ENR101 Essential Mathematics 1 Assignment 2: Algebra and Trigonometry

This document presents the solutions to ENR101 Essential Mathematics 1 Assignment 2, focusing on algebra and trigonometry. The assignment includes problems on finding the equation of a circle, determining the slope of a line, and solving for perpendicular lines. It further explores trigonometric concepts such as verifying identities and using sum and difference formulas. The assignment also involves sketching graphs of trigonometric functions, finding intercepts, and determining maximum and minimum values. Furthermore, the document provides solutions for solving systems of equations using the elimination method and graphing. Lastly, it covers finding all trigonometric functions of an angle, and evaluating expressions using trigonometric identities and formulas. The solutions are detailed, showing all necessary steps for a comprehensive understanding of the concepts.

Assignment - 2
ENR101 Essential Mathematics 1: Algebra and Trigonometry
MAY 15, 2019
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Question 1
a) From the given equation of the circle, find the centre and the radius.
x2−6 x + y2+ 10 y =−30
Sol.
x2−6 x + y2+ 10 y =−30
Completing the square for x2−6 x
(x−3)2−9+ y2+10 y =−30
(x−3)2 + y2 +10 y=−21
Completingthe square for y2−10 y
(x−3)2 +( y −5)2−25=−21
(x−3)2 +( y −5)2=4
Comparing the equation withthe form
( x−h ) 2+ ( y −k ) 2=r2
r =2
h=3
k =5
The centre of the ˚is given by ( h , k ) = ( 3,5 )
The radius of the ˚is r =2
b) A line passes through the points (−1 ,−6) and (3,5). Find slope and hence, show that
the equation of the line is y= 11
4 x− 13
4
Sol.
If the line passes through two points ( x 1 , y 1 )∧ ( x 2, y 2 ) ,
The slope of the line mis givenby
m= y 2− y 1
x 2−x 1
Let ( x 1 , y 1 )= (−1 ,−6 )∧ ( x 2 , y 2 )= ( 3,5 )
m= 5−(−6)
3−(−1)
m= 11
4
c) Find the slope of the line that is perpendicular to 2 x+3 y =1and give a reason for your
answer.
Sol.
2 x+3 y =1
3 y=1−2 x
y= 1
3 − 2
3 x
y= (−2
3 ) x+( 1
3 )
Comparing withthe equation y=mx +c , where m isthe slope

m=−2
3
The slope of the line perpendicular ¿ theabove line will be− 1
m
m' = 3
2
Reason :The relation between the slopes of the perpendicular lines m∧m' is givenby
m m' =−1
d) Find the equation of the line perpendicular to 2 x+3 y =1 that passes through the point
(2,5). Give your answer in standard form.
Sol.
The slopeis found∈part ( c ) m' = 3
2
The line can be written as
y= 3
2 x +c
The line passesthrough the point P ( 2,5 )
5=3
2 2+c
c=2
The equation of the line is
y= 3
2 x +2
2 y=3 x+4
2 y−3 x−4=0
Question 2
The triangle PQRhas angle 25 ° at Q, while sides PQ and QR have lengths 13 cm and 8 cm
respectively.
a) Draw the diagram of the triangle (Your drawing does not need to be in scale.)
Sol.
P
Q R
13 cm

b) Find the sizes of all the angles and lengths of all sides of the triangle PQR .
Sol.
sin P
QR = sin Q
PR = sin R
PQ
sin P
8 = sin 25 °
PR = sin R
13
P+Q+ R=180°
P+R=180−25
P+ R=155°
R=155−P
sin P
8 = sin 25 °
PR = sin155−P
13
sin P
8 = sin 25 °
PR = sin( π− ( 25+ P ) )
13
sin P
8 = sin 25 °
PR = sin(25+ P)
13
Question 3
Verify the following trigonometric identities. In each case you might like to start with the
most complicated side, and use algebraic manipulation to show that it can be written like the
other side.
a) (tan θ)2=(tanθ)2 (csc θ)2−1
Sol.
tan θ= sin θ
cos θ
cosec θ= 1
sin θ
Simplifying R . H . S
( tan θ ) 2 ( csc θ ) 2−1
( sin θ
cos θ )
2
( 1
sin θ )
2
−1
( 1
cos θ )2
−1
( secθ ) 2−1
( tanθ )2
RHS=LHS
Hence verified
b) 1−sin ϕ
cos ϕ = cos ϕ
1+ sin ϕ
8 cm

Sol.
We know that
(sin ϕ )2+(cos ϕ)2=1
( cos ϕ ) 2=1− ( sin ϕ ) 2
(cos ϕ)2=(1+sin ϕ )(1−sin ϕ )
1−sin ϕ
cos ϕ = cos ϕ
1+ sin ϕ
Hence verified
Question 4
Consider the function y=4 sin (2 x− π
3 ) for −π ≤ x ≤ π. Follow the steps below to sketch the
graph of y.
a) State the amplitude and period in the graph of this function.
Sol.
Amplitude=4
Period= 2 π
2
Period=π
b) Solve y=4 sin (2 x− π
3 ) for −π ≤ x ≤ π to find the horizontal intercepts (x-intercepts)
of the function.
Sol.
y=4 sin ( 2 x− π
3 )
0=sin(2 x− π
3 )
2 x− π
3 =0 , π
2 x=π+ π
3
x= 4 π
6
x= 2 π
3
c) Find the values of x for −π ≤ x ≤ π for which the maximum and the minimum values
of the function occur.
Sol.
−π ≤ x ≤ π

−1 ≤sin ( 2 x − π
3 ) ≤1
−4 ≤ 4 sin (2 x− π
3 )≤ 4
Minimum=−4∧Maximum=4
d) State the range of the function.
Sol.
Range is(−4,4 )
e) Using the information obtained in (a)–(d), draw the graph of y=4 sin (2 x− π
3 ) for
−π ≤ x ≤ π
Sol.
-160 -140 -120 -100 -80 -60 -40 -20 0
-4
-3
-2
-1
0
1
2
3
4
y
Question 5
a) Solve the system of equations by using the elimination method. Marks will be
awarded for checking your solution.
4 x+2 y=14
5 x+ 2 y =16
Sol.
4 x+2 y=14 … … …(1)
5 x+2 y =16 …… … ( 2 )
Eq ( 2 )−Eq ( 1 )
x=2
4∗2+ 2 y =14
2 y=14−8
2 y=6
y=3

b) Graph the two equations on the same Cartesian axes and hence solve the system.
2 x+3 y =12
x− y =1
Solution P(3,2)
Question 6
a) Find all trigonometric functions of θ, given that cos ( π
2 −θ )=3
5 ∧cos θ= 4
5
Sol.
cos ( π
2 −θ )=3
5 ∧cos θ= 4
5
sin ( θ ) =3
5 ∧cos θ= 4
5
tan θ= sin θ
cos θ
tanθ= 3
4
cosec θ= 1
sin θ = 5
3
sec θ= 1
cos θ = 5
4
cot θ= 1
tan θ = 4
3
b) Using the values from part (a), find cos 2 θ
Sol.
cos 2 θ=1−2 ( sin θ ) 2
cos 2 θ=1− 2∗9
25

cos 2 θ=1− 18
25
cos 2 θ= 7
25
Question 7
a) Use the sum or difference formulas to find the exact values for sin 7 π
12 ∧cos 7 π
12
Sol.
cos 7 π
12 =cos ( 3 π
12 + 4 π
12 )
¿ cos ( π
4 + π
3 )
cos ( a+b ) =cos a cos b−sin asin b
cos 7 π
12 =cos π
4 cos π
3 −sin π
4 sin π
3
cos 7 π
12 = 1
√ 2
1
2 − 1
√ 2
√ 3
2
cos 7 π
12 =1− √3
2 √ 2
sin 7 π
12 =sin ( 3 π
12 + 4 π
12 )
¿ sin ( π
4 + π
3 )
sin ( a+ b ) =sin a cos b+cos a sin b
sin 7 π
12 =sin π
4 cos π
3 +cos π
4 sin π
3
sin 7 π
12 = 1
√ 2
1
2 + 1
√ 2
√ 3
2
sin 7 π
12 = 1+ √3
2 √2
b) Use the result of the previous part of this question exactly to find tan 7 π
12 .
Sol.
tan 7 π
12 =
sin 7 π
12
cos 7 π
12
tan 7 π
12 = 1+ √3
1− √3
c) Without using a calculator show that
( sin 7 π
12 )
2
+ ( cos 7 π
12 )
2
=1 exactly

Sol.
( 1+ √ 3
2 √2 )2
+
( 1− √3
2 √2 )2
1+3+2 √ 3+1+3−2 √ 3
(2 √ 2)2
8
8
¿ 1
Hence Proved