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# Properties of Inverse Trigonometric Functions

In this section, we investigate some properties of inverse trigonometric functions. The properties to be discussed are valid within the principal value branches of the corresponding inverse trigonometric functions and where they are defined.

Properties of Inverse Trigonometric Functions

In this section, we investigate some properties of inverse trigonometric functions. The properties to be discussed are valid within the principal value branches of the corresponding inverse trigonometric functions and where they are defined.

### Proof

All the above results follow from the definition of the respective inverse functions.

For instance, (i) let sinθ = x ;   θ [- π/2 , π/2 ]

Now, sin θ = x gives θ = sin-1 x , by definition of inverse sine function.

Thus, sin-1 (sinθ ) = θ .

### Property-II

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

(ii)  cos (cos-1  x) = x , if   x [-1, 1]

(iii) tan (tan-1  x) = x ,   if   x R

(iv)  cosec (cosec-1  x) = x ,   if   x R \ (-1, 1)

(v)  sec(sec-1  x) = x ,   if  x R \ (-1, 1)

(vi)  cot (cot-1  x) = x , if  x R

### Proof

(i) For x [-1, 1] , sin-1 x is well defined.

Let sin-1 x = θ . Then, by definition θ [- π/2 , π/2 ] and sin θ = x

Thus,   sinθ = implies   sin (sin-1  x) = x .

Similarly, other results are proved.

Note

For any trigonometric function y = f (x), we have f ( f -1 (x)) = x for all x in the range of f . This  follows  from  the  definition  of   f -1 (x) .  When  we  have,    f (g-1 (x)),  where    g -1 (x) = sin-1 x or cos-1 x, it will usually be necessary to draw a triangle defined by the inverse trigonometric function to solve the problem. For instance, to find  cot (sin-1 x) , we have to draw a triangle using sin-1 x. However, we have to be a little more careful with expression of the form f -1 ( f (x)).

(ii) Evaluation of f -1[ f (x)] , where f is any one of the six trigonometric functions.

(a) If x is in the restricted domain (principal domain) of  f , then f -1[ f (x)] = x .

(b) If x is not in the restricted domain of f , then find x1 within the restricted domain of f such that f (x) = f (x1) . Now, f -1[ f (x)] = x1. For instance,

### Property-III (Reciprocal inverse identities)

(i) sin-1 ( 1/x ) = cosec x ,  if x R \ (-1, 1) .

(ii) cos-1 ( 1/x ). = sec x , if x R \ (-1, 1) .

(iii)

### Proof

(i) If x R \ (-1, 1) , then 1/x [-1, 1] and x ≠ 0 . Thus, sin-1 ( 1/x ) is well defined.

Let sin-1 ( 1/x ) = θ . Then, by definition θ [- π/2 , π/2 ] \ {0} and sinθ = 1/x .

Thus, cosecθ = x , which in turn gives θ = cosec-1 x .

Now, sin-1 ( 1/x ) = θ = cosec-1 x . Thus, sin-1 ( 1/x ) = cosec-1 x , x R \ (-1, 1).

Similarly, other results are proved.

### Property-IV (Reflection identities)

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

(ii) tan-1 (-x) = - tan-1 x , if x R .

(iii) cosec-1 (-x) = - cosec-1 x , if |x| ≥ 1 or x R \ (-1, 1).

(iv) cos-1 (-x) = π - cos-1 x , if x [-1, 1] .

(v) sec-1 (-x) = π - sec-1 x , if |x| ≥ 1 or x R \ (-1, 1) .

(vi) cot-1 (-x) = π - cot-1 x , if x R .

Proof

(i) If x [-1, 1] , then -x [-1, 1] . Thus, sin-1 (-x) is well defined

Let sin-1 (-x) = θ . Then θ [- π/2 , π/2 ] and sinθ = -x .

Now, sinθ = -x gives x = -sinθ = sin(-θ )

From x = sin(-θ ) , we must have sin-1 x = -θ , which in turn gives θ = -sin-1 x .

Hence, sin-1 (-x) = -sin-1 x .

(iv) If x [-1, 1] , then -x [-1, 1] . Thus, cos-1 (-x) is well defined

Let cos-1 (-x) = θ . Then θ [0, π ] and cos θ = -x .

Now, cosθ = -x implies x = -cosθ = cos (π -θ ) .

Thus, π -θ = cos-1 x, which gives θ = π - cos-1 x .

Hence, cos-1 (-x) = π - cos-1 x .

Similarly, other results are proved.

Note

(i) The inverse function of an one-to-one and odd function is also an odd function. For instance, y = sin-1 x is an odd function, since sine function is both one-to-one and odd in the restricted domain [- π/2 , π/2 ] .

(ii) Is the inverse function of an even function also even? It turns out that the question does not make sense, because an even function cannot be one-to-one if it is defined anywhere other than 0. Observe that cos-1 x and sec-1 x are not even functions.

### Property-V ( Cofunction inverse identities )

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

(ii) tan-1 x + cot-1 x = π/2 , x R.

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

### Proof

Similarly, (iii) can be proved.

Property-VI

Proof

Property-VII

Proof

(i) By taking y x = in Property-VI (v) , we get the desired result

Similarly, other results are proved.

Property-VIII

Proof

### Proof

(i) Let  sin-1  x = θ . Then, sin θ  x . Since 0 ≤ x ≤1 , we get   0 ≤ θ π/2 .

(ii) Suppose that -1 ≤ x ≤ 0 and sin-1 x = θ . Then - π /2 ≤ θ < 0

Similarly, other results are proved.

### Property-X

(i) 3sin-1 x = sin-1 (3x - 4x3 ) ,  x [- 1/2 , 1/2 ] .

(ii) 3cos-1 x = cos-1 (4x3 - 3x) , x [ 1/2 , 1]

Proof

(i) Let x = sin θ . Thus, θ = sin-1x .

Now, 3x - 4x3 = 3sinθ - 4 sin3 θ = sin 3θ

Thus, sin-1 (3x - 4x3 ) = 3θ = 3sin-1x .

The other result is proved in a similar way.

### Remark

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12th Mathematics : UNIT 4 : Inverse Trigonometric Functions : Properties of Inverse Trigonometric Functions |