Solving Trigonometric Equations and Identities, Math 22 Assignment

This document presents a comprehensive solution set for a trigonometry assignment, likely for a university-level mathematics course (Math 22). The solutions cover a wide range of trigonometric concepts, including proving identities, solving trigonometric equations, and analyzing trigonometric functions. The assignment addresses proving trigonometric identities such as (2 - 2sinθcosθ = 1), (tanθcosecθ = secθ), and (cosecθ(1-cosθ)(1+cosθ) = sinθ). It also provides solutions for solving trigonometric equations, including finding general solutions and solutions within specified ranges. Graphical analysis is also included, with plots of functions like y = sec x and y = cosec x. The document also includes inequality problems and provides detailed step-by-step solutions and explanations for each problem, making it a valuable resource for students studying trigonometry.

MATHEMATICS 1
ENGINEERING MATHEMATICS
by Student’s Name
Course Name
Professor’s Name
University Name
City, State
Date

MATHEMATICS 2
QUESTION 3
a) (sin θ+cos θ ¿2 -2 sin θ cos θ = 1
Solution
Manipulating left side
(sin θ+cos θ ¿2 -2 sin θ cos θ
Apply perfect square formula: (a +b)2 = a2+2ab+b2
a = cos θ and b = sin θ
= cos2θ + 2cos θ sin θ + sin2θ
= cos2θ + 2cos θ sin θ + sin2θ -2cos θ sin θ
Add similar elements: 2cos θ sin θ -2cos θ sin θ =0
= cos2θ + sin2θ
Use the following identity cos2x + sin2x = 1
Hence cos2θ + sin2θ = 1
∴ (sin θ+cos θ ¿2 -2 sin θ cos θ = 1 is True
b) tanθ cosecθ = secθ
Solution
Manipulating the left side
tanθ cosecθ
Using the basic trigonometric identity cosec x = 1
sin x

MATHEMATICS 3
= tanθ * 1
sinθ
Use the following identity: tan x = sinx
cos x
= sinθ
cos θ * 1
sinθ
Cancel the like terms
= 1
cos θ from identity: 1
cos θ = secθ
= secθ which is true
Therefore tanθ cosecθ = secθ
c) cosecθ(1-cosθ)(1+cosθ) = sinθ
Solution
Manipulating left side
cosecθ(1-cosθ)(1+cosθ)
Use the following identity: cosec x = 1
sin x
= 1
sin θ (1-cosθ)(1+cosθ)
Use the following identity: (1-cosθ)(1+cosθ) = 1-cos2θ
= 1
sin θ (1-cos2θ)
Simplify (1-cos2θ) = sin2θ
= 1
sin θ sin2θ

MATHEMATICS 4
= sinθ which is true
Therefore cosecθ(1-cosθ)(1+cosθ) = sinθ
d) cotθ secθ tan θ√ (1−sin2 θ) = 1
Solution
Manipulating left side
cotθ secθ tan θ√(1−sin2 θ)
Simplify
√ (1−sin2 θ) = cosθ
= 1
tanθ * 1
cosθ * tanθ * cosθ
Cancel the like terms
= 1 which is true
Therefore cotθ secθ tan θ √(1−sin2 θ) = 1
e) 1
sec2 θ + 1
cosec2 θ =1
Solution
Manipulating left side
1
sec2 θ + 1
cosec2 θ
Express with sin, cos
Using the basic trigonometric identity cosec x = 1
sin x

MATHEMATICS 5
=
1
( 1
sinθ ) 2 +
1
( 1
cosθ )2
Simplify by apply exponent rule and fraction rule
=
1
( 1
cosθ )2 = cos2
θ and
1
( 1
sinθ ) 2 = sin2
θ
= cos2θ + sin2
θ
Use the following identity: cos2θ + sin2
θ = 1
= 1 two sides could take the same form
Hence 1
sec2 θ + 1
cosec2 θ =1
f) secθ – cosθ = sinθtanθ
Solution
Manipulating left side
secθ – cosθ
Express with sin, cos
Using the basic trigonometric identity sec x = 1
cos x
= - cos θ + 1
cos θ
Simplify and applying exponent rule
= −cos θcosθ
cos θ + 1
cos θ
Since denominators are equal, combine the fraction:

MATHEMATICS 6
= −cos2 θ+1
cos θ
= 1−cos2 θ
cos θ
Use the following identity: 1−cos2 x = sin2 x
= sin2 θ
cos θ
Use the following identity: sinx
cos x = tan x
= sin θtanθ which is true
Hence secθ – cosθ = sin θtanθ
g) sin2θ + 2cos2θ = 2- sin2θ
Solution
Manipulating left side
sin2θ + 2cos2θ
Use the following identity: cos2 x = 1- sin2 x
= sin2θ + 2(1- sin2 θ)
Expand: 2(1- sin2 θ)
= 2- 2 sin2 θ
= sin2θ + 2- 2 sin2 θ
Simplify and group like terms
= sin2θ - 2sin2 θ+2

MATHEMATICS 7
Add similar elements:sin2θ - 2sin2 θ = - sin2θ
= - sin2θ + 2
= 2- sin2θ which is true
Therefore sin2θ + 2cos2θ = 2- sin2θ
h) tanθ + cotθ = 1
sin θcosθ
Solution
Manipulating the left side
tanθ + cotθ
Express with sin, cos:
= cosθ
sin θ + tanθ
Using the basic trigonometric identity tan x = sin x
cos x
= cosθ
sin θ + sin θ
cos θ
Simplify above
cosθ
sin θ = cos2 θ
sin θcosθ and sin θ
cos θ = sin2 θ
sin θcosθ
= cos2 θ
sin θcosθ + sin2 θ
sin θcosθ
Since the denominator are equal, combine the fraction
= cos2 θ+sin2 θ
sin θcosθ

MATHEMATICS 8
Use the following identity: cos2 x +sin2 x = 1
= 1
sin θcosθ which is true
Therefore tanθ + cotθ = 1
sin θcosθ
QUESTION 4
a) sec x = 2
Solution
sec x = 2, 0 ≤ x ≤ 3600
General solution for sec x = 2
x=600 +3600 n, x=3000 +3600 n
Solutions for range 0 ≤ x ≤ 3600
x=600, x=3000
b) cot x = √3
Solution
cot x = √3, 0 ≤ x ≤ 3600
General solution for sec x = √3
x=300 +1800 n,
Solutions for range 0 ≤ x ≤ 3600

MATHEMATICS 9
x=300, x=2100
c) cosec x = √ 2
Solution
cosec x = √ 2, 0 ≤ x ≤ 2π
cosec x can’t be greater than one for real solutions
No solution for x ∈ R
d) sec x = 1.2
Solution
sec x = 1.2, 0 ≤ x ≤ 3600
General solution for sec x =1.2
sec x=a → x = arcsec a + 3600 n, x= 3600 –arcsec a + 3600 n
x = arcsec 1.2 + 3600 n, x= 3600 –arcsec 1.2 + 3600 n
Solutions for range 0 ≤ x ≤ 3600
x=0.58568, x=5.69749
e) cot x =3
Solution
cot x = 3, 0 ≤ x ≤ 3600
General solution for cot x =3
cot x=a → x = arccot a + 1800 n
x = arccot 3 + 1800 n

MATHEMATICS 10
Solutions for range 0 ≤ x ≤ 3600
x= arccot 3, x= arccot 3 + 1800 n
Solution in decimal form
x=0.3215, x=0.32175…..+180
f) cosec x =1
Solution
cosec x =1, 0 ≤ x ≤ 2π
General solution for cosec x =1
Solve ec x = 0 + 2πn
x= 2 πn
ec
Solutions for range 0 ≤ x ≤ 2π
No solution for x ∈ R
QUESTION 5
a) cosecθ- sinθ = cotθ cosθ
Solution
Manipulating left side
cosecθ - sinθ
Using the basic Trigonometric identity cosec x = 1
sin x
= 1
sin θ – sinθ

MATHEMATICS 11
Simplify above
Since sinθ = sinθsinθ
sin θ
= 1
sin θ – sinθsinθ
sin θ since denominators are equal we combine
= 1−sinθsinθ
sin θ
= 1−sin2 θ
sin θ apply following identity 1-sin2x = cos2x
= cos2 θ
sin θ use the following identity cosx
sinx =cot x
= cosθ cotθ which is true
∴ cosecθ- sinθ = cotθ cosθ is true
b) cos2θ-sin2θ=2 cos2θ-1
Solution
Manipulating left side
cos2θ-sin2θ
Using the following identity factor sin2x =1-cos2x
= - (1- cos2θ) + cos2θ
Distribute parentheses
= -1-(- cos2θ)
Apply minus-plus rule
-(-a)=a, -(a)=-a

MATHEMATICS 12
= -1+ cos2θ
= -1+ cos2θ + cos2θ
Add similar elements
= -1+2 cos2θ which is true
∴ cos2θ-sin2θ=2 cos2θ-1
c) cosec2 θ+sec2 θ= cosec2 θsec2 θ
Solution
Manipulating left side
cosec2 θ+sec2 θ
= 1
sin 2θ + 1
cos 2 θ
= cos 2 θ+sin 2θ
sin 2 θcos 2 θ = 1
sin 2θcos 2 θθ
= cosec2 θsec2 θ which is true
Therefore cosec2 θ+sec2 θ= cosec2 θsec2 θ
d) sin2 θ
1−cosθ =1+cosθ
Solution
Manipulating left side
sin2 θ
1−cosθ
Using the following identity factor sin2x =1-cos2x