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```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.

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:

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:

(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].

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

∴ 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

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.

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

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.

⇒ (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

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.

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).

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