Inverse Circular Function Inverse Trigonometric Functions

Inverse of trigonometric ratios  

We know that y = sin x means y is the value of sine of angle x if we consider domain and co-domain both as set R of a real numbers. Sine ratio as seen from the fig. is many-one into function.

   sinx-curve 

But it is clear that if we restrict the domain to [-π/2,π/2] and range to 

[–1, 1], then. Y = sin x is one-one onto and hence it is invertible. 

So, y = sin x                          x ∈ [-π/2,π/2], y ∈ [–1, 1] 

⇒ x = sin–1 y                          y ∈ [–1, 1]. x ∈ [-π/2,π/2]     

This value of x is called the principle value, i.e. belonging to [-π/2,π/2] and [-π/2,π/2] range and it is called principal value range. 

Note: The smallest numerical angle is called principal value. 

In general the inverse circular functions with their domain and range can be tabulated as: 

inverse-circular-functions

Note: Principal value range of all Inverse Circular function is very important as the function is defined only in this range. 

Pause: Odd function are defined as f(–x) = –f(x) and even function as f(–x) = f(x). 

The inverse circular functions are defined as below:- 

1. sin–1 (–x) = –sin–1 x,               –1 < x < 1 Odd function 

2. cos–1 (–x) = π –cos–1 x,          –1 < x < 1 Neither odd nor even 

3. tan–11 (–x) = –tan–1 x,             x ∈ R Odd function 

4. cot–1 (–x) = π – cot–1 x,           x ∈ R Neither odd nor even 

5. cosec–1 (–x) = –cosec–1 x,        x < –1 or x > 1 Odd function 

6. sec–1 (–x) = π –sec–1 x,           x < –1 or x > 1 Neither odd nor even

Let us see the proof of any one of the above. 

Proof 2: 

Let cos-1 (–x) = θ, then cos θ = –x 

or, – cos θ = x or cos (π – θ) = x 

or, π – θ = cos-1 x or cos-1 (–x) = π –cos-1 x 

Similarly we can prove other results. 

Caution: Instead of taking cos (π – θ) equal to – cos θ, we could have taken cos (π + θ). We opt for cos (π – θ) because (π – θ) lies in a principal value range i.e. 0 ≤ cos-1 x ≤ π. 

Illustration: 

Find cot-1 (–1) 

Solution: 

cot-1 (–1) = π – cot-1 (1) 

= π -π/4 [? cot-1 (–x) = π – cot-1 x] 

= 3π/4. 

Properties of Inverse Circular Function

1. Self adjusting Properties 

(a) sin (sin-1 x) = x, x ∈ [–1, 1] 

Let sin-1 x = θ, θ &isin (-π/2,π/2) ……… (1) 

Then x = sin θ 

From (1) putting the value of θ in (2), we get, ……… (2) 

sin (sin-1 x) = sin (θ) = x 

(b) sin-1 sin x = x, x ∈ [-π/2,π/2]. 

Let sin x = y, y ∈ [–1, 1] ……… (1) 

Then x = sin-1 y ……… (2) 

Putting the value of y in (2) from (1), we get, 

sin-1 sin x = x 

Illustration: 

Evaluate sin-1 sin 5, 5 is radian 

Solution: 

We know sin-1 sin x = x, x ∈ [-π/2,π/2] or x ∈ [–1.57, 1.57] 

We can write, 5 = 2 π + 5 – 2 π 

sin 5 = sin (2 π + (5 – 2π)) 

= sin (5 – 2 π), 5 – 2 π ∈ [–1.57, 1.57] 

 ∴ sin-1 sin 5 = sin-1 sin (5 – 2π) 

  = 5 – 2π 

2. Reciprocal Property 

sin-1 (1/x) = cosec-1 x, x ≤ –1, or x ≥ 1 

let cosec-1 x = θ ⇒ cosec θ, θ ∈ [-π/2,π/2] – {0} 

⇒ LHS = sin-1 (1/(cosec θ )) = θ 

⇒ sin-1 (sin θ) = θ which is true for above domain of θ 

 = cosec-1 x = RHS. 

Note: cot-1 (1/x) = tan-1 x holds only for x > 0. This is because of principal value range of Inverse function cot and tan are different. 

When x < 0, the principal value range is π/2 < cot-1 (1/x) < π, whereas -π/2 < tan-1x< 0. Hence equality will never hold in this case. 

cot-1 (-1/2) = tan-1 (–x) 

π – cot-1 (1/2) = –tan-1 x 

cot-1 (1/2) = π + tan-1 x 

Hence cot-1 (1/2) = {   tan-1 x,       x>0                              

 π+tan-1 x,    x< 0

Relation between Inverse Trigonometric Functions 

1. sin-1 x + cos-1 x = π/2    x ∈ [–1, 1] 

Let sin-1 x = θ, θ ∈ [-π/2,π/2] 

⇒ x = sin θ, 

⇒ x = cos (π/2-θ ),(π/2-θ ) ∈ [0, π] 

⇒ cos-1 x = π/2 – θ 

⇒ cos-1 x + sin-1 x = π/2   x ∈ [–1, 1]. 

right-angle-triangle 

We can express any inverse circular function in terms of the other by drawing a right-angled triangle. Thus, if we have to explain tan-1 x in terms of the other, then take the perpendicular of the triangle as x and base as 1. Now, hypotenuse is √(1+x2 ) 

  θ = tan-1 x = cot-1 1/x = sin-1 x/√(1+x2 )=cos-11/√(1+x2 )=cosec-1√(1+x2 )/x= sec-1√(1+x2 ) 

Illustration: 

Prove that sin cot-1 tan cos-1 x = x. 

Solution: 

Note:  In such problems we proceed with the last term of LHS. 

Here, LHS = sin cot-1 tan cos-1 x 

Let cos-1 x = θ, x = cos θ 

right-angle-triangle2 

∴ tan θ = √(1+x2 )/x 

tan cos-1 x= √(1+x2 )/x 

(1) Reduces to, sin cot-1 (√(1+x2 )/x) …… (2) 

Again let cot-1 √(1+x2 )/x = φ 

√(1+x2 )/x = cot φ 

 ∴ sin φ = x 

right-angle-triangle3 

Hence, (2) becomes 

sin cot-1 √(1+x2 )/x = x = R.H.S. 

Illustration: 

Find the value of cos (2 cos-1 x + sin-1 x) at x = 1/5, Where 0 ≤ cos-1 x ≤ π, –&pi/2 ≤ sin-1 x ≤ π/2 

Solution: 

cos (2 cos-1 x + sin-1 x) = cos (cos-1 x + (cos-1 x + sin-1 x)) 

= cos (cos-1 x+ π/2) 

= –sin cos-1 x …… (1) 

Let cos-1 x = θ 

⇒ x = cos θ 

sin θ = √(1-x2 ) 

cos-1 x = θ – sin-1 √(1-x2 ) 

Now, (1) becomes 

(2 cos-1 x + sin-1 x) = –sin cos-1 x = –sin-1 √(1-x2 ) 

 = –√(1-x2 ) as –1 ≤ √(1-x2 ) ≤ 1 

= – √(1-1/25) at x = 1/5 

= – 2/5 √6.

theorem

Since x ≥ 0, y ≥ 0 so RHS ∈ (0, π/2) but if x and y are such that 

LHS > π/2 then above equation is not valid. So above equation is valid only when 

sin-1 x + sin-1 y ≤ π/2 

⇒ sin-1 x ≤ π/2 – sin-1 y 

 ⇒ sin-1 x ≤ cos–1 y 

taking sine ( ) both sides 

sin (sin-1 x) < sin (cos–1 y) 

(∴sine is an increasing function in [-π/2,π/2] ) 

⇒ x < √(1-y2 ) 

⇒ x2 + y2 < 1 

So above relation is valid only when x2 + y2 ≤ 1 

Now, if x2 + y2 > 1 then 

Let us assume LHS = θ and RHS = φ 

Then sin θ = sin φ, which is true only when θ = φ or θ = π – φ 

⇒ sin-1 x + sin-1 y = π – sin-1 (x√(1-y2 )+y√(1-x2 )). 

Illustration: 

If sin-1 x + (sin-1 y + sin-1 z) = π/2 

Solution: 

Find out x2 + y2 + z2 + 2xyz. 

Let sin-1 x = A, sin-1 y = B and sin-1 z = C 

⇒ A + B + C = π/2 

A + B = π/2 – C. 

Let’s operate cos on both sides. 

Note: That we have chosen to operate cos because it becomes easier to solve. 

cos (A + B0 = cos (π/2 – C) 

cos A cos B – sin A sin B = sin C 

√((1-x2 ) ) √((1-y2 ) )-xy=z 

xy + z = √((1-x2 )(1-y2 ) ) 

Squaring both sides 

x2 y2 + z2 + 2xyz = 1 + x2 y2 – x2 – y2 

x2 + y2 + z2 = 2xyz = 1

solution 

Hence proved. 

Theorem: tan-1 x + tan-1 y = tan-1 {(x + y)/(1 – xy)}           if xy < 1 

= π –tan-1 (x + y)/(xy – 1) if xy > 1 

= π/2 if xy = 1. 

Proof: 

Let tan-1 x = α so that tan α = x 

and tan-1 y = β so that tan β = y 

Also let tan-1 ((x+y)/(1-xy)) = γ so that tan γ = (x+y)/(1-xy) 

We have then to prove that 

α + β = γ 

Now tan (α + β) = (tanα +tanβ )/(1-tanα tanβ )=(x+y)/(1-xy) = tan γ 

So, the relation is proved 

tan-1 x + tan-1 y = tan-1 ((x+y)/(1-xy)) …… (1)