Inverse Trigonometric Functions - Properties

Inverse trigonometric functions are also called “Arc Functions,” since for a given value of a trigonometric function; they produce the length of arc needed to obtain that particular value. The range of an inverse function is defined as the range of values the inverse function can attain with the defined domain of the function. Domain of a function is defined as the set of every possible independent variable where the function exists. Inverse Trigonometric Functions are defined in a certain interval.

Considering the domain and range of the inverse functions, following formulas are important to be noted:

  • \(\sin (\sin ^{-1}x) = x\), if -1 ≤ x ≤ 1
  • \(\cos (\sin ^{-1}x) = x\), if -1 ≤ x ≤ 1
  • \(\tan (\sin ^{-1}x) = x\), if -∞ ≤ x ≤∞
  • \(\cot (\sin ^{-1}x) = x\), if -∞≤ x ≤∞
  • \( sec (\sin ^{-1}x) = x\), if -∞ ≤ x ≤ -1 or 1 ≤ x ≤ ∞
  • \( cosec (\sin ^{-1}x) = x\), if -∞ ≤ x ≤ -1 or 1 ≤ x ≤ ∞

Also, the following formulas are defined for inverse trigonometric functions.

  • \(\sin^{-1} (\sin y) = y\) if -π/2 ≤ y ≤ π/2
  • \(\cos^{-1}(\cos y)= y\) if 0 ≤ y ≤ π
  • \(\tan^{-1}(\tan y)= y\) if -π/2 <y< π/2
  • \(\cot^{-1}(\cot y)= y\) if 0<y< π
  • \(\sec^{-1}(\sec y)= y\) if 0 ≤ y ≤ π, y ≠ π/2
  • \(cosec^{-1}(cosec y)= y\) if -π/2 ≤ y ≤ π/2, y ≠ 0

Inverse Trigonometric Functions & Properties

Now, let us discuss few important properties related to inverse trigonometric functions.

1). \(\sin ^{-1} \left ( \frac{1}{x} \right ) = cosec ^{-1} x, x≥1 \; or \; x≤-1\)

Proof: Let \(cosec ^{-1} x = y\), i.e., \(x=cosec y\)

\(⇒ \left(\frac{1}{x}\right) =\sin y\)

Thus,

\( \sin^{-1}\left (\frac{1}{x}\right)= y\)

Or,

\(\sin^{-1} \left(\frac{1}{x}\right)=  cosec^{-1} x\)

Similarly using the same concept following results can be obtained:

  • \(\cos ^{-1}\left ( \frac{1}{x} \right ) = \sec ^{-1}x, x≥1 \; or \; x≤-1\)
  • \(\cos ^{-1}\left ( \frac{1}{x} \right ) = \sec ^{-1}x,  x>0\)

2). \(\sin^{-1}(-x)=-\sin^{-1}x, x ϵ [-1,1]\)

Proof: Let \(\sin^{-1}(-x)= y, i.e., -x =\sin y\)

⇒ x = -sin y

Thus,

x = sin (-y)

Or,

\(\sin^{-1}(x)= -y = -\sin^{-1}(-x)\)

Therefore,

\(\sin^{-1}(-x) = -\sin^{-1}(x)\)

Similarly using the same concept following results can be obtained:

  • \( cosec^{-1}(-x) =-cosec^{-1}x,|x| ≥ 1\)

  • \(\tan^{-1}(-x)=-\tan^{-1} x, \; x ϵ R\)

3). \(\cos^{-1}(-x)= \pi –\cos^{-1}x, x ϵ [-1,1]\)

Proof: \(\cos^{-1}(-x) = y\) i.e., -x = cos y

\(x = -\cos y = \cos (\pi – y)\)

Thus,

\(\cos^{-1} (x) = \pi – y \)

Or,

\(\cos^{-1}(x) = \pi –\cos^{-1}(-x)\)

Therefore,

\(\cos^{-1}(-x) = \pi –\cos^{-1}(x)\)

Similarly using the same concept following results can be obtained:

  • \(\sec^{-1}(-x) = \pi –\sec^{-1}x, \; |x| ≥ 1\)

  • \(\cot^{-1}(-x) = \pi –\cot^{-1}x, \; x ϵ R\)

4). \(\sin^{-1}(x) + \cos^{-1}(x) = \left(\frac{\pi}{2}\right), x ϵ [-1,1]\)

Proof: Let \(\sin^{-1}(x) = y, i.e., x = \sin y = \cos (\left(\frac{\pi}{2}\right)-y)\)

\(\cos^{-1} (x) =\left(\frac{\pi}{2}\right) – y = \left(\frac{\pi}{2}\right) -\sin^{-1}(x)\)

Thus,

\(\sin^{-1}(x) + \cos^{-1}(x) = \left(\frac{\pi}{2}\right)\)

Similarly using the same concept following results can be obtained:

  • \(\tan^{-1}(x) + \cot^{-1}(x) = \left(\frac{\pi}{2}\right), x ϵ R\)
  • \( cosec^{-1}(x) + \sec^{-1}(x) = \left(\frac{\pi}{2}\right), |x| ≥ 1\)

5). \(\tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1} \left(\frac{x+y}{1-xy}\right), xy<1\)

Proof: Let \(\tan^{-1}(x) = \alpha \) and \(\tan^{-1}(y) = \beta \), i.e., \( x = \tan (\alpha)\) and \( y = \tan (\beta)\)

\( \tan (\alpha + \beta) = \frac {\tan \alpha + \tan \beta}{1 – \tan \alpha \tan\beta}\)

Thus,

\( (\alpha) + (\beta) = \tan^{-1}( \frac{x+y}{1-xy})\)

Therefore,

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

Similarly using the same concept following results can be obtained:

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

  • \(2 \tan^{-1}(x) = \tan^{-1} \left( \frac {2x}{1 – x^2}\right), |x|<1\)

6).  \(2 \tan^{-1}x = \sin ^{-1}\left ( \frac{2x}{1+x^{2}} \right ), |x|<1\)

Proof: Let, \(\tan ^{-1}x = y\) i.e., \(x = tan y \)

\(\Rightarrow \sin^{-1} \left ( \frac{2x}{1+x^{2}} \right ) = \sin ^{-1}\left ( \frac{2 \tan y}{1+ \tan^{2}y} \right )\)

Thus,

\(\Rightarrow \sin ^{-1}\left ( \frac{2 \tan y}{1+ \tan^{2}y} \right ) = \sin ^{-1} (\sin 2y) = 2y = 2 \tan ^{-1}x\)

Similarly using the same concept following results can be concluded:

  • \(2 \tan ^{-1}x = \cos ^{-1}\left ( \frac{1 – x^{2}}{1 + x^{2}} \right ), x≥0\)
  • \(2 \tan ^{-1}x = \tan ^{-1}\left ( \frac{2x}{1 – x^{2}} \right ), -1<x<1\)<

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Practise This Question

if a function f(x) is discontinuous in the interval (a,b) then baf(x)dx never exists.