Inverse circular function (Inverse Trigonometric Functions) 


Inverse of trigonometric ratios exists. 

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 

          cos (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) 

Case I:     When x ≥ 0, y ≥ 0 and xy < 1, (x+y)/(1-xy) is positive 

               ∴ tan-1 [(x+y)/(1-xy)] will be positive angle 

Case II:   When x ≥ 0, y ≥ 0 and xy > 1, (x+y)/(1-xy) is negative 

              ∴ tan-1 ((x+y)/(1-xy)) will be negative angle. 

               Hence we add π to make it positive 

             ∴ tan-1 x + tan-1 y = π + tan-1 [(x+y)/(1-xy)]. 

Case III: When xy = 1, (x+y)/(1-xy) = ∞ then 

             ∴ tan-1 x + tan-1 y = tan-1 ∞ = π/2



Illustration: 

          Show that the equation 

    tan–1 {((x+1))/((x-1) )} + tan–1 {((x-1))/x} = tan–1 (–7) has no solution. 

           equation 

         ⇒ (x–2)2 = 0 ⇒ x = 2 

         Putting x = 2 in the equation, the equation becomes 

          tan–1 3 + tan–1 1/2 = tan–1 (–7) 

         Since x y = 3 . 1/2 > 1 

         So, tan–1 x + tan–1 y = π + tan–1 ((x+y)/(1-xy)) 

          tan–1 3 + tan–1 1/2 = 3 + tan–1 ((3+1/2)/(1-3/2)) 

          = π + tan–1 (–7) ≠ tan–1 (–7) = R.H.S. 

Hence given equation has no roots as x = 2 does not satisfy the given equation.

 

 

 

Illustration: 

        Solve the equation 

       tan-1 2x + tan-1 3x = nπ + π/4

Solution: 

         We know that 

       tanx 

So we have the following cases. 

Case I:     2x . 3x < 1 ⇒ x2 < 1/6 ⇒ -1/√6 < x < 1/√6 

For this 

      ⇒ L.H.S. = tan-1 {(2x+3x)/(1-6x^2 )} = nπ + π/4 

Taking tan of both the sides 

       5x/(1-6x2 ) = tan (nπ + π/4) = 1 

       ⇒ 6x2 + 5x – 1 = 0 

       ⇒ 6x2 + 6x – x – 1 = 0 

       ⇒ (x + 1) (6x – 1) = 0 

       ⇒ x = 1/6 or x = –1 

       Since –1 ∉ (-1/√6,1/√6) 

Hence x = 1/6 is a solution. 


Case II:        2x . 3x > 1 ⇒ x ∈ (-∞ ,-1/√6) ∪ (1/√6,∞ ) 

For this we have 

      LHS = π – tan-1 (5x/(6x2-1)) = nπ + π/4 

     ⇒ tan (π -tan-1 (5x/(6x2-1)) )=tan(nπ + π/4) 

     ⇒ –tan (tan-1 (5x/(6x2-1)) )=1 

     = 5x = 1 – 6x2 

     ⇒ 6x2+ 5x – 1 = 0 

     ⇒ (x + 1) (x – 1/6) = 0 

     ⇒ x = –1, x = 1/6, since 1/6 ∉ (-∞ ,-1/√6)∪(1/√6,∞ ). 

Hence x = –1 is another solution. 

    ⇒ x = –1, 1/6 are solutions of the given equation. 



Theorem:     2 tan-1 x = sin-1 2x/(1+x2 ), where |x| < 1 

                               = cos–1 (1-x2)/(1+x2 ), where x > 0 

                               = tan-1 2x/(1-x2 ), where |x| < 1. 

Proof:        Let tan-1 x = θ, then tan θ = x 

    sin 2 θ = (2 tan θ )/(1+tan2 θ )=2x/(1+x2 ) 

   ⇒ 2 θ = sin-1 (2x/(1+x2 )) 

   ⇒ 2 tan-1 x = sin-1 (2x/(1+x2 )) 

  Both sides are balanced for |x| < 1. 


Caution:     Restriction on x is |x|< 1. To verify it suppose the formulae holds good even for x > 1. 

       Then L.H.S. = 2 tan-1 x 

       = 2 tan-1 (a no. 1 > 1) [for x > 1] 

      > π/2 [∵ tan-1 (>1) > π/4], [for principal values] 

R.H.S. = sin–1 2x/(1+x2 ) which lies between -π/2 and π/2 for all values of x, [for principal values] 

      Thus, we see that for x > 1, 

      L.H.S. > π/2, whereas R.H.S. ≤ π/2. 

     Hence two cannot be equal. 

   solution2 

Both sides are balanced for |x| < 1. 

Note:     For |x| > 1 

     2 tan-1 x = π – sin–1 2x/(1+x2 ) 

     = π – cos–1 (1-x2)/(1+x2 ) 

     = π – tan-1 2x/(1-x2 )

Illustration: 

        Show that the function y = 2 tan-1 x + sin-1 2x/(1-x2 ) is a constant for x ≥ 1. Find the value of this constant. 

Solution: 

           We know, sin-1 2x/(1+x2 ) = 2 tan-1 x if |x| < 1 

        Here given x ≥ 1. So if we convert in above condition then we can get solution. 

        x ≥ 1 can be written as 1/x < 1. 

       So, for 1/x < 1 

        sin-1 2x/(1+x2 ) = sin-1 (2/x)/(1/x2 +1) = sin-1 2(1/x)/((1/x)2+1) = 2 tan-1(1/x) 

       Now, y = 2 tan-1 x + 2 tan-1 (1/x) 

        = 2 [tan-1 x + cot–1 x] 

        = π = const. 


Illustration: 

        Show that 2tan-1 1/2 + tan-1 1/7 = tan-1 31/17. 

Solution: 

        We have 2tan-1 1/2 = tan-1 (2-1/2)/(1-1/4) = tan-1 4/3 

        ⇒ 2tan-1 ((4/3+1/7)/(1-4/21)) = tan-1 31/17 



Illustration: 

        Solve 2tan-1(cosx) = tan-1(2cosecx). 

Solution: 

        = tan-1[2cot x cos ecx] = tan-1(2 cosec x) (given) 

        ⇒ cotx = 1 ⇒ x = nπ + π/4. 


Illustration: 

        Show that tan-1 (x/√(a2-x2 )) = sin-1 (x/a), a > 0. 

Solution: 

        Let sin-1 x/a = θ, then x = a sin θ, -π/2 < θ < π/2 as (x ≠ a) 

        ⇒ x/√(a2-x2 )=(a sin θ )/√(a2-a2 sin2 θ ) = tan θ ⇒ tan-1 (x/√(a2-x2 )) 

        = tan-1 (tanθ) = θ = sin-1(x/a). 



illustration

 

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