine ratio is
It is called sine inverse of it is also denoted as sin thus:
Sinѳ =
Ѳ = sin-¹

 = arc sin
Similarly, if acosѳ + b = 0, (a ≠ 0)
then    cosѳ =
ѳ = cos^-1

 Evaluation
Sketch the graph of:
(i) y = sin2x                        (ii) y = cos x
(iii) y = sec x                        (iv) cosec x
all at intervals of 30◦ range 0≤ x ≤ 360.

 General Evaluation
(1) Draw the graph of y = 2cosx – 1 in the range 0◦ ≤ x ≤ 360◦ at intervals of 30◦.
(2) Draw the graph of y = 3sin x – 1 in the range of 0◦ ≤ x ≤ 360◦ at intervals of 30◦
(3) Prove that sec²ѳ + cosec²ѳ = (tanѳ + cotѳ) ².

 Weekend Assignment
Given that 4cos x + 3sin x = 5, find the value of
(1) Sin x                
(2) Cos x                
(3) Tan ѳ                
(4) Cot ѳ                
(5) Sin x + cos x            

 Theory
(1) Draw the graph of inverse trig function for sin x –
(2) Find the inverse of the following and their domains
(a) y = sin x            (b) y = cos x            (c) y = tan x
(d) y = cosec x        (e) y = sec x            (e) y = cot x