Trigonometry
.
System of measuring angles
There are three systems of measuring angles.
1 Sexagesimal system
2 Centisimal system
3 Circular system
Sexagesimal system : In this unit of measurement of an angle is 'degree'
Degree : (1/360) ᵗʰ part of one complete rotation is called one degree, denoted as 1°.
Minute : A degree is divided into 60 equal parts and each part is called one minute denoted 1ˡ
1°=60ˡ
Second : A minute is divided into 60 equal parts and each part is called one second, denoted by 1ˡˡ
1ˡ=60ˡˡ
In this system One right angle = 90°
Centesimal system : In this unit of measurement of an angle is 'grade'.
Grade : (1/400)ᵗʰ part of one complete rotation is called one grade, denoted as 1g.
Minute : A grade is divided into 100 equal parts and each part is called one minute, denoted 1ˡ
1g=100ˡ
Second : A minute is divided into 100 equal parts and each part is called one second, denoted by 1ˡˡ
1ˡ=100ˡˡ
In this system One right angle = 100g.
In this system One right angle = 100g
Note: sexagesimal system minute, centisimal system minute values are different.
Circular system: In this unit of measurement of an angle is 'radian'.
Radian:The angle subtended by an arc of length is equal to the radius of the circle at its centre is called radian, it is denoted as 1c.
In this system one right angle = πc/2
1c=57°16ˡ (approx.)
1°=0.01746c
The formula connecting the three systems is
Here, 'D' denotes degrees, 'G' denotes grades and 'C' denotes radians
From sector
Length of arc l = rθ
Area of sector (A) = (½) r² θ
Where θ is measured in radians
✦ 180° = π radians
✦ 1° = π/180°=0.01745 radians (approx.)
✦ π =
✦ Each interior angle of a regular polygon of 'n' sides =
✦ sinθcosecθ=1 ; sinθ = 1/cosecθ; cosecθ=1/sinθ
Also, –1≤ sinθ≥ 1; cosecθ≤ – 1 or cosecθ ≥1
✦ cosθsecθ=1 ; cosθ = 1/secθ; secθ=1/cosθ
Also, –1≤ cosθ≥ 1; secθ≤ – 1 or secθ ≥1
✦ tanθcotθ=1 ; tanθ = 1/cotθ; cotθ=1/tanθ
Also, – ∞ ≤ tanθ ≥ ∞ ; – ∞ ≤ cotθ ≥ ∞
.
System of measuring angles
There are three systems of measuring angles.
1 Sexagesimal system
2 Centisimal system
3 Circular system
Sexagesimal system : In this unit of measurement of an angle is 'degree'
Degree : (1/360) ᵗʰ part of one complete rotation is called one degree, denoted as 1°.
Minute : A degree is divided into 60 equal parts and each part is called one minute denoted 1ˡ
1°=60ˡ
Second : A minute is divided into 60 equal parts and each part is called one second, denoted by 1ˡˡ
1ˡ=60ˡˡ
In this system One right angle = 90°
Centesimal system : In this unit of measurement of an angle is 'grade'.
Grade : (1/400)ᵗʰ part of one complete rotation is called one grade, denoted as 1g.
Minute : A grade is divided into 100 equal parts and each part is called one minute, denoted 1ˡ
1g=100ˡ
Second : A minute is divided into 100 equal parts and each part is called one second, denoted by 1ˡˡ
1ˡ=100ˡˡ
In this system One right angle = 100g.
In this system One right angle = 100g
Note: sexagesimal system minute, centisimal system minute values are different.
Circular system: In this unit of measurement of an angle is 'radian'.
Radian:The angle subtended by an arc of length is equal to the radius of the circle at its centre is called radian, it is denoted as 1c.
In this system one right angle = πc/2
1c=57°16ˡ (approx.)
1°=0.01746c
The formula connecting the three systems is
D
90
=
G
100
=
2C
π
Here, 'D' denotes degrees, 'G' denotes grades and 'C' denotes radians
From sector
Length of arc l = rθ
Area of sector (A) = (½) r² θ
Where θ is measured in radians
✦ 180° = π radians
✦ 1° = π/180°=0.01745 radians (approx.)
✦ π =
Circumference of the Circle
Diameter of the Circle
= 22/7=3.1416(approx)✦ Each interior angle of a regular polygon of 'n' sides =
(n–2)×180°
n
✦ sinθcosecθ=1 ; sinθ = 1/cosecθ; cosecθ=1/sinθ
Also, –1≤ sinθ≥ 1; cosecθ≤ – 1 or cosecθ ≥1
✦ cosθsecθ=1 ; cosθ = 1/secθ; secθ=1/cosθ
Also, –1≤ cosθ≥ 1; secθ≤ – 1 or secθ ≥1
✦ tanθcotθ=1 ; tanθ = 1/cotθ; cotθ=1/tanθ
Also, – ∞ ≤ tanθ ≥ ∞ ; – ∞ ≤ cotθ ≥ ∞
.
Pythagoras theorem :
In a right angled triangle
(Hypotenuse)²= (Base)² +(Height)²
Trigonometric Identities:
✦ sin²θ+cos²θ =1, where θ ∈ R
✦ sec²θ – tan²θ =1, where θ is not odd multiple of π/2
✦ cosec²θ – cot²θ =1, where θ is not integral multiple of π
Trigonometry Formulae
Compound Angles
✦ sin(A+B)=sinA cosB + cosA sinB
✦ sin(A – B)=sinA cosB – cosA sinB
✦ cos(A+B)=cosA cosB – sinA sinB
✦ cos(A – B)=cosA cosB + sinA sinB
✦ tan(A+B)=
✦ tan(A – B)=
✦ cot(A+B)=
✦ cot(A–B)=
✦ sin(A+B).sin(A–B)= sin²A – sin²B=cos²B–cos²A
✦ cos(A+B).cos(A–B)= cos²A – sin²B=cos²B–sin²A
Trigonometry Formulae
Multiple Angles
✦ sin2A=2 sinA cosA=
✦ cos2A=cos²A – sin²A=2cos²A –1=1 –2sin²A=
✦ tan2A=
✦ cot2A=
✦ sin3A = 3sinA – 4sin³A
✦ cos3A = 4cos³A – 3cosA
✦ tan3A =
✦ cot3A =
Trigonometry Formulae
Sub multiple Angles
✦ sinA = 2sinᴬ/₂ cosᴬ/₂
✦ cosA = cos² ᴬ/₂ – sin² ᴬ/₂=2cos² ᴬ/₂ –1= 1– 2sin² ᴬ/₂=
✦ tanA=
✦ cotA=
Trigonometry Formulae
Transformations
∀ A, B, C, D ∈ R
✦ sin(A+B) +sin(A –B)=2sinA cosB
✦ sin(A+B) –sin(A –B)=2cosA sinB
✦ cos(A+B) +cos(A –B)=2cosA cosB
✦ cos(A+B) –cos(A –B)= –2sinA sinB (or) cos(A –B) –cos(A +B)= 2sinA sinB
✦ sinC+sinD=2sin[(C+D)/2] cos[(C –D)/2]
✦ sinC –sinD=2cos[(C+D)/2] sin[(C –D)/2]
✦ cosC+cosD=2cos[(C+D)/2] cos[(C –D)/2]
✦ cosC –cosD= – 2sin[(C+D)/2] sin[(C –D)/2]
....
✦ sin(A – B)=sinA cosB – cosA sinB
✦ cos(A+B)=cosA cosB – sinA sinB
✦ cos(A – B)=cosA cosB + sinA sinB
✦ tan(A+B)=
(tanA + tanB)
(1–tanA.tanB)
✦ tan(A – B)=
(tanA – tanB)
(1+tanA.tanB)
✦ cot(A+B)=
(cotB.cotA–1)
(cotB + cotA)
✦ cot(A–B)=
(cotB.cotA+1)
(cotB – cotA)
✦ sin(A+B).sin(A–B)= sin²A – sin²B=cos²B–cos²A
✦ cos(A+B).cos(A–B)= cos²A – sin²B=cos²B–sin²A
Trigonometry Formulae
Multiple Angles
✦ sin2A=2 sinA cosA=
2tanA
1+tan²A
✦ cos2A=cos²A – sin²A=2cos²A –1=1 –2sin²A=
1–tan²A
1+tan²A
✦ tan2A=
2tanA
1–tan²A
, where A and 2A are not odd multiple of π/₂✦ cot2A=
cot²A–1
2cotA
, where A and 2A are not integral multiple of π✦ sin3A = 3sinA – 4sin³A
✦ cos3A = 4cos³A – 3cosA
✦ tan3A =
3tanA – tan³A
1 – 3tan²A
, where A and 2A are not odd multiple of π/₂✦ cot3A =
3cotA – cot³A
1 –3cot²A
, where A and 2A are not integral multiple of πTrigonometry Formulae
Sub multiple Angles
✦ sinA = 2sinᴬ/₂ cosᴬ/₂
✦ cosA = cos² ᴬ/₂ – sin² ᴬ/₂=2cos² ᴬ/₂ –1= 1– 2sin² ᴬ/₂=
(1 –tan² ᴬ/₂)
(1+tan² ᴬ/₂)
✦ tanA=
(2tanᴬ/₂)
(1 – tan² ᴬ/₂)
, where A and ᴬ/₂ are not odd multiple of π/₂✦ cotA=
(cot² ᴬ/₂ – 1)
(2cotᴬ/₂)
, where A and ᴬ/₂ are not integral multiple of πTrigonometry Formulae
Transformations
∀ A, B, C, D ∈ R
✦ sin(A+B) +sin(A –B)=2sinA cosB
✦ sin(A+B) –sin(A –B)=2cosA sinB
✦ cos(A+B) +cos(A –B)=2cosA cosB
✦ cos(A+B) –cos(A –B)= –2sinA sinB (or) cos(A –B) –cos(A +B)= 2sinA sinB
✦ sinC+sinD=2sin[(C+D)/2] cos[(C –D)/2]
✦ sinC –sinD=2cos[(C+D)/2] sin[(C –D)/2]
✦ cosC+cosD=2cos[(C+D)/2] cos[(C –D)/2]
✦ cosC –cosD= – 2sin[(C+D)/2] sin[(C –D)/2]
....
....
....
(P) Prove that
Solution:
LHS/-
⇒ (cotθ – tanθ) ×
⇒[
⇒
⇒
⇒
(P) Find the value of tan9°–tan27°–tan63°+tan81°
Solution :
Given tan9° – tan27° –tan63° + tan81°
⇒tan9° – tan27° – tan(90°– 27°) + tan(90°–9°)
⇒tan9° – tan27° – cot27° + cot9° [∵tan(90°–θ)]
⇒ tan9° + cot9° – ( tan27° + cot27°)
⇒2cosec2(9°)–[2cosec2(27°)] [∵cotA+tanA=2cosec2A]
⇒ 2cosec18° – 2cosec54°
= 2[√5 +1 – (√5 – 1)]
= 2[√5 +1 – √5 + 1]
= 2(2)
= 4
(P) If cosA=¾ , then show that 32sinᴬ/₂sin⁵ᴬ/₂ =11
Solution :
Given cosA=¾
L.H.S /- 32sinᴬ/₂sin⁵ᴬ/₂
= 16(2sinᴬ/₂sin⁵ᴬ/₂)
= 16[cos(ᴬ/₂ – ⁵ᴬ/₂) – cos(ᴬ/₂ + ⁵ᴬ/₂) ] {∵ 2sinA.sinB = cos(A –B) – cos(A+B)}
= 16[cos⁴ᴬ/₂ – cos⁶ᴬ/₂]
= 16(cos2A – cos3A)
= 16[2cos²A –1 – (4cos³A –3cosA)] {∵ cos2A=2cos²A –1 ; co3A=4cos³A – 3cosA}
= 16[2cos²A –1 – 4cos³A +3cosA]
= 16[2(¾)² – 1 – 4 (¾)³ +3 (¾)]
= 16[2(9/16) – 1 – 4 (27/64) +(9/4)]
= 16[(18/16) – 1 –(27/16)+(9/4)]
= 16[(18 – 16 – 27 +36)/ 16]
= 54 – 43
= 11
(P) Prove that 3(sinx – cosx)⁴ +6(sinx + cosx)² + 4(sin⁶x + cos⁶x)=13
Solution :
L.H.S/-
3(sinx – cosx)⁴ +6(sinx + cosx)² + 4(sin⁶x + cos⁶x)
⇒ 3(sin⁴x – 4sin³xcosx+6sin²xcos²x – 4sinxcos³x+cos⁴x)+6(sin²x + cos²x+2sinxcosx)+ 4[(sin²x)³+ (cos²x)³]
[∵ (x –y)ⁿ = ⁿC₀xⁿy⁰ – ⁿC₁xⁿ⁻¹y¹ +ⁿC₂xⁿ⁻²y² – ⁿC₃xⁿ⁻³y³ +.......]
⇒ 3(sin⁴x – 4sin³xcosx+6sin²xcos²x – 4sinxcos³x+cos⁴x)+6(1+2sinxcosx)+ 4[(sin²x+ cos²x)(sin⁴x –sin²xcos²x+cos⁴x)] {∵ sin²θ+ cos²θ=1}
⇒ 3[sin⁴x +cos⁴x– 4sinxcosx(sin²x+cos²x)+6sin²xcos²x] +6+12sinxcosx + 4[(1)(sin⁴x –sin²xcos²x+cos⁴x)]
⇒3[sin⁴x +cos⁴x– 4sinxcosx(1)+2sin²xcos²x+4sin²xcos²x] +6+12sinxcosx + 4[(1)(sin⁴x +2sin²xcos²x+cos⁴x – 3sin²xcos²x)]
⇒3[sin⁴x +cos⁴x+2sin²xcos²x– 4sinxcosx(1)+4sin²xcos²x] +6+12sinxcosx + 4[(sin⁴x +2sin²xcos²x+cos⁴x – 3sin²xcos²x)]
⇒3[(cos²x+sin²x)²– 4sinxcosx(1)+4sin²xcos²x] +6+12sinxcosx + 4[(cos²x+sin²x)² – 3sin²xcos²x)]
⇒3[(1)²– 4sinxcosx(1)+4sin²xcos²x] +6+12sinxcosx + 4[(1)² – 3sin²xcos²x)]
⇒3– 12sinxcosx(1)+12sin²xcos²x] +6+12sinxcosx + 4 – 12sin²xcos²x)]
⇒3+6+4=13
(cotθ – tanθ)
1 – 2sin²θ
=secθcosecθSolution:
LHS/-
(cotθ – tanθ)
1 – 2sin²θ
⇒ (cotθ – tanθ) ×
1
1 – 2sin²θ
⇒[
cosθ
sinθ
–
sinθ
cosθ
] ×
1
cos2θ
{∵cos2θ=1 – 2sin²θ}⇒
(cos²θ – sin²θ)
cosθsinθ
×
1
cos2θ
⇒
cos2θ
cosθsinθ
×
1
cos2θ
{∵ cos2θ=cos²θ – sin²θ}⇒
1
cosθsinθ
=secθcosecθ(P) Find the value of tan9°–tan27°–tan63°+tan81°
Solution :
Given tan9° – tan27° –tan63° + tan81°
⇒tan9° – tan27° – tan(90°– 27°) + tan(90°–9°)
⇒tan9° – tan27° – cot27° + cot9° [∵tan(90°–θ)]
⇒ tan9° + cot9° – ( tan27° + cot27°)
⇒2cosec2(9°)–[2cosec2(27°)] [∵cotA+tanA=2cosec2A]
⇒ 2cosec18° – 2cosec54°
= 2[√5 +1 – (√5 – 1)]
= 2[√5 +1 – √5 + 1]
= 2(2)
= 4
(P) If cosA=¾ , then show that 32sinᴬ/₂sin⁵ᴬ/₂ =11
Solution :
Given cosA=¾
L.H.S /- 32sinᴬ/₂sin⁵ᴬ/₂
= 16(2sinᴬ/₂sin⁵ᴬ/₂)
= 16[cos(ᴬ/₂ – ⁵ᴬ/₂) – cos(ᴬ/₂ + ⁵ᴬ/₂) ] {∵ 2sinA.sinB = cos(A –B) – cos(A+B)}
= 16[cos⁴ᴬ/₂ – cos⁶ᴬ/₂]
= 16(cos2A – cos3A)
= 16[2cos²A –1 – (4cos³A –3cosA)] {∵ cos2A=2cos²A –1 ; co3A=4cos³A – 3cosA}
= 16[2cos²A –1 – 4cos³A +3cosA]
= 16[2(¾)² – 1 – 4 (¾)³ +3 (¾)]
= 16[2(9/16) – 1 – 4 (27/64) +(9/4)]
= 16[(18/16) – 1 –(27/16)+(9/4)]
= 16[(18 – 16 – 27 +36)/ 16]
= 54 – 43
= 11
(P) Prove that 3(sinx – cosx)⁴ +6(sinx + cosx)² + 4(sin⁶x + cos⁶x)=13
Solution :
L.H.S/-
3(sinx – cosx)⁴ +6(sinx + cosx)² + 4(sin⁶x + cos⁶x)
⇒ 3(sin⁴x – 4sin³xcosx+6sin²xcos²x – 4sinxcos³x+cos⁴x)+6(sin²x + cos²x+2sinxcosx)+ 4[(sin²x)³+ (cos²x)³]
[∵ (x –y)ⁿ = ⁿC₀xⁿy⁰ – ⁿC₁xⁿ⁻¹y¹ +ⁿC₂xⁿ⁻²y² – ⁿC₃xⁿ⁻³y³ +.......]
⇒ 3(sin⁴x – 4sin³xcosx+6sin²xcos²x – 4sinxcos³x+cos⁴x)+6(1+2sinxcosx)+ 4[(sin²x+ cos²x)(sin⁴x –sin²xcos²x+cos⁴x)] {∵ sin²θ+ cos²θ=1}
⇒ 3[sin⁴x +cos⁴x– 4sinxcosx(sin²x+cos²x)+6sin²xcos²x] +6+12sinxcosx + 4[(1)(sin⁴x –sin²xcos²x+cos⁴x)]
⇒3[sin⁴x +cos⁴x– 4sinxcosx(1)+2sin²xcos²x+4sin²xcos²x] +6+12sinxcosx + 4[(1)(sin⁴x +2sin²xcos²x+cos⁴x – 3sin²xcos²x)]
⇒3[sin⁴x +cos⁴x+2sin²xcos²x– 4sinxcosx(1)+4sin²xcos²x] +6+12sinxcosx + 4[(sin⁴x +2sin²xcos²x+cos⁴x – 3sin²xcos²x)]
⇒3[(cos²x+sin²x)²– 4sinxcosx(1)+4sin²xcos²x] +6+12sinxcosx + 4[(cos²x+sin²x)² – 3sin²xcos²x)]
⇒3[(1)²– 4sinxcosx(1)+4sin²xcos²x] +6+12sinxcosx + 4[(1)² – 3sin²xcos²x)]
⇒3– 12sinxcosx(1)+12sin²xcos²x] +6+12sinxcosx + 4 – 12sin²xcos²x)]
⇒3+6+4=13
