trigonometry trigonometric identities trigonometric
TRANSCRIPT
Trigonometry
TrigonometricIdentities
TrigonometricEquations
A Level Maths Boost - Trigonometry
Dr Rhian Taylor
5 May 2021
Trigonometry
TrigonometricIdentities
TrigonometricEquations
We know thatsin2 x + cos2 x = 1
We can derive the following identities:
sin2 x
sin2 x+
cos2 x
sin2 x=
1
sin2 x
1 +1
tan2 x=
1
sin2 x1 + cot2 x = cosec2x
sin2 x
cos2 x+
cos2 x
cos2 x=
1
cos2 x
tan2 x + 1 =1
cos2 xtan2 x + 1 = sec2 x
Trigonometry
TrigonometricIdentities
TrigonometricEquations
We know thatsin2 x + cos2 x = 1
We can derive the following identities:
sin2 x
sin2 x+
cos2 x
sin2 x=
1
sin2 x
1 +1
tan2 x=
1
sin2 x1 + cot2 x = cosec2x
sin2 x
cos2 x+
cos2 x
cos2 x=
1
cos2 x
tan2 x + 1 =1
cos2 xtan2 x + 1 = sec2 x
Trigonometry
TrigonometricIdentities
TrigonometricEquations
We know thatsin2 x + cos2 x = 1
We can derive the following identities:
sin2 x
sin2 x+
cos2 x
sin2 x=
1
sin2 x
1 +1
tan2 x=
1
sin2 x
1 + cot2 x = cosec2x
sin2 x
cos2 x+
cos2 x
cos2 x=
1
cos2 x
tan2 x + 1 =1
cos2 xtan2 x + 1 = sec2 x
Trigonometry
TrigonometricIdentities
TrigonometricEquations
We know thatsin2 x + cos2 x = 1
We can derive the following identities:
sin2 x
sin2 x+
cos2 x
sin2 x=
1
sin2 x
1 +1
tan2 x=
1
sin2 x1 + cot2 x = cosec2x
sin2 x
cos2 x+
cos2 x
cos2 x=
1
cos2 x
tan2 x + 1 =1
cos2 xtan2 x + 1 = sec2 x
Trigonometry
TrigonometricIdentities
TrigonometricEquations
We know thatsin2 x + cos2 x = 1
We can derive the following identities:
sin2 x
sin2 x+
cos2 x
sin2 x=
1
sin2 x
1 +1
tan2 x=
1
sin2 x1 + cot2 x = cosec2x
sin2 x
cos2 x+
cos2 x
cos2 x=
1
cos2 x
tan2 x + 1 =1
cos2 xtan2 x + 1 = sec2 x
Trigonometry
TrigonometricIdentities
TrigonometricEquations
We know thatsin2 x + cos2 x = 1
We can derive the following identities:
sin2 x
sin2 x+
cos2 x
sin2 x=
1
sin2 x
1 +1
tan2 x=
1
sin2 x1 + cot2 x = cosec2x
sin2 x
cos2 x+
cos2 x
cos2 x=
1
cos2 x
tan2 x + 1 =1
cos2 x
tan2 x + 1 = sec2 x
Trigonometry
TrigonometricIdentities
TrigonometricEquations
We know thatsin2 x + cos2 x = 1
We can derive the following identities:
sin2 x
sin2 x+
cos2 x
sin2 x=
1
sin2 x
1 +1
tan2 x=
1
sin2 x1 + cot2 x = cosec2x
sin2 x
cos2 x+
cos2 x
cos2 x=
1
cos2 x
tan2 x + 1 =1
cos2 xtan2 x + 1 = sec2 x
Trigonometry
TrigonometricIdentities
TrigonometricEquations
We are given that
sin(A± B) = sinA cosB ± cosA sinB
cos(A± B) = cosA cosB ∓ sinA sinB
tan(A± B) =tanA± tanB
1 ∓ tanA tanB
We can derive the following double angle identities:
sin(2θ) = sin(θ + θ)
= sin θ cos θ + cos θ sin θ
= 2 sin θ cos θ
Trigonometry
TrigonometricIdentities
TrigonometricEquations
We are given that
sin(A± B) = sinA cosB ± cosA sinB
cos(A± B) = cosA cosB ∓ sinA sinB
tan(A± B) =tanA± tanB
1 ∓ tanA tanB
We can derive the following double angle identities:
sin(2θ) = sin(θ + θ)
= sin θ cos θ + cos θ sin θ
= 2 sin θ cos θ
Trigonometry
TrigonometricIdentities
TrigonometricEquations
We are given that
sin(A± B) = sinA cosB ± cosA sinB
cos(A± B) = cosA cosB ∓ sinA sinB
tan(A± B) =tanA± tanB
1 ∓ tanA tanB
We can derive the following double angle identities:
sin(2θ) = sin(θ + θ)
= sin θ cos θ + cos θ sin θ
= 2 sin θ cos θ
Trigonometry
TrigonometricIdentities
TrigonometricEquations
We are given that
sin(A± B) = sinA cosB ± cosA sinB
cos(A± B) = cosA cosB ∓ sinA sinB
tan(A± B) =tanA± tanB
1 ∓ tanA tanB
We can derive the following double angle identities:
sin(2θ) = sin(θ + θ)
= sin θ cos θ + cos θ sin θ
= 2 sin θ cos θ
Trigonometry
TrigonometricIdentities
TrigonometricEquations
cos(2θ) = cos(θ + θ)
= cos θ cos θ − sin θ sin θ
= cos2 θ − sin2 θ
= 2 cos2 θ − 1
= 1 − 2 sin2 θ
tan(2θ) = tan(θ + θ)
=tan θ + tan θ
1 − tan θ tan θ
=2 tan θ
1 − tan2 θ
Trigonometry
TrigonometricIdentities
TrigonometricEquations
cos(2θ) = cos(θ + θ)
= cos θ cos θ − sin θ sin θ
= cos2 θ − sin2 θ
= 2 cos2 θ − 1
= 1 − 2 sin2 θ
tan(2θ) = tan(θ + θ)
=tan θ + tan θ
1 − tan θ tan θ
=2 tan θ
1 − tan2 θ
Trigonometry
TrigonometricIdentities
TrigonometricEquations
cos(2θ) = cos(θ + θ)
= cos θ cos θ − sin θ sin θ
= cos2 θ − sin2 θ
= 2 cos2 θ − 1
= 1 − 2 sin2 θ
tan(2θ) = tan(θ + θ)
=tan θ + tan θ
1 − tan θ tan θ
=2 tan θ
1 − tan2 θ
Trigonometry
TrigonometricIdentities
TrigonometricEquations
cos(2θ) = cos(θ + θ)
= cos θ cos θ − sin θ sin θ
= cos2 θ − sin2 θ
= 2 cos2 θ − 1
= 1 − 2 sin2 θ
tan(2θ) = tan(θ + θ)
=tan θ + tan θ
1 − tan θ tan θ
=2 tan θ
1 − tan2 θ
Trigonometry
TrigonometricIdentities
TrigonometricEquations
cos(2θ) = cos(θ + θ)
= cos θ cos θ − sin θ sin θ
= cos2 θ − sin2 θ
= 2 cos2 θ − 1
= 1 − 2 sin2 θ
tan(2θ) = tan(θ + θ)
=tan θ + tan θ
1 − tan θ tan θ
=2 tan θ
1 − tan2 θ
Trigonometry
TrigonometricIdentities
TrigonometricEquations
cos(2θ) = cos(θ + θ)
= cos θ cos θ − sin θ sin θ
= cos2 θ − sin2 θ
= 2 cos2 θ − 1
= 1 − 2 sin2 θ
tan(2θ) = tan(θ + θ)
=tan θ + tan θ
1 − tan θ tan θ
=2 tan θ
1 − tan2 θ
Trigonometry
TrigonometricIdentities
TrigonometricEquations
cos(2θ) = cos(θ + θ)
= cos θ cos θ − sin θ sin θ
= cos2 θ − sin2 θ
= 2 cos2 θ − 1
= 1 − 2 sin2 θ
tan(2θ) = tan(θ + θ)
=tan θ + tan θ
1 − tan θ tan θ
=2 tan θ
1 − tan2 θ
Trigonometry
TrigonometricIdentities
TrigonometricEquations
cos(2θ) = cos(θ + θ)
= cos θ cos θ − sin θ sin θ
= cos2 θ − sin2 θ
= 2 cos2 θ − 1
= 1 − 2 sin2 θ
tan(2θ) = tan(θ + θ)
=tan θ + tan θ
1 − tan θ tan θ
=2 tan θ
1 − tan2 θ
Trigonometry
TrigonometricIdentities
TrigonometricEquations
1 Show thatcos x
1 − sin x− cos x
1 + sin x= 2 tan x
cos x
1 − sin x− cos x
1 + sin x=
cos x(1 + sin x) − cos x(1 − sin x)
(1 − sin x)(1 + sin x)
=cos x + cos x sin x − cos x + cos x sin x
1 − sin2 x
=2 sin x cos x
cos2 x
=2 sin x
cos x= 2 tan x
Trigonometry
TrigonometricIdentities
TrigonometricEquations
2 Show thatcos x
1 − sin x− cos x
1 + sin x= 2 tan x
cos x
1 − sin x− cos x
1 + sin x=
cos x(1 + sin x) − cos x(1 − sin x)
(1 − sin x)(1 + sin x)
=cos x + cos x sin x − cos x + cos x sin x
1 − sin2 x
=2 sin x cos x
cos2 x
=2 sin x
cos x= 2 tan x
Trigonometry
TrigonometricIdentities
TrigonometricEquations
3 Show thatcos x
1 − sin x− cos x
1 + sin x= 2 tan x
cos x
1 − sin x− cos x
1 + sin x=
cos x(1 + sin x) − cos x(1 − sin x)
(1 − sin x)(1 + sin x)
=cos x + cos x sin x − cos x + cos x sin x
1 − sin2 x
=2 sin x cos x
cos2 x
=2 sin x
cos x= 2 tan x
Trigonometry
TrigonometricIdentities
TrigonometricEquations
4 Show thatcos x
1 − sin x− cos x
1 + sin x= 2 tan x
cos x
1 − sin x− cos x
1 + sin x=
cos x(1 + sin x) − cos x(1 − sin x)
(1 − sin x)(1 + sin x)
=cos x + cos x sin x − cos x + cos x sin x
1 − sin2 x
=2 sin x cos x
cos2 x
=2 sin x
cos x= 2 tan x
Trigonometry
TrigonometricIdentities
TrigonometricEquations
5 Show thatcos x
1 − sin x− cos x
1 + sin x= 2 tan x
cos x
1 − sin x− cos x
1 + sin x=
cos x(1 + sin x) − cos x(1 − sin x)
(1 − sin x)(1 + sin x)
=cos x + cos x sin x − cos x + cos x sin x
1 − sin2 x
=2 sin x cos x
cos2 x
=2 sin x
cos x
= 2 tan x
Trigonometry
TrigonometricIdentities
TrigonometricEquations
6 Show thatcos x
1 − sin x− cos x
1 + sin x= 2 tan x
cos x
1 − sin x− cos x
1 + sin x=
cos x(1 + sin x) − cos x(1 − sin x)
(1 − sin x)(1 + sin x)
=cos x + cos x sin x − cos x + cos x sin x
1 − sin2 x
=2 sin x cos x
cos2 x
=2 sin x
cos x= 2 tan x
Trigonometry
TrigonometricIdentities
TrigonometricEquations
7 Show thatcos2 x =
cosecx cos x
tan x + cot x
cosecx cos x
tan x + cot x=
1sin x cos x
sin xcos x + cos x
sin x
=cos x
sin x÷(
sin x
cos x+
cos x
sin x
)=
cos x
sin x÷(
sin2 x + cos2 x
sin x cos x
)=
cos x
sin x÷(
1
sin x cos x
)=
cos x
sin x×(
sin x cos x
1
)=
cos2 x sin x
sin x= cos2 x
Trigonometry
TrigonometricIdentities
TrigonometricEquations
8 Show thatcos2 x =
cosecx cos x
tan x + cot x
cosecx cos x
tan x + cot x=
1sin x cos x
sin xcos x + cos x
sin x
=cos x
sin x÷(
sin x
cos x+
cos x
sin x
)=
cos x
sin x÷(
sin2 x + cos2 x
sin x cos x
)=
cos x
sin x÷(
1
sin x cos x
)=
cos x
sin x×(
sin x cos x
1
)=
cos2 x sin x
sin x= cos2 x
Trigonometry
TrigonometricIdentities
TrigonometricEquations
9 Show thatcos2 x =
cosecx cos x
tan x + cot x
cosecx cos x
tan x + cot x=
1sin x cos x
sin xcos x + cos x
sin x
=cos x
sin x÷(
sin x
cos x+
cos x
sin x
)
=cos x
sin x÷(
sin2 x + cos2 x
sin x cos x
)=
cos x
sin x÷(
1
sin x cos x
)=
cos x
sin x×(
sin x cos x
1
)=
cos2 x sin x
sin x= cos2 x
Trigonometry
TrigonometricIdentities
TrigonometricEquations
10 Show thatcos2 x =
cosecx cos x
tan x + cot x
cosecx cos x
tan x + cot x=
1sin x cos x
sin xcos x + cos x
sin x
=cos x
sin x÷(
sin x
cos x+
cos x
sin x
)=
cos x
sin x÷(
sin2 x + cos2 x
sin x cos x
)
=cos x
sin x÷(
1
sin x cos x
)=
cos x
sin x×(
sin x cos x
1
)=
cos2 x sin x
sin x= cos2 x
Trigonometry
TrigonometricIdentities
TrigonometricEquations
11 Show thatcos2 x =
cosecx cos x
tan x + cot x
cosecx cos x
tan x + cot x=
1sin x cos x
sin xcos x + cos x
sin x
=cos x
sin x÷(
sin x
cos x+
cos x
sin x
)=
cos x
sin x÷(
sin2 x + cos2 x
sin x cos x
)=
cos x
sin x÷(
1
sin x cos x
)
=cos x
sin x×(
sin x cos x
1
)=
cos2 x sin x
sin x= cos2 x
Trigonometry
TrigonometricIdentities
TrigonometricEquations
12 Show thatcos2 x =
cosecx cos x
tan x + cot x
cosecx cos x
tan x + cot x=
1sin x cos x
sin xcos x + cos x
sin x
=cos x
sin x÷(
sin x
cos x+
cos x
sin x
)=
cos x
sin x÷(
sin2 x + cos2 x
sin x cos x
)=
cos x
sin x÷(
1
sin x cos x
)=
cos x
sin x×(
sin x cos x
1
)
=cos2 x sin x
sin x= cos2 x
Trigonometry
TrigonometricIdentities
TrigonometricEquations
13 Show thatcos2 x =
cosecx cos x
tan x + cot x
cosecx cos x
tan x + cot x=
1sin x cos x
sin xcos x + cos x
sin x
=cos x
sin x÷(
sin x
cos x+
cos x
sin x
)=
cos x
sin x÷(
sin2 x + cos2 x
sin x cos x
)=
cos x
sin x÷(
1
sin x cos x
)=
cos x
sin x×(
sin x cos x
1
)=
cos2 x sin x
sin x
= cos2 x
Trigonometry
TrigonometricIdentities
TrigonometricEquations
14 Show thatcos2 x =
cosecx cos x
tan x + cot x
cosecx cos x
tan x + cot x=
1sin x cos x
sin xcos x + cos x
sin x
=cos x
sin x÷(
sin x
cos x+
cos x
sin x
)=
cos x
sin x÷(
sin2 x + cos2 x
sin x cos x
)=
cos x
sin x÷(
1
sin x cos x
)=
cos x
sin x×(
sin x cos x
1
)=
cos2 x sin x
sin x= cos2 x
Trigonometry
TrigonometricIdentities
TrigonometricEquations
15 Show that
(1 − sin x)(1 + cosecx) = cos x cot x
(1 − sin x)(1 + cosecx) = 1 + cosecx − sin x − sin xcosecx
= 1 +1
sin x− sin x − sin x
1
sin x
=1
sin x− sin x
=1 − sin2 x
sin x
=cos2 x
sin x
=cos x cos x
sin x= cos x cot x
Trigonometry
TrigonometricIdentities
TrigonometricEquations
16 Show that
(1 − sin x)(1 + cosecx) = cos x cot x
(1 − sin x)(1 + cosecx) = 1 + cosecx − sin x − sin xcosecx
= 1 +1
sin x− sin x − sin x
1
sin x
=1
sin x− sin x
=1 − sin2 x
sin x
=cos2 x
sin x
=cos x cos x
sin x= cos x cot x
Trigonometry
TrigonometricIdentities
TrigonometricEquations
17 Show that
(1 − sin x)(1 + cosecx) = cos x cot x
(1 − sin x)(1 + cosecx) = 1 + cosecx − sin x − sin xcosecx
= 1 +1
sin x− sin x − sin x
1
sin x
=1
sin x− sin x
=1 − sin2 x
sin x
=cos2 x
sin x
=cos x cos x
sin x= cos x cot x
Trigonometry
TrigonometricIdentities
TrigonometricEquations
18 Show that
(1 − sin x)(1 + cosecx) = cos x cot x
(1 − sin x)(1 + cosecx) = 1 + cosecx − sin x − sin xcosecx
= 1 +1
sin x− sin x − sin x
1
sin x
=1
sin x− sin x
=1 − sin2 x
sin x
=cos2 x
sin x
=cos x cos x
sin x= cos x cot x
Trigonometry
TrigonometricIdentities
TrigonometricEquations
19 Show that
(1 − sin x)(1 + cosecx) = cos x cot x
(1 − sin x)(1 + cosecx) = 1 + cosecx − sin x − sin xcosecx
= 1 +1
sin x− sin x − sin x
1
sin x
=1
sin x− sin x
=1 − sin2 x
sin x
=cos2 x
sin x
=cos x cos x
sin x= cos x cot x
Trigonometry
TrigonometricIdentities
TrigonometricEquations
20 Show that
(1 − sin x)(1 + cosecx) = cos x cot x
(1 − sin x)(1 + cosecx) = 1 + cosecx − sin x − sin xcosecx
= 1 +1
sin x− sin x − sin x
1
sin x
=1
sin x− sin x
=1 − sin2 x
sin x
=cos2 x
sin x
=cos x cos x
sin x= cos x cot x
Trigonometry
TrigonometricIdentities
TrigonometricEquations
21 Show that
(1 − sin x)(1 + cosecx) = cos x cot x
(1 − sin x)(1 + cosecx) = 1 + cosecx − sin x − sin xcosecx
= 1 +1
sin x− sin x − sin x
1
sin x
=1
sin x− sin x
=1 − sin2 x
sin x
=cos2 x
sin x
=cos x cos x
sin x
= cos x cot x
Trigonometry
TrigonometricIdentities
TrigonometricEquations
22 Show that
(1 − sin x)(1 + cosecx) = cos x cot x
(1 − sin x)(1 + cosecx) = 1 + cosecx − sin x − sin xcosecx
= 1 +1
sin x− sin x − sin x
1
sin x
=1
sin x− sin x
=1 − sin2 x
sin x
=cos2 x
sin x
=cos x cos x
sin x= cos x cot x
Trigonometry
TrigonometricIdentities
TrigonometricEquations
1 Solve 2 tan2 θ + 4 sec θ + sec2 θ = 2 for 0 ≤ θ < 360.
2 tan2 θ + 4 sec θ + sec2 θ = 2
2(sec2 θ − 1)θ + 4 sec θ + sec2 θ = 2
2 sec2 θ − 2θ + 4 sec θ + sec2 θ = 2
3 sec2 θ + 4 sec θ − 4 = 0
(sec θ + 2)(3 sec θ − 2) = 0
We have two possible solutions
sec θ = 2 and sec θ =2
3
Trigonometry
TrigonometricIdentities
TrigonometricEquations
2 Solve 2 tan2 θ + 4 sec θ + sec2 θ = 2 for 0 ≤ θ < 360.
2 tan2 θ + 4 sec θ + sec2 θ = 2
2(sec2 θ − 1)θ + 4 sec θ + sec2 θ = 2
2 sec2 θ − 2θ + 4 sec θ + sec2 θ = 2
3 sec2 θ + 4 sec θ − 4 = 0
(sec θ + 2)(3 sec θ − 2) = 0
We have two possible solutions
sec θ = 2 and sec θ =2
3
Trigonometry
TrigonometricIdentities
TrigonometricEquations
3 Solve 2 tan2 θ + 4 sec θ + sec2 θ = 2 for 0 ≤ θ < 360.
2 tan2 θ + 4 sec θ + sec2 θ = 2
2(sec2 θ − 1)θ + 4 sec θ + sec2 θ = 2
2 sec2 θ − 2θ + 4 sec θ + sec2 θ = 2
3 sec2 θ + 4 sec θ − 4 = 0
(sec θ + 2)(3 sec θ − 2) = 0
We have two possible solutions
sec θ = 2 and sec θ =2
3
Trigonometry
TrigonometricIdentities
TrigonometricEquations
4 Solve 2 tan2 θ + 4 sec θ + sec2 θ = 2 for 0 ≤ θ < 360.
2 tan2 θ + 4 sec θ + sec2 θ = 2
2(sec2 θ − 1)θ + 4 sec θ + sec2 θ = 2
2 sec2 θ − 2θ + 4 sec θ + sec2 θ = 2
3 sec2 θ + 4 sec θ − 4 = 0
(sec θ + 2)(3 sec θ − 2) = 0
We have two possible solutions
sec θ = 2 and sec θ =2
3
Trigonometry
TrigonometricIdentities
TrigonometricEquations
5 Solve 2 tan2 θ + 4 sec θ + sec2 θ = 2 for 0 ≤ θ < 360.
2 tan2 θ + 4 sec θ + sec2 θ = 2
2(sec2 θ − 1)θ + 4 sec θ + sec2 θ = 2
2 sec2 θ − 2θ + 4 sec θ + sec2 θ = 2
3 sec2 θ + 4 sec θ − 4 = 0
(sec θ + 2)(3 sec θ − 2) = 0
We have two possible solutions
sec θ = 2 and sec θ =2
3
Trigonometry
TrigonometricIdentities
TrigonometricEquations
6 Solve 2 tan2 θ + 4 sec θ + sec2 θ = 2 for 0 ≤ θ < 360.
2 tan2 θ + 4 sec θ + sec2 θ = 2
2(sec2 θ − 1)θ + 4 sec θ + sec2 θ = 2
2 sec2 θ − 2θ + 4 sec θ + sec2 θ = 2
3 sec2 θ + 4 sec θ − 4 = 0
(sec θ + 2)(3 sec θ − 2) = 0
We have two possible solutions
sec θ = 2 and sec θ =2
3
Trigonometry
TrigonometricIdentities
TrigonometricEquations
7 Solve 2 tan2 θ + 4 sec θ + sec2 θ = 2 for 0 ≤ θ < 360.
2 tan2 θ + 4 sec θ + sec2 θ = 2
2(sec2 θ − 1)θ + 4 sec θ + sec2 θ = 2
2 sec2 θ − 2θ + 4 sec θ + sec2 θ = 2
3 sec2 θ + 4 sec θ − 4 = 0
(sec θ + 2)(3 sec θ − 2) = 0
We have two possible solutions
sec θ = 2 and sec θ =2
3
Trigonometry
TrigonometricIdentities
TrigonometricEquations
1
cos θ= 2 and
1
cos θ=
2
3
cos θ =1
2and cos θ =
3
2
θ = 120
We also have another solution between 0 and 360, which is360 − 120 = 240.
θ = 120, 240
Trigonometry
TrigonometricIdentities
TrigonometricEquations
1
cos θ= 2 and
1
cos θ=
2
3
cos θ =1
2and cos θ =
3
2
θ = 120
We also have another solution between 0 and 360, which is360 − 120 = 240.
θ = 120, 240
Trigonometry
TrigonometricIdentities
TrigonometricEquations
1
cos θ= 2 and
1
cos θ=
2
3
cos θ =1
2and cos θ =
3
2
θ = 120
We also have another solution between 0 and 360, which is360 − 120 = 240.
θ = 120, 240
Trigonometry
TrigonometricIdentities
TrigonometricEquations8 • Show that the equation 4 cos2 x + 9 sin x − 6 = 0 can be written as
4 sin2 x − 9 sin x + 2 = 0.
4 cos2 x + 9 sin x − 6 = 0
4(1 − sin2 x) + 9 sin x − 6 = 0
4 sin2 x − 9 sin x + 2 = 0
Trigonometry
TrigonometricIdentities
TrigonometricEquations Show that the equation 4 cos2 x + 9 sin x − 6 = 0 can be written as
4 sin2 x − 9 sin x + 2 = 0.
4 cos2 x + 9 sin x − 6 = 0
4(1 − sin2 x) + 9 sin x − 6 = 0
4 sin2 x − 9 sin x + 2 = 0
Trigonometry
TrigonometricIdentities
TrigonometricEquations Show that the equation 4 cos2 x + 9 sin x − 6 = 0 can be written as
4 sin2 x − 9 sin x + 2 = 0.
4 cos2 x + 9 sin x − 6 = 0
4(1 − sin2 x) + 9 sin x − 6 = 0
4 sin2 x − 9 sin x + 2 = 0
Trigonometry
TrigonometricIdentities
TrigonometricEquations Show that the equation 4 cos2 x + 9 sin x − 6 = 0 can be written as
4 sin2 x − 9 sin x + 2 = 0.
4 cos2 x + 9 sin x − 6 = 0
4(1 − sin2 x) + 9 sin x − 6 = 0
4 sin2 x − 9 sin x + 2 = 0
Trigonometry
TrigonometricIdentities
TrigonometricEquations
Hence solve 4 cos2 x + 9 sin x − 6 = 0 for 0 ≤ x ≤ 720, giving youranswers to one decimal place.
4 sin2 x − 9 sin x + 2 = 0
(4 sin x − 1)(sin x − 2) = 0
We have two possible solutions,
sin x =1
4and sin x = 2
Solving for x , we havex = 14.5
Our next solutions between 0 and 720 will be• • • •• 180− 14.5 = 165.5,
• 360 + 14.5 = 374.5,• 360 + 165.5 = 525.5,
We havex = 14.5, 165.5, 374.5, 525.5
Trigonometry
TrigonometricIdentities
TrigonometricEquations
Hence solve 4 cos2 x + 9 sin x − 6 = 0 for 0 ≤ x ≤ 720, giving youranswers to one decimal place.
4 sin2 x − 9 sin x + 2 = 0
(4 sin x − 1)(sin x − 2) = 0
We have two possible solutions,
sin x =1
4and sin x = 2
Solving for x , we havex = 14.5
Our next solutions between 0 and 720 will be•• 180− 14.5 = 165.5,• 360 + 14.5 = 374.5,• 360 + 165.5 = 525.5,
We havex = 14.5, 165.5, 374.5, 525.5
Trigonometry
TrigonometricIdentities
TrigonometricEquations
Hence solve 4 cos2 x + 9 sin x − 6 = 0 for 0 ≤ x ≤ 720, giving youranswers to one decimal place.
4 sin2 x − 9 sin x + 2 = 0
(4 sin x − 1)(sin x − 2) = 0
We have two possible solutions,
sin x =1
4and sin x = 2
Solving for x , we havex = 14.5
Our next solutions between 0 and 720 will be•• 180− 14.5 = 165.5,• 360 + 14.5 = 374.5,• 360 + 165.5 = 525.5,
We havex = 14.5, 165.5, 374.5, 525.5
Trigonometry
TrigonometricIdentities
TrigonometricEquations
Hence solve 4 cos2 x + 9 sin x − 6 = 0 for 0 ≤ x ≤ 720, giving youranswers to one decimal place.
4 sin2 x − 9 sin x + 2 = 0
(4 sin x − 1)(sin x − 2) = 0
We have two possible solutions,
sin x =1
4and sin x = 2
Solving for x , we havex = 14.5
Our next solutions between 0 and 720 will be•• 180− 14.5 = 165.5,• 360 + 14.5 = 374.5,• 360 + 165.5 = 525.5,
We havex = 14.5, 165.5, 374.5, 525.5
Trigonometry
TrigonometricIdentities
TrigonometricEquations
Hence solve 4 cos2 x + 9 sin x − 6 = 0 for 0 ≤ x ≤ 720, giving youranswers to one decimal place.
4 sin2 x − 9 sin x + 2 = 0
(4 sin x − 1)(sin x − 2) = 0
We have two possible solutions,
sin x =1
4and sin x = 2
Solving for x , we havex = 14.5
Our next solutions between 0 and 720 will be•• 180− 14.5 = 165.5,• 360 + 14.5 = 374.5,• 360 + 165.5 = 525.5,
We havex = 14.5, 165.5, 374.5, 525.5
Trigonometry
TrigonometricIdentities
TrigonometricEquations
Hence solve 4 cos2 x + 9 sin x − 6 = 0 for 0 ≤ x ≤ 720, giving youranswers to one decimal place.
4 sin2 x − 9 sin x + 2 = 0
(4 sin x − 1)(sin x − 2) = 0
We have two possible solutions,
sin x =1
4and sin x = 2
Solving for x , we havex = 14.5
Our next solutions between 0 and 720 will be
•• 180− 14.5 = 165.5,• 360 + 14.5 = 374.5,• 360 + 165.5 = 525.5,
We havex = 14.5, 165.5, 374.5, 525.5
Trigonometry
TrigonometricIdentities
TrigonometricEquations
Hence solve 4 cos2 x + 9 sin x − 6 = 0 for 0 ≤ x ≤ 720, giving youranswers to one decimal place.
4 sin2 x − 9 sin x + 2 = 0
(4 sin x − 1)(sin x − 2) = 0
We have two possible solutions,
sin x =1
4and sin x = 2
Solving for x , we havex = 14.5
Our next solutions between 0 and 720 will be•• 180− 14.5 = 165.5,
• 360 + 14.5 = 374.5,• 360 + 165.5 = 525.5,
We havex = 14.5, 165.5, 374.5, 525.5
Trigonometry
TrigonometricIdentities
TrigonometricEquations
Hence solve 4 cos2 x + 9 sin x − 6 = 0 for 0 ≤ x ≤ 720, giving youranswers to one decimal place.
4 sin2 x − 9 sin x + 2 = 0
(4 sin x − 1)(sin x − 2) = 0
We have two possible solutions,
sin x =1
4and sin x = 2
Solving for x , we havex = 14.5
Our next solutions between 0 and 720 will be•• 180− 14.5 = 165.5,• 360 + 14.5 = 374.5,
• 360 + 165.5 = 525.5,
We havex = 14.5, 165.5, 374.5, 525.5
Trigonometry
TrigonometricIdentities
TrigonometricEquations
Hence solve 4 cos2 x + 9 sin x − 6 = 0 for 0 ≤ x ≤ 720, giving youranswers to one decimal place.
4 sin2 x − 9 sin x + 2 = 0
(4 sin x − 1)(sin x − 2) = 0
We have two possible solutions,
sin x =1
4and sin x = 2
Solving for x , we havex = 14.5
Our next solutions between 0 and 720 will be•• 180− 14.5 = 165.5,• 360 + 14.5 = 374.5,• 360 + 165.5 = 525.5,
We havex = 14.5, 165.5, 374.5, 525.5
Trigonometry
TrigonometricIdentities
TrigonometricEquations
Hence solve 4 cos2 x + 9 sin x − 6 = 0 for 0 ≤ x ≤ 720, giving youranswers to one decimal place.
4 sin2 x − 9 sin x + 2 = 0
(4 sin x − 1)(sin x − 2) = 0
We have two possible solutions,
sin x =1
4and sin x = 2
Solving for x , we havex = 14.5
Our next solutions between 0 and 720 will be•• 180− 14.5 = 165.5,• 360 + 14.5 = 374.5,• 360 + 165.5 = 525.5,
We havex = 14.5, 165.5, 374.5, 525.5
Trigonometry
TrigonometricIdentities
TrigonometricEquations
9 Solve cos x = 3 tan x for 0 ≤ x ≤ 3π, giving your answers to 3significant figures.
cos x = 3 tan x
cos x = 3sin x
cos xcos2 x = 3 sin x
1 − sin2 x = 3 sin x
sin2 x + 3 sin x − 1 = 0
We have two possible solutions
sin x = 0.30278 and sin x = −3.30278
Trigonometry
TrigonometricIdentities
TrigonometricEquations
10 Solve cos x = 3 tan x for 0 ≤ x ≤ 3π, giving your answers to 3significant figures.
cos x = 3 tan x
cos x = 3sin x
cos xcos2 x = 3 sin x
1 − sin2 x = 3 sin x
sin2 x + 3 sin x − 1 = 0
We have two possible solutions
sin x = 0.30278 and sin x = −3.30278
Trigonometry
TrigonometricIdentities
TrigonometricEquations
11 Solve cos x = 3 tan x for 0 ≤ x ≤ 3π, giving your answers to 3significant figures.
cos x = 3 tan x
cos x = 3sin x
cos x
cos2 x = 3 sin x
1 − sin2 x = 3 sin x
sin2 x + 3 sin x − 1 = 0
We have two possible solutions
sin x = 0.30278 and sin x = −3.30278
Trigonometry
TrigonometricIdentities
TrigonometricEquations
12 Solve cos x = 3 tan x for 0 ≤ x ≤ 3π, giving your answers to 3significant figures.
cos x = 3 tan x
cos x = 3sin x
cos xcos2 x = 3 sin x
1 − sin2 x = 3 sin x
sin2 x + 3 sin x − 1 = 0
We have two possible solutions
sin x = 0.30278 and sin x = −3.30278
Trigonometry
TrigonometricIdentities
TrigonometricEquations
13 Solve cos x = 3 tan x for 0 ≤ x ≤ 3π, giving your answers to 3significant figures.
cos x = 3 tan x
cos x = 3sin x
cos xcos2 x = 3 sin x
1 − sin2 x = 3 sin x
sin2 x + 3 sin x − 1 = 0
We have two possible solutions
sin x = 0.30278 and sin x = −3.30278
Trigonometry
TrigonometricIdentities
TrigonometricEquations
14 Solve cos x = 3 tan x for 0 ≤ x ≤ 3π, giving your answers to 3significant figures.
cos x = 3 tan x
cos x = 3sin x
cos xcos2 x = 3 sin x
1 − sin2 x = 3 sin x
sin2 x + 3 sin x − 1 = 0
We have two possible solutions
sin x = 0.30278 and sin x = −3.30278
Trigonometry
TrigonometricIdentities
TrigonometricEquations
15 Solve cos x = 3 tan x for 0 ≤ x ≤ 3π, giving your answers to 3significant figures.
cos x = 3 tan x
cos x = 3sin x
cos xcos2 x = 3 sin x
1 − sin2 x = 3 sin x
sin2 x + 3 sin x − 1 = 0
We have two possible solutions
sin x = 0.30278 and sin x = −3.30278
Trigonometry
TrigonometricIdentities
TrigonometricEquations
Our first solution for x is
x = 0.3076
Our next solutions will be• π − 0.3076 = 2.83,• 2π + 0.3076 = 6.59,• 2π + 2.83 = 9.12
Our solutions are
x = 0.308, 2.83, 6.59, 9.12
Trigonometry
TrigonometricIdentities
TrigonometricEquations
Our first solution for x is
x = 0.3076
Our next solutions will be• π − 0.3076 = 2.83,
• 2π + 0.3076 = 6.59,• 2π + 2.83 = 9.12
Our solutions are
x = 0.308, 2.83, 6.59, 9.12
Trigonometry
TrigonometricIdentities
TrigonometricEquations
Our first solution for x is
x = 0.3076
Our next solutions will be• π − 0.3076 = 2.83,• 2π + 0.3076 = 6.59,
• 2π + 2.83 = 9.12
Our solutions are
x = 0.308, 2.83, 6.59, 9.12
Trigonometry
TrigonometricIdentities
TrigonometricEquations
Our first solution for x is
x = 0.3076
Our next solutions will be• π − 0.3076 = 2.83,• 2π + 0.3076 = 6.59,• 2π + 2.83 = 9.12
Our solutions are
x = 0.308, 2.83, 6.59, 9.12
Trigonometry
TrigonometricIdentities
TrigonometricEquations
Our first solution for x is
x = 0.3076
Our next solutions will be• π − 0.3076 = 2.83,• 2π + 0.3076 = 6.59,• 2π + 2.83 = 9.12
Our solutions are
x = 0.308, 2.83, 6.59, 9.12
Trigonometry
TrigonometricIdentities
TrigonometricEquations
16 Solve 5 + 2 cot2 α = 7(1 + cosecα), where α = 2x + π3 , in the
interval 0 ≤ x ≤ 2π. Give your answers to 2 decimal places.
5 + 2(cosec2α− 1) = 7(1 + cosecα)
5 + 2cosec2α− 2 = 7 + 7cosecα
2cosec2α− 7cosecα− 4 = 0
(2cosecα + 1)(cosecα− 4) = 0
We have two possible solutions
cosecα = −1
2and cosecα = 4
Trigonometry
TrigonometricIdentities
TrigonometricEquations
17 Solve 5 + 2 cot2 α = 7(1 + cosecα), where α = 2x + π3 , in the
interval 0 ≤ x ≤ 2π. Give your answers to 2 decimal places.
5 + 2(cosec2α− 1) = 7(1 + cosecα)
5 + 2cosec2α− 2 = 7 + 7cosecα
2cosec2α− 7cosecα− 4 = 0
(2cosecα + 1)(cosecα− 4) = 0
We have two possible solutions
cosecα = −1
2and cosecα = 4
Trigonometry
TrigonometricIdentities
TrigonometricEquations
18 Solve 5 + 2 cot2 α = 7(1 + cosecα), where α = 2x + π3 , in the
interval 0 ≤ x ≤ 2π. Give your answers to 2 decimal places.
5 + 2(cosec2α− 1) = 7(1 + cosecα)
5 + 2cosec2α− 2 = 7 + 7cosecα
2cosec2α− 7cosecα− 4 = 0
(2cosecα + 1)(cosecα− 4) = 0
We have two possible solutions
cosecα = −1
2and cosecα = 4
Trigonometry
TrigonometricIdentities
TrigonometricEquations
19 Solve 5 + 2 cot2 α = 7(1 + cosecα), where α = 2x + π3 , in the
interval 0 ≤ x ≤ 2π. Give your answers to 2 decimal places.
5 + 2(cosec2α− 1) = 7(1 + cosecα)
5 + 2cosec2α− 2 = 7 + 7cosecα
2cosec2α− 7cosecα− 4 = 0
(2cosecα + 1)(cosecα− 4) = 0
We have two possible solutions
cosecα = −1
2and cosecα = 4
Trigonometry
TrigonometricIdentities
TrigonometricEquations
20 Solve 5 + 2 cot2 α = 7(1 + cosecα), where α = 2x + π3 , in the
interval 0 ≤ x ≤ 2π. Give your answers to 2 decimal places.
5 + 2(cosec2α− 1) = 7(1 + cosecα)
5 + 2cosec2α− 2 = 7 + 7cosecα
2cosec2α− 7cosecα− 4 = 0
(2cosecα + 1)(cosecα− 4) = 0
We have two possible solutions
cosecα = −1
2and cosecα = 4
Trigonometry
TrigonometricIdentities
TrigonometricEquations
21 Solve 5 + 2 cot2 α = 7(1 + cosecα), where α = 2x + π3 , in the
interval 0 ≤ x ≤ 2π. Give your answers to 2 decimal places.
5 + 2(cosec2α− 1) = 7(1 + cosecα)
5 + 2cosec2α− 2 = 7 + 7cosecα
2cosec2α− 7cosecα− 4 = 0
(2cosecα + 1)(cosecα− 4) = 0
We have two possible solutions
cosecα = −1
2and cosecα = 4
Trigonometry
TrigonometricIdentities
TrigonometricEquations
sinα = −2 and sinα =1
4
α = 0.25268
Note that since 0 ≤ x ≤ 2π, our limits for α areπ3 ≤ α ≤ 4π + π
3 or 1.047 ≤ α ≤ 13.614. Our next solutionswill be
• π − 0.25268 = 2.8889,• 2π + 0.25268 = 6.53587,• 2π + 2.8889 = 9.172,• 2π + 6.53587 = 12.819,
Since α = 2x + π3 , our final solutions for x are
0.92, 2.74, 4.06, 5.89
Trigonometry
TrigonometricIdentities
TrigonometricEquations
sinα = −2 and sinα =1
4
α = 0.25268
Note that since 0 ≤ x ≤ 2π, our limits for α areπ3 ≤ α ≤ 4π + π
3 or 1.047 ≤ α ≤ 13.614. Our next solutionswill be
• π − 0.25268 = 2.8889,• 2π + 0.25268 = 6.53587,• 2π + 2.8889 = 9.172,• 2π + 6.53587 = 12.819,
Since α = 2x + π3 , our final solutions for x are
0.92, 2.74, 4.06, 5.89
Trigonometry
TrigonometricIdentities
TrigonometricEquations
sinα = −2 and sinα =1
4
α = 0.25268
Note that since 0 ≤ x ≤ 2π, our limits for α areπ3 ≤ α ≤ 4π + π
3 or 1.047 ≤ α ≤ 13.614.
Our next solutionswill be
• π − 0.25268 = 2.8889,• 2π + 0.25268 = 6.53587,• 2π + 2.8889 = 9.172,• 2π + 6.53587 = 12.819,
Since α = 2x + π3 , our final solutions for x are
0.92, 2.74, 4.06, 5.89
Trigonometry
TrigonometricIdentities
TrigonometricEquations
sinα = −2 and sinα =1
4
α = 0.25268
Note that since 0 ≤ x ≤ 2π, our limits for α areπ3 ≤ α ≤ 4π + π
3 or 1.047 ≤ α ≤ 13.614. Our next solutionswill be
• π − 0.25268 = 2.8889,
• 2π + 0.25268 = 6.53587,• 2π + 2.8889 = 9.172,• 2π + 6.53587 = 12.819,
Since α = 2x + π3 , our final solutions for x are
0.92, 2.74, 4.06, 5.89
Trigonometry
TrigonometricIdentities
TrigonometricEquations
sinα = −2 and sinα =1
4
α = 0.25268
Note that since 0 ≤ x ≤ 2π, our limits for α areπ3 ≤ α ≤ 4π + π
3 or 1.047 ≤ α ≤ 13.614. Our next solutionswill be
• π − 0.25268 = 2.8889,• 2π + 0.25268 = 6.53587,
• 2π + 2.8889 = 9.172,• 2π + 6.53587 = 12.819,
Since α = 2x + π3 , our final solutions for x are
0.92, 2.74, 4.06, 5.89
Trigonometry
TrigonometricIdentities
TrigonometricEquations
sinα = −2 and sinα =1
4
α = 0.25268
Note that since 0 ≤ x ≤ 2π, our limits for α areπ3 ≤ α ≤ 4π + π
3 or 1.047 ≤ α ≤ 13.614. Our next solutionswill be
• π − 0.25268 = 2.8889,• 2π + 0.25268 = 6.53587,• 2π + 2.8889 = 9.172,
• 2π + 6.53587 = 12.819,
Since α = 2x + π3 , our final solutions for x are
0.92, 2.74, 4.06, 5.89
Trigonometry
TrigonometricIdentities
TrigonometricEquations
sinα = −2 and sinα =1
4
α = 0.25268
Note that since 0 ≤ x ≤ 2π, our limits for α areπ3 ≤ α ≤ 4π + π
3 or 1.047 ≤ α ≤ 13.614. Our next solutionswill be
• π − 0.25268 = 2.8889,• 2π + 0.25268 = 6.53587,• 2π + 2.8889 = 9.172,• 2π + 6.53587 = 12.819,
Since α = 2x + π3 , our final solutions for x are
0.92, 2.74, 4.06, 5.89
Trigonometry
TrigonometricIdentities
TrigonometricEquations
sinα = −2 and sinα =1
4
α = 0.25268
Note that since 0 ≤ x ≤ 2π, our limits for α areπ3 ≤ α ≤ 4π + π
3 or 1.047 ≤ α ≤ 13.614. Our next solutionswill be
• π − 0.25268 = 2.8889,• 2π + 0.25268 = 6.53587,• 2π + 2.8889 = 9.172,• 2π + 6.53587 = 12.819,
Since α = 2x + π3 , our final solutions for x are
0.92, 2.74, 4.06, 5.89