Inverse Trigonometry Function

Definition and Formulas

Introduction: - Inverse of a function f, denoted by f -1, exists if f is one-one and onto. There are many functions which are not one-one, onto or both and hence we can not talk of their inverses. Trigonometric functions are not one-one and onto over their natural domains and ranges and hence their inverses do not exist. In this chapter, we shall study the restrictions on domains and ranges of trigonometric functions that ensure the existence of their inverses and observe their behavior through graphical representations.

 if f: X→Y such that f(x) = y is one-one and onto 

then a unique function g: Y→X such that g(y) = x, where x ∈ X and y ∈ Y.

Domain and Range: - Simply domain is the x values and Range is a subset of co-domain or values of y that x takes.

S.NO.	Inverse trigonometry function	Domain	Range
1.	sin-1x	[-1,1]	[-π/2,  π/2]
2.	cos-1x	[-1,1]	[0 , π]
3.	tan-1x	R	(-π/2,  π/2)
4.	cot-1x	R	(0, π)
5.	sec-1x	R-(-1,1)	[0 , π] - {π/2}
6.	cosec-1x	R-(-1,1)	[-π/2,  π/2]-{0}

   

 Graphs of Inverse trigonometry function: -

1.  sin-1x

    

2. cos-1x

       

3. tan-1x

 4. cot-1x 

 5. sec-1x and cosec-1x

          

NOTE:-

1. sin-1x is not equal to (sin x)-1.

Proof: - (sin x)-1 = 1/sinx = cosec x

But sin-1x is an inverse function of sin x.

Principal Value:- The value of an inverse trigonometric function which lies in the range of inverse trigonometry function called the principal value of the inverse functions. 

Properties of Inverse trigonometry function:-

1.(i) sin-1(1/x) = cosec-1x , x³1 or x∈ – 1    

 (ii)cos-1 (1/x) = sec-1x , x³1 or x ∈ – 1

(iii)tan-1 (1/x) = cot-1x , x>0  

2.(i) sin-1 (-x) = - sin-1x

(ii)tan-1 (-x) = -tan-1x

(iii)cosec-1 (-x) = -cosec-1x

3.(i) cos-1 (-x) = π - cos-1x , x ∈ [-1,1] 

(ii)sec-1 (-x) = π - sec-1x , |x| ≥ 1   

(iii)cot-1 (-x) = π - cot-1x , x ∈ R

4.(i) sin-1x + cos-1x = π/2 , x ∈ [-1,1] 

(ii) tan-1x + cot-1x = π/2 , x ∈ R

(iii)sec-1x + cosec-1x = π/2 , |x| ≥ 1

5.(i) tan-1x + tan-1y = tan-1 [(x + y) / 1-xy] , xy<1

(ii) tan-1x - tan-1y = tan-1 [(x - y) / 1+xy] , xy>-1

6.(i) cot-1x + cot-1y = cot-1[(xy-1) / x+y]

(ii) cot-1x - cot-1y = cot-1[(xy+1) / y-x]

7.(i) 2tan-1x = sin-1 (2x/1+x2) , |x|≤1

(ii) 2tan-1x = cos-1 [(1-x2)/1+x2] , x≥0

(iii) 2tan-1x = tan-1 (2x/1-x2) , -1< x < 1

8.(i) sin(sin-1x) = x , x ∈ [-1,1] and

sin-1(sin x) = x , x ∈ [-π/2 ,  π/2]

(ii) cos(cos-1x) = x , x ∈ [-1,1] and

cos-1(cos x) = x , x ∈ [0 , π]

(iii) tan(tan-1x) = x , x ∈ R and

tan-1(tan x) = x , x ∈ (-π/2 ,  π/2)

(iv) cot(cot-1x) = x , x ∈ R and

cot-1(cot x) = x , x ∈ (0 , π)

(v) sec(sec-1x) = x , x ∈ R- (-1,1) and

sec-1(sec x) = x , x ∈ [0 , π] - {π/2}

(vi) cosec(cosec-1x) = x , x ∈ R- (-1,1) and

cosec-1(cosec x) = x ,x ∈  [-π/2 ,  π/2]-{0}  

9.(i) sin-1x + sin-1y = sin-1[x√1-y2 + y√1-x2]

(ii) sin-1x - sin-1 y = sin-1[x√1-y2 - y√1-x2]

10.(i) cos-1x + cos-1y = cos-1[xy- √1-x2 √1-y2]

(ii) cos-1x - cos-1y = cos-1[xy+ √1-x2 √1-y2]

Properties of Inverse trigonometry function(with proof):-

1. (i) sin-1 (1/x) = cosec-1x , x³1 or x∈– 1

Proof:- Let y = cosec-1x  ……(1)

⇒ cosec y = x

⇒   x = cosec y

⇒   x = 1/sin y

(Here we can simply do cross multiply)

⇒   sin y = 1/x

⇒   y = sin-1 (1/x)

From Eq. 1:- Put the value of y

⇒ sin-1 (1/x) = cosec-1x (Hence Proved….)

(ii)cos-1 (1/x) = sec-1x , x³1 or x∈– 1

Proof:- Let y = sec-1x ……(1)

⇒ sec y = x

⇒   x = sec y

⇒   x = 1/cos y

(Here we can simply do cross multiply)

⇒   cos y = 1/x

⇒   y = cos-1 (1/x)

From Eq. 1:- Put the value of y

⇒ cos-1 (1/x) = sec-1x (Hence Proved….)

(iii) tan-1 (1/x) = cot-1x , x>0  

Proof:- Let y = cot-1x ………(1)

⇒ cot y = x

⇒   x = cot y

⇒   x = 1/tan y

(Here we can simply do cross multiply)

⇒   tan y = 1/x

⇒   y = tan-1 (1/x)

From Eq. 1:- Put the value of y

⇒ tan-1 (1/x) = cot-1x (Hence Proved….)

2. (i) sin-1 (-x) = - sin-1x , x ∈ [-1,1] 

Proof:- Let y =  sin-1 (-x) ………(1)

⇒ sin y = -x

⇒ x = -sin y       

We know -sin y = sin (-y)

⇒ x = sin (-y)

⇒ sin-1x = -y

⇒ y = - sin-1x

From Eq. 1:- Put the value of y

⇒ sin-1 (-x) = - sin-1x  (Hence Proved….)

(ii) tan-1 (-x) = -tan-1x , x ∈ R

Proof:- Let y =  tan-1 (-x) ……(1)

⇒ tan y = -x

⇒ x = -tan y

We know -tan y = tan (-y)

⇒ x = tan (-y)

⇒ tan-1x = -y

⇒ y = - tan-1x

From Eq. 1:- Put the value of y

⇒ tan-1 (-x) = - tan-1x  (Hence Proved….)

(iii) cosec-1 (-x) = -cosec-1x , |x| ≥ 1  

Proof:- Let y =  cosec-1 (-x)  ………(1)

⇒ cosec y = -x

⇒ x = -cosec y

We know -cosec y = cosec (-y)

⇒ x = cosec (-y)

⇒ cosec-1x = -y

⇒ y = - cosec-1x

From Eq. 1:- Put the value of y

⇒ cosec-1 (-x) = - cosec-1x (Hence Proved….)

3. (i)cos-1 (-x) = π - cos-1x , x ∈ [-1,1]

Proof:- Let y = cos-1 (-x) ………(1)

cos y = -x

x = -cos y    

We know cos(π-y) = -cos y

x = cos (π-y)

cos-1x = π-y

y = π-cos-1x

From Eq. 1:- Put the value of y

cos-1 (-x) = π-cos-1x (Hence Proved….)

(ii)sec-1 (-x) = π - sec-1x , |x| ≥ 1   

Proof:- Let y = sec-1 (-x) …………(1)

sec y = -x

x = -sec y

We know sec(π-y) = -sec y

x = sec (π-y)

sec-1x = π-y

y = π-sec-1x

From Eq. 1:- Put the value of y

sec-1 (-x) = π-sec-1x (Hence Proved….)

(iii)cot-1 (-x) = π - cot-1x , x ∈ R

Proof:- Let y = cot-1 (-x) ………(1)

cot y = -x

x = -cot y         

We know cot(π-y) = -cot y

x = cot (π-y)

cot-1x = π-y

y = π-cot-1x

From Eq. 1:- Put the value of y

cot-1 (-x) = π-cot-1x (Hence Proved….)

4.(i) sin-1x + cos-1x = π/2 , x ∈ [-1,1] 

Proof:- Let y = sin-1x…………………………….(1)

⇒ x = sin y

⇒ x = cos(π/2 - y)  

We know cos (π/2 - y) = sin y

⇒ cos-1x = π/2 – y

From Eq. 1:- Put the value of y

⇒ cos-1x = π/2 - sin-1x

⇒ sin-1x + cos-1x = π/2 (Hence Proved….)

(ii) tan-1x + cot-1x = π/2 , x ∈ R

Proof:- Let y = tan-1x………(1)

⇒ x = tan y

⇒ x = cot (π/2 - y)  

We know cot (π/2 - y) = tan y

⇒ cot-1x = π/2 – y

From Eq. 1:- Put the value of y

⇒ cot-1x = π/2 - tan-1x

⇒ tan-1x + cot-1x = π/2 (Hence Proved….)

(iii) sec-1x + cosec-1x = π/2 , |x| ≥ 1

Proof:- Let y = sec-1x……………(1)

⇒ x = sec y

⇒ x = cosec (π/2 - y)       

We know cosec (π/2 - y) = sec y

⇒ cosec-1x = π/2 – y

From Eq. 1:- Put the value of y

⇒ cosec-1x = π/2 - sec-1x

⇒ sec-1x + cosec-1x = π/2 (Hence Proved….)

5.(i) tan-1x + tan-1y = tan-1 [(x + y) / 1-xy] , xy<1

Proof:- Let tan-1x = m and tan-1y = n………(1)

⇒ x = tan m and y =tan n………(2)

⇒ tan (m + n) = [(tan m + tan n) / 1-(tan m*tan n)]

From Eq. 2:- Put the values

⇒ tan (m + n) = [(x + y) / 1-xy]

⇒ m + n = tan-1[(x + y) / 1-xy]

 From Eq. 1:- Put the values

⇒ tan-1x + tan-1y = tan-1[(x + y) / 1-xy] (Hence Proved….)

(ii) tan-1x - tan-1y = tan-1 [(x - y) / 1+xy] , xy>-1

Proof:- Let tan-1x = m and tan-1y = n…….(1)

⇒ x = tan m and y =tan n…………. (2)

⇒ tan (m - n) = [(tan m - tan n) / 1+(tan m*tan n)]

From Eq. 2:- Put the values

⇒ tan (m - n) = [(x - y) / 1+xy]

⇒ m - n = tan-1[(x - y) / 1+xy]

From Eq. 1:- Put the values

⇒ tan-1x - tan-1y = tan-1[(x - y) / 1+xy] (Hence Proved….)

6.(i) cot-1x + cot-1y = cot-1[(xy-1) / x+y]

Proof:- Let cot-1x = m and cot-1y = n……….(1)

⇒ x = cot m and y = cot n………….(2)

We know cot (m + n) = [(cot m*cot n -1) / cot m + cot n]

From Eq. 2:- Put the values

⇒ cot (m + n) = [(xy-1) / x + y]

⇒ m + n = cot-1[(xy-1) / x + y]

From Eq. 1:- Put the values

⇒ cot-1x + cot-1y = cot-1[(xy-1) / x + y] (Hence Proved….)

(ii) cot-1x - cot-1y = cot-1[(xy+1) / y-x]

Proof:- Let cot-1x = m and cot-1y = n…….(1)

⇒ x = cot m and y = cot n……….(2)

We know cot (m - n) = [(cot m*cot n +1) / cot n - cot m]

From Eq. 2:- Put the values

⇒ cot (m - n) = [(xy+1) / y - x]

⇒ m - n = cot-1[(xy+1) / y - x]

From Eq. 1:- Put the values

⇒ cot-1x - cot-1y = cot-1[(xy+1) / y - x] (Hence Proved….)

7.(i) 2tan-1x = sin-1 (2x/1+x2) , |x|≤1

Proof:- Let y = tan-1x ….(1)      

⇒ x = tan y ….(2)

We know sin 2y = 2tan y / 1+tan2y

From Eq. 2:- Put the value

⇒ sin 2y = 2x / 1+x2

⇒ 2y = sin-1 (2x / 1+x2)

From Eq. 1:- Put the value of y

⇒ 2 tan-1x = sin-1 (2x / 1+x2) (Hence Proved….)

(ii) 2tan-1x = cos-1 [(1-x2)/1+x2] , x≥0

Proof:- Let y = tan-1x …….(1)   

⇒ x = tan y …………(2)

We know cos 2y = 1-tan2y / 1+tan2y

From Eq. 2:- Put the value

⇒ cos 2y = 1-x2 / 1+x2

⇒ 2y = cos-1 (1-x2 / 1+x2)

From Eq. 1:- Put the value of y

⇒ 2 tan-1x = cos-1 (1-x2 / 1+x2) (Hence Proved….)

(iii) 2tan-1x = tan-1 (2x/1-x2) , -1< x < 1

Proof:- Let y = tan-1x …….(1) 

⇒ x = tan y …….(2)

We know tan 2y = 2tan y / 1-tan2y

From Eq. 2:- Put the value

⇒ tan 2y = 2x / 1-x2

⇒ 2y = tan-1 (2x / 1-x2)

From Eq. 1:- Put the value of y

⇒ 2 tan-1x = tan-1 (2x / 1-x2) (Hence Proved….)

8.(i) sin(sin-1x) = x , x ∈ [-1,1] and sin-1(sin x) = x , x ∈ [-π/2 ,  π/2]

Proof:- Let y = sin-1x….(1)

⇒ x = sin y………….(2)

From Eq. 1:- Put the values

⇒ x = sin(sin-1x)

⇒ sin(sin-1x) = x (Hence Proved….)

Eq. 1 is :- y = sin-1x

From Eq. 2:- Put the values

⇒ y = sin-1(sin y)

Replace y by x:-

⇒ sin-1(sin x) = x (Hence Proved….)

9.(i) sin-1x + sin-1y = sin-1[x√1-y2 + y√1-x2]

Proof:- Let sin-1x  =A and sin-1y = B ….(1)

⇒ x = sin A and y = sin B …….(2)

We know sin (A + B) = sin A*cos B + cos A*sinB

{{Hint:- sin2B + cos2B =1   

⇒ cos2B = 1-sin2B    

⇒ cos B = √1-sin2B 

Similarly, cos A = √1-sin2A}} 

From Eq. 2:- Put the values

⇒ sin (A + B) = [x√1-y2 + y√1-x2]

⇒ A + B = sin-1[x√1-y2 + y√1-x2]      

From Eq. 1:- Put the values     

⇒ sin-1x + sin-1y = sin-1[x√1-y2 + y√1-x2] (Hence Proved….)    

(ii) sin-1x - sin-1 y = sin-1[x√1-y2 - y√1-x2]

Proof:- Let sin-1x  =A and sin-1y = B …….(1)

⇒ x = sin A and y = sin B …….(2)

We know sin (A - B) = sin A*cos B - cos A*sinB

{{Hint:- sin2B + cos2B =1   

⇒ cos2B = 1-sin2B    

⇒ cos B = √1-sin2B 

Similarly, cos A = √1-sin2A}} 

From Eq. 2:- Put the values

⇒ sin (A - B) = [x√1-y2 - y√1-x2]

⇒ A - B = sin-1[x√1-y2 - y√1-x2]      

From Eq. 1:- Put the values     

⇒ sin-1x - sin-1y = sin-1[x√1-y2 - y√1-x2] (Hence Proved….)     

10.(i) cos-1x + cos-1y = cos-1[xy- √1-x2 √1-y2]

Proof:- Let cos-1x  =A and cos-1y = B …….(1)

⇒ x = cos A and y = cos B ……….(2)

We know cos (A + B) = cos A*cos B - sin A*sinB 

{{Hint:- sin2B + cos2B =1   

⇒ sin2B = 1-cos2B    

⇒ sin B = √1-cos2B 

Similarly, sin A = √1-cos2A}} 

From Eq. 2:- Put the values

⇒ cos (A + B) = [xy -√1-x2√1-y2]

⇒ A + B = cos-1[xy -√1-x2√1-y2]      

From Eq. 1:- Put the values            

⇒ cos-1x + cos-1y = cos-1[xy -√1-x2√1-y2] (Hence Proved….)

(ii) cos-1x - cos-1y = cos-1[xy+ √1-x2 √1-y2]

Proof:- Let cos-1x  =A and cos-1y = B ……….(1)

⇒ x = cos A and y = cos B …….(2)

We know cos (A - B) = cos A*cos B + sin A*sinB

{{Hint:- sin2B + cos2B =1   

⇒ sin2B = 1-cos2B    

⇒ sin B = √1-cos2B 

Similarly, sin A = √1-cos2A}} 

From Eq. 2:- Put the values

⇒ cos (A - B) = [xy +√1-x2√1-y2]

⇒ A - B = cos-1[xy +√1-x2√1-y2]      

From Eq. 1:- Put the values     

⇒ cos-1x - cos-1y = cos-1[xy +√1-x2√1-y2] (Hence Proved….)     

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