Properties of Inverse Trigonometric Functions

The inverse trigonometric functions are also called arcus functions or anti trigonometric functions. These are the inverse functions of the trigonometric functions with suitably restricted domains. Specifically, they are the inverse functions of the sine, cosine, tangent, cotangent, secant, and cosecant functions, and are used to obtain an angle from any of the angle’s trigonometric ratios. Inverse trigonometric functions are widely used in engineering, navigation, physics, and geometry.

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Properties of Trigonometric Inverse Functions

Here are the properties of the inverse trigonometric functions with proof.

Property 1

1. sin^-1 (1/x) = cosec^-1x , x ≥ 1 or x ≤ -1
2. cos^-1 (1/x) = sec^-1x , x ≥ 1 or x ≤ -1
3. tan^-1 (1/x) = cot^-1x , x > 0

Proof : sin^-1 (1/x) = cosec^-1x , x ≥ 1 or x ≤ -1,
Let  \(\sin^{-1}x =y\)
i.e. x = cosec y
\( \frac{1}{x}=\sin y \)
\( \sin^{-1}\frac{1}{x})=y \)
\(\sin^{-1}\frac{1}{x})=cosec^{-1}x\)
\(\sin^{-1}(\frac{1}{x})=cosec^{-1}x\)
Hence, \(\sin^{-1} \frac{1}{x}=cosec^{-1}x\) where, x ≥ 1 or x ≤ -1.

Browse more Topics Under Inverse Trigonometric Functions

• Basic concepts of Inverse Trigonometric Functions
• Properties of Inverse Trigonometric Functions

Property 2

1. sin^-1(-x) = – sin^-1(x),    x ∈ [-1,1]
2. tan^-1(-x) = -tan^-1(x),   x ∈ R
3. cosec^-1(-x) = -cosec^-1(x), |x| ≥ 1

Proof: sin^-1(-x) = -sin^-1(x),    x ∈ [-1,1]
Let,  \(\sin^{-1} \left ( -x \right )=y\)
Then \(-x=\sin y\)
\(x=-\sin y\)
\(x=\sin \left ( -y \right )\)
\(\sin^{-1}=\sin^{-1} \left ( \sin \left ( -y \right ) \right )\)
\(\sin^{-1}x=y\)
\(\sin^{-1} x=-\sin^{-1} \left ( -x \right )\)
Hence,\(\sin^{-1} \left ( -x \right )=-\sin^{-1}\) x ∈ [-1,1]

Property 3

1. cos^-1(-x) = π – cos^-1 x, x ∈ [-1,1]
2. sec^-1(-x) = π – sec^-1x, |x| ≥ 1
3. cot^-1(-x) = π – cot^-1x, x ∈ R

Proof : cos^-1(-x) = π – cos^-1 x, x ∈ [-1,1]
Let \(\cos^{-1} \left ( -x \right )=y\)
\( \cos y=-x \)   \( x = -\cos y\)
\( x = \cos \left (\pi -y \right )\)
Since,  \(cos{\pi -q} = -cos{q} \)
\(\cos^{-1} x=\pi -y\)
\(\cos^{-1}x=\pi – \cos^{-1} –x\)
Hence, \(\cos^{-1} -x = \pi – \cos^{-1} x\)

Property 4

1. sin^-1x + cos^-1x = π/2, x ∈ [-1,1]
2. tan^-1x + cot^-1x = π/2, x ∈ R
3. cosec^-1x + sec^-1x = π/2, |x| ≥ 1

Proof : sin^-1x + cos^-1x = π/2, x ∈ [-1,1]
Let \(\sin^{-1} x=y\) or \( x = \sin y = \cos(\frac{\pi }{2}-y)\)
\(\cos^{-1} x= \cos^{-1} \left ( \cos \left ( \frac{\pi }{2}-y \right ) \right )\)
\(\cos^{-1} x= \frac{\pi }{2}-y\)
\(\cos^{-1} x=\frac{\pi }{2}-\sin^{-1} x\)
\(\sin^{-1} + \cos^{-1} x=\frac{\pi }{2}\)
Hence, sin^-1x + cos^-1x = π/2, x ∈ [-1,1]

Property 5

1. tan^-1x + tan^-1y = tan^-1((x+y)/(1-xy)), xy < 1.
2. tan^-1x – tan^-1y = tan^-1((x-y)/(1+xy)), xy > -1.

Proof : tan^-1x + tan^-1y = tan^-1((x+y)/(1-xy)), xy < 1.
Let \(\tan^{-1} x=A\)
And \(\tan^{-1} y=B\)
Then, \(\tan A=x\)
\(\tan B=y\)
Now, \(\tan(A+B) = (\tan A + \tan B)/(1-\tan A\tan B)\)
\(\tan(A+B) = \frac{x+y}{1-xy} \)
\(\tan^{-1} \left (  \frac{x+y}{1-xy}\right ) = A + B \)
Hence, \(\tan^{-1} \left (  \frac{x+y}{1-xy}\right )= \tan^{-1} x +\tan^{-1} y\)

Property 6

1. 2tan^-1x = sin^-1 (2x/(1+x²)), |x| ≤ 1
2. 2tan^-1x = cos^-1((1-x²)/(1+x²)), x ≥ 0
3. 2tan^-1x = tan^-1(2x/(1 – x²)), -1 < x <1

Proof : 2tan^-1x = sin^-1 (2x/(1+x²)), |x| ≤ 1
Let \(\tan^{-1} x = y\) and  \(x = \tan y\)
Consider RHS. \(\sin^{-1}(\frac{2x}{1+x^{2}})\)
\( = \sin^{-1} \left ( \frac{2\tan y}{1+{\tan ^{2}}{y}} \right )\)
\( = \sin^{-1} \left ( \sin 2y \right ) \)
Since, sin2θ=2tanθ/(1+tan²θ),
\( = 2y \)
\( = 2\tan^{-1} x\) which is our LHS
Hence 2 tan^-1x = sin^-1 (2x/(1+x²)), |x| ≤ 1

Solved Example

Question 1. Prove that “sin^-1(-x) = – sin^-1(x),    x ∈ [-1,1]”

Ans: Let, \(\sin^{-1} \left ( -x \right )=y\)
Then \(-x=\sin y\)
\(x=-\sin y\)
\(x=\sin \left ( -y \right )\)
\(\sin^{-1} x=\arcsin \left ( \sin \left ( -y \right ) \right )\)
\(\sin^{-1} x=y\)
\(\sin^{-1} x=-\sin^{-1} \left ( -x \right )\)
Hence, \(\sin^{-1} \left ( -x \right )=-\sin^{-1} x\), x ∈ [-1,1]

This concludes our discussion on the topic of trigonometric inverse functions.

Question 2: How can we find out the inverse of any function?

Answer: We can find out the inverse of a function by following these steps:

1. First of all, replace the ‘f(x)’ with the ‘y’.
2. Then, we will be replacing every ‘x’ with ‘a’ and ‘y’ and every ‘y’ with an ‘x’.
3. Solving the equation from Step number 2 for ‘y’.
4. Replacing the ‘y’ with ‘f − 1(x)f − 1(x)’.
5. Finally we have to verify our work by checking that; (f∘f− 1) (x) = x ( f∘f− 1) (x) = ‘x’ and (f − 1∘f) (x) = x (f − 1∘f) (x) = ‘x’ are both true.

Question 3: What is the example of the inverse function?

Answer: The example for the inverse function is; if f takes ‘a’ to ‘b’, then the inverse; ‘f − 1 f^{-1} f − 1f’, start the superscript, minus, one (1), end the superscript, must take ‘b’ to ‘a’.

Question 4: What is the inverse of ‘sin’?

Answer: The inverse of ‘sin’ is the arcsine function, however, ‘sin’ itself won’t be invertible because it is not injective, and therefore, it is not invertible. For achieving the arcsine function we will be restricting the domain of ‘sin’ to ‘[−π2, π2]’.

Question 5: What is the inverse of ‘1’?

Answer: A multiplicative inverse (reciprocal) of a number ‘x’, denoted by ‘1/x’ or ‘x−1’, is a numerical value which when multiplied by ‘x’ yields ‘1’ (the multiplicative identity).