By:- Pranav Khandelwal
INVERSE
TRIGONOMETRIC
FUNCTION
INTRODUCTION
▶ Onto function : (aka surjective function)
- Range and Co-Domain of a function are same, for eg:
For a function to be inverse trigo function, it needs to be 1-1 and onto
function (aka bijective function)
▶ One to one function: (aka injective function)
-No two elements have same output, for eg:
𝒔𝒊𝒏−𝟏 𝒙
▶ Domain : [-
1,1]
▶ Range : [-
π
,π
]
2 2
▶ Increasing
function
▶ Odd function
▶ sin−1( −𝑥) = -
sin−1 𝑥
▶ sin−1
𝑥=
𝑑 1
𝑑𝑥
1−𝑥2
𝒄𝒐𝒔−𝟏 𝒙
▶ Domain : [-1,1]
▶ Range : [0, π]
▶ Decreasing function
▶ Neither odd nor even
▶ cos−1( −𝑥 )= π -
cos−1 𝑥
𝑑
𝑥
▶
𝑑
cos−1 𝑥
= -
1
1−𝑥2
𝒕𝒂𝒏−𝟏 𝒙
▶ Domain : All Real
no.s π
▶ Range : (-π
,)
2
2
▶ Increasing function
▶ Odd function
▶ tan−1 −𝑥 =
−tan−1 𝑥
𝑑
𝑥
▶
𝑑
tan−1 𝑥
=
1
1+𝑥2
𝒄𝒐𝒕−𝟏 𝒙
▶ Domain : All Real No.s
▶ Range : (0, π)
▶ Decreasing function
▶ Neither odd nor even
function
▶ cot−1(−𝑥)= π - cot−1 𝑥
𝑑
𝑥
▶
𝑑
cot−1(𝑥)=
−
1
1+𝑥2
𝒔𝒆𝒄−𝟏 𝒙
▶ Domain : −∞, −1 𝑈 [1,
∞)
2
2
▶ Range : [0,π
) U
(𝜋
,π]
▶ Neither even nor odd
function
▶ sec−1(−𝑥) = π - sec−1 𝑥
𝑑
𝑥
▶
𝑑
sec−1 𝑥
=
1
𝑥
𝑥2−1
𝒄𝒐𝒔𝒆𝒄−𝟏 𝒙
▶ Domain : −∞. −1 U [1,
∞)
▶ Range : [
�
�
2
− , 0) U
(0,
2
𝜋
]
▶ Odd function
▶ csc−1 −𝑥=
−csc−1 𝑥
𝑑
𝑥
▶
𝑑
csc−1 𝑥
= -
1
𝑥
𝑥2−1
NOTE:
▶ None of the inverse trigo functions are periodic functions.
▶ 𝑓(f −1 𝑥 ) = x for every x belonging to its domain
▶ Sign of an inequality changes if we apply a decreasing function
on both sides of an inequality. Sign of inequality remains same if
it's an increasing function.
▶ 𝑦 = sin−1(sin 𝑥) = nπ + (−1)𝑛 𝑥
▶ 𝑦 = cs𝑐−1 (csc 𝑥) = nπ +
(−1)𝑛 𝑥
▶ 𝑦 = cos−1 (cos 𝑥) =
2nπ ± 𝑥
▶ 𝑦 = sec−1 (sec 𝑥) =
2nπ ± 𝑥
▶ 𝑦 = tan−1(tan 𝑥) =
𝑛𝜋 + 𝑥
▶ 𝑦 = cot−1(𝑐𝑜𝑡 𝑥) =
𝑛𝜋 + 𝑥
PROPERTIES:
▶ sin−1 𝑥 + cos−1 𝑥
=
�
�
▶ tan−1 𝑥 +
co𝑡−1 𝑥 =
2
�
�
2
▶ s𝑒𝑐−1 𝑥 +
cosec−1 𝑥 =
�
�
2
�
�
▶ sin−1 1
=
csc−1 𝑥
�
�
▶ cos−1 1
=
sec−1 𝑥
▶
tan
−1 1
𝑥
cot−1
𝑥
= ቊ
−𝜋 +
cot−1 𝑥
; 𝑥 >
0
; 𝑥 <
0
�
�
�
�
▶ tan−1 𝑥 + tan−1 1
=
ቐ2 −
𝜋
2
; 𝑥 >
0
; 𝑥 <
▶ tan−1 𝑥 + tan−1 𝑦
=
tan−
1
𝑥+
𝑦
1−𝑥
𝑦
; 𝑥 > 0, 𝑦 > 0,
𝑥𝑦 < 1
𝜋 +
tan−1
𝑥+
𝑦
1−𝑥
𝑦
; 𝑥 > 0, 𝑦 > 0,
𝑥𝑦 > 1
▶ tan−1 𝑥 − tan−1 𝑦 =
tan−1
𝑥−
𝑦
1+𝑥
𝑦
; 𝑥 > 0, 𝑦
> 0
THANKYOU

ITF-converted.pptx class 12 cbse project

  • 1.
  • 2.
    INTRODUCTION ▶ Onto function: (aka surjective function) - Range and Co-Domain of a function are same, for eg: For a function to be inverse trigo function, it needs to be 1-1 and onto function (aka bijective function) ▶ One to one function: (aka injective function) -No two elements have same output, for eg:
  • 3.
    𝒔𝒊𝒏−𝟏 𝒙 ▶ Domain: [- 1,1] ▶ Range : [- π ,π ] 2 2 ▶ Increasing function ▶ Odd function ▶ sin−1( −𝑥) = - sin−1 𝑥 ▶ sin−1 𝑥= 𝑑 1 𝑑𝑥 1−𝑥2
  • 4.
    𝒄𝒐𝒔−𝟏 𝒙 ▶ Domain: [-1,1] ▶ Range : [0, π] ▶ Decreasing function ▶ Neither odd nor even ▶ cos−1( −𝑥 )= π - cos−1 𝑥 𝑑 𝑥 ▶ 𝑑 cos−1 𝑥 = - 1 1−𝑥2
  • 5.
    𝒕𝒂𝒏−𝟏 𝒙 ▶ Domain: All Real no.s π ▶ Range : (-π ,) 2 2 ▶ Increasing function ▶ Odd function ▶ tan−1 −𝑥 = −tan−1 𝑥 𝑑 𝑥 ▶ 𝑑 tan−1 𝑥 = 1 1+𝑥2
  • 6.
    𝒄𝒐𝒕−𝟏 𝒙 ▶ Domain: All Real No.s ▶ Range : (0, π) ▶ Decreasing function ▶ Neither odd nor even function ▶ cot−1(−𝑥)= π - cot−1 𝑥 𝑑 𝑥 ▶ 𝑑 cot−1(𝑥)= − 1 1+𝑥2
  • 7.
    𝒔𝒆𝒄−𝟏 𝒙 ▶ Domain: −∞, −1 𝑈 [1, ∞) 2 2 ▶ Range : [0,π ) U (𝜋 ,π] ▶ Neither even nor odd function ▶ sec−1(−𝑥) = π - sec−1 𝑥 𝑑 𝑥 ▶ 𝑑 sec−1 𝑥 = 1 𝑥 𝑥2−1
  • 8.
    𝒄𝒐𝒔𝒆𝒄−𝟏 𝒙 ▶ Domain: −∞. −1 U [1, ∞) ▶ Range : [ � � 2 − , 0) U (0, 2 𝜋 ] ▶ Odd function ▶ csc−1 −𝑥= −csc−1 𝑥 𝑑 𝑥 ▶ 𝑑 csc−1 𝑥 = - 1 𝑥 𝑥2−1
  • 9.
    NOTE: ▶ None ofthe inverse trigo functions are periodic functions. ▶ 𝑓(f −1 𝑥 ) = x for every x belonging to its domain ▶ Sign of an inequality changes if we apply a decreasing function on both sides of an inequality. Sign of inequality remains same if it's an increasing function. ▶ 𝑦 = sin−1(sin 𝑥) = nπ + (−1)𝑛 𝑥 ▶ 𝑦 = cs𝑐−1 (csc 𝑥) = nπ + (−1)𝑛 𝑥
  • 10.
    ▶ 𝑦 =cos−1 (cos 𝑥) = 2nπ ± 𝑥 ▶ 𝑦 = sec−1 (sec 𝑥) = 2nπ ± 𝑥
  • 11.
    ▶ 𝑦 =tan−1(tan 𝑥) = 𝑛𝜋 + 𝑥 ▶ 𝑦 = cot−1(𝑐𝑜𝑡 𝑥) = 𝑛𝜋 + 𝑥
  • 12.
    PROPERTIES: ▶ sin−1 𝑥+ cos−1 𝑥 = � � ▶ tan−1 𝑥 + co𝑡−1 𝑥 = 2 � � 2 ▶ s𝑒𝑐−1 𝑥 + cosec−1 𝑥 = � � 2 � � ▶ sin−1 1 = csc−1 𝑥 � � ▶ cos−1 1 = sec−1 𝑥 ▶ tan −1 1 𝑥 cot−1 𝑥 = ቊ −𝜋 + cot−1 𝑥 ; 𝑥 > 0 ; 𝑥 < 0 � � � � ▶ tan−1 𝑥 + tan−1 1 = ቐ2 − 𝜋 2 ; 𝑥 > 0 ; 𝑥 <
  • 13.
    ▶ tan−1 𝑥+ tan−1 𝑦 = tan− 1 𝑥+ 𝑦 1−𝑥 𝑦 ; 𝑥 > 0, 𝑦 > 0, 𝑥𝑦 < 1 𝜋 + tan−1 𝑥+ 𝑦 1−𝑥 𝑦 ; 𝑥 > 0, 𝑦 > 0, 𝑥𝑦 > 1 ▶ tan−1 𝑥 − tan−1 𝑦 = tan−1 𝑥− 𝑦 1+𝑥 𝑦 ; 𝑥 > 0, 𝑦 > 0
  • 14.