Inverse Trigonometric Identities

Inverse trigonometric identities are mathematical relationships that involve the inverse trigonometric functions: arcsine (sin^-1), arccosine (cos^-1), arctangent (tan^-1), arccosecant (cosec^-1), arcsecant (sec^-1), and arccotangent (cot^-1). These functions determine the angles (or arcs) corresponding to given trigonometric ratios.

Here is the list of all the inverse trigonometric identities and their types. 

List

Basic Identities

A trigonometric function and its inverse cancel each other out, returning the original input within a specific range. These identities explain how trigonometric functions interact with their inverses.

• sin^-1(sin x) = x, for x ∈ ${\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2}\right]}$
• cos^-1(cos x) = x, for x ∈ [0, π]
• tan^-1(tan x) = x, for x ∈ ${\left( -\dfrac{\pi }{2},\dfrac{\pi }{2}\right)}$

Similarly, for inverse functions applied to trigonometric values:

• sin(sin^-1(x)) = x, for x ∈ [-1, 1]
• cos⁡(cos⁡^-1(x)) = x, for x ∈ [-1, 1]
• tan⁡(tan⁡^-1(x)) = x, for x ∈ ℝ
• cosec(cosec^-1 x) = x, for -1 ≤ x ≤ ∞ or -∞ < x ≤ 1
• sec(sec^-1 x) = x, for 1 ≤ x ≤ ∞ or -∞ < x ≤ 1
• cot(cot^-1 x) = x, for -∞ < x < ∞

Negative Identities

Negative identities describe how inverse trigonometric functions behave when their input is negative. Some functions return a simple negation, while others involve a transformation using π:

For sine, tangent, and cosecant functions, a negative input results in the negation of the function’s value:

• sin^-1(-x) = -sin^-1 x, for x ∈ [-1, 1]
• tan^-1(-x) = -tan^-1 x, for x ∈ ℝ
• cosec^-1(-x) = -cosec^-1 x, for |x| ≥ 1

For cosine, secant, and cotangent functions, the transformation involves subtracting the inverse function’s value from π, ensuring the result remains within the principal range:

• cos^-1(-x) = π – cos^-1 x, for x ∈ [-1, 1]
• sec^-1(-x) = π – sec^-1x, for |x| ≥ 1
• cot^-1(-x) = π – cot^-1x, for x ∈ ℝ

Reciprocal Identities

Reciprocal identities establish direct relationships between inverse trigonometric functions and their reciprocal counterparts:

• ${\sin ^{-1}\left( \dfrac{1}{x}\right) =\text{cosec}\, ^{-1}x}$, for x ≥ 1 or x ≤ -1
• ${\cos ^{-1}\left( \dfrac{1}{x}\right) =\sec ^{-1}x}$, for x ≥ 1 or x ≤ -1
• ${\tan ^{-1}\left( \dfrac{1}{x}\right) =\cot ^{-1}x}$, for x > 0

Complementary Identities

The sum of two complementary inverse trigonometric functions always results in a constant value of ${\dfrac{\pi }{2}}$:

• sin^-1 x + cos^-1 x = ${\dfrac{\pi }{2}}$, for x ∈ [-1,1]
• tan^-1 x + cot^-1 x = ${\dfrac{\pi }{2}}$, for x ∈ ℝ
• cosec^-1 x + sec^-1 x = ${\dfrac{\pi }{2}}$, for |x| ≥ 1

Sum and Difference Identities

These identities are used where two inverse tangent functions are added or subtracted:

• tan^-1 x + tan^-1 y = ${\tan ^{-1}\left( \dfrac{x+y}{1-xy}\right)}$, for xy < 1
• tan^-1 x – tan^-1 y = ${\tan ^{-1}\left( \dfrac{x-y}{1+xy}\right)}$, for xy > -1

Double-Angle Identities

These relate inverse trigonometric functions to double angles:

• 2 tan^-1 x = ${\sin ^{-1}\left( \dfrac{2x}{1+x^{2}}\right)}$, for |x| ≤ 1
• 2 tan^-1 x = ${\cos ^{-1}\left( \dfrac{1-x^{2}}{1+x^{2}}\right)}$, for x ≥ 0
• 2 tan^-1 x = ${\tan ^{-1}\left( \dfrac{2x}{1-x^{2}}\right)}$, for -1 < x < 1

Here is a summary of all the identities.

We will now use the above identities to solve or simplify some expressions involving inverse trigonometric functions.

Proofs 

Now, let us prove a few inverse trigonometric identities.

sin^-1(sin x) = x

Let y = sin^-1(sinx)

⇒ sin y = sin x

Since the range of the inverse sine function is ${\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2}\right]}$, y must lie in this interval.

Thus, ${-\dfrac{\pi }{2}\leq y\leq \dfrac{\pi }{2}}$

⇒ y = x

⇒ sin^-1(sinx) = x

Thus, sin^-1(sin x) = x, for x ∈ ${\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2}\right]}$

sin^-1(-x) = -sin^-1 x

Let sin^-1(-x) = y

⇒ -x = sin y

⇒ x = -sin y

⇒ x = sin(-y)

⇒ sin^-1 x = sin^-1(sin(-y))

⇒ sin^-1 x = -y

⇒ -sin^-1 x = y

⇒ -sin^-1 x = sin^-1(-x)

Thus, sin^-1(-x) = -sin^-1 x, for x ∈ [-1, 1]

cos^-1(-x) = π – cos^-1 x

Let cos^-1(-x) = y

⇒ -x = cos y 

⇒ x = -cos y

Since cos(π – q) = -cos q

⇒ x = cos(π – y)

⇒ cos^-1 x = π – y

⇒ cos^-1 x = π – cos^-1(-x)

⇒ cos^-1(-x) = π – cos^-1 x

Thus, cos^-1(-x) = π – cos^-1 x, for x ∈ [-1, 1]

${\sin ^{-1}\left( \dfrac{1}{x}\right) =\text{cosec}\, ^{-1}x}$

Let cosec^-1 x = y

⇒ x = cosec y

⇒ ${\dfrac{1}{x}=\sin y}$

⇒ ${\sin ^{-1}\left( \dfrac{1}{x}\right)}$ = y

⇒ ${\sin ^{-1}\left( \dfrac{1}{x}\right)}$ = ${\text{cosec}\, ^{-1}x}$

Thus, ${\sin ^{-1}\left( \dfrac{1}{x}\right) =\text{cosec}\, ^{-1}x}$, for x ≥ 1 or x ≤ -1

sin^-1 x + cos^-1 x = ${\dfrac{\pi }{2}}$

Let sin^-1 x = y 

⇒ x = sin y

⇒ x = ${\cos \left( \dfrac{\pi }{2}-y\right)}$

⇒ cos^-1 x = ${\cos ^{-1}\left( \cos \left( \dfrac{\pi }{2}-y\right) \right)}$

⇒ cos^-1 x = ${\dfrac{\pi }{2}-y}$

⇒ cos^-1 x = ${\dfrac{\pi }{2}}$ – sin^-1 x

⇒ sin^-1 x + cos^-1 x = ${\dfrac{\pi }{2}}$

Thus, sin^-1 x + cos^-1 x = ${\dfrac{\pi }{2}}$, for x ∈ [-1, 1]

tan^-1 x + tan^-1 y = ${\tan ^{-1}\left( \dfrac{x+y}{1-xy}\right)}$

Let tan^-1 x = A and tan^-1 y = B

⇒ x = tan A and y = tan B

Since tan(A + B) = ${\dfrac{\tan A+\tan B}{1-\tan A\tan B}}$

⇒ tan(A + B) = ${\dfrac{x+y}{1-xy}}$

⇒ A + B = ${\tan ^{-1}\left( \dfrac{x+y}{1-xy}\right)}$

⇒ tan^-1 x + tan^-1 y = ${\tan ^{-1}\left( \dfrac{x+y}{1-xy}\right)}$

Thus, tan^-1 x + tan^-1 y = ${\tan ^{-1}\left( \dfrac{x+y}{1-xy}\right)}$, for xy < 1

2 tan^-1 x = ${\sin ^{-1}\left( \dfrac{2x}{1+x^{2}}\right)}$

Let tan^-1 x = y 

⇒ x = tan y

Now, R.H.S. = ${\sin ^{-1}\left( \dfrac{2x}{1+x^{2}}\right)}$

= ${\sin ^{-1}\left( \dfrac{2\tan y}{1+\tan ^{2}y}\right)}$

Since sin 2θ = ${\dfrac{2\tan \theta }{1+\tan ^{2}\theta }}$

= ${\sin ^{-1}\left( \sin 2y\right)}$

= 2y

= 2 tan^-1 x = L.H.S.

Thus, 2 tan^-1 x = ${\sin ^{-1}\left( \dfrac{2x}{1+x^{2}}\right)}$, for |x| ≤ 1

Solved Examples