Elementary properties of inverse trigonometric functions

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The elementary properties of inverse trigonometric functions are valid within the principal value branches of the corresponding inverse function wherever they are defined. However, these properties are valid for a limited section of the domain of the inverse functions. Recall that, if y = sin$^{-1}$x and x= sin y then y = sin$^{-1}$x. This means that:

Sin (sin$^{-1}$x) = x such that x is an element of $[$-1, 1$]$ and sin$^{-1}$(sin x) = x such that x is an element of $[-\frac{\pi }{2},\frac{\pi }{2}]$

This property is true for other inverse trigonometric functions. The proof of these properties are provided below:

  1. Sin$^{-1}$$\frac{1}{x}$= cosec$^{-1}$x, x ≥ 1 or x ≤ -1
    Let cosec$^{-1}$x = y , i.e. x = cosec yThis means that

    $\frac{1}{x}=\sin y$

    $\sin ^{-1}\frac{1}{x}=cosec^{-1}x$

  2. Cos$^{-1}$$\frac{1}{x}$= sec$^{-1}$x, x ≥ 1 or x ≤ -1Let sec$^{-1}$x = y , i.e. x = sec y

    This means that

    $\frac{1}{x}=\cos y$

    $Cos^{-1}\frac{1}{x}=\sec ^{-1}x$

  3. Tan$^{-1}$$\frac{1}{x}$ = cot$^{-1}$x, x $>$ 0Let cot$^{-1}$x = y , i.e. x = cot y

    This means that

    $\frac{1}{x}=\tan y$

    $\tan ^{-1}\frac{1}{x}=\cot ^{-1}x$

  4. $\sin ^{-1}(-x)=-\sin ^{-1}x$, $x \varepsilon [-1, 1]$Let $\sin ^{-1}(-x)=y, i.e., -x=\sin y$ such that x = -sin y

    Therefore, x = sin (-y)

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

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

    The other two identities can be proven in a similar way.

  5. $\tan ^{-1}(-x)=-\tan ^{-1}x$, $x \varepsilon R$
  6. $cosec^{-1}(-x)=-cosec^{-1}x$, $\vert x\vert \ge 1$
  7. $\cos ^{-1}(-x)=\pi -\cos ^{-1}x$, $x \varepsilon \lbrack -1, 1\rbrack $Let $\cos ^{-1}(-x)=y, i.e., -x=\cos y$ such that x = -cos y = cos (\pi – y )

    Therefore,

    $\cos ^{-1}x=\pi -y= \pi -\cos ^{-1}(-x)$

    $\cos ^{-1}(-x)=\pi -\cos ^{-1}x$

    We can prove the following 2 properties in a similar way.

  8. $\sec ^{-1}(-x)=\pi -\sec ^{-1}x$,$\vert x\vert \ge 1$
  9. $\cot ^{-1}(-x)=\pi -\cot ^{-1}x$, $x \varepsilon R$
  10. $\sin ^{-1}x+\cos ^{-1}x=\frac{\pi }{2}, x \varepsilon \lbrack -1, 1\rbrack $
    Let $\sin ^{-1}x=y$, x = sin y = $\cos (\frac{\pi }{2}-y)$$\cos ^{-1}x=\frac{\pi }{2}-y=\frac{\pi }{2}- \sin ^{-1}x$

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

    We can prove the following 2 properties in a similar way.

  11. $tan ^{-1}x+\cot ^{-1}x=\frac{\pi }{2}, x \varepsilon R $
  12. $cosec ^{-1}x+\sec ^{-1}x=\frac{\pi }{2}, ,\vert x\vert \ge 1$
  13. $\tan ^{-1}x+\tan ^{-1}y=\tan ^{-1}\frac{x+y}{1-xy}$ ; xy $<$ 1
    Let $\tan ^{-1}x= \theta $ and $\tan ^{-1}y= \varphi $. This means that x = tan θ, y = tan Φ$\tan (\theta +\varphi )=\frac{\tan \theta +\tan \varphi }{1-\tan \theta \tan \varphi }=\frac{x+y}{1-xy}$

    $\theta +\varphi =\tan ^{-1}\frac{x+y}{1-xy}$

    $\tan ^{-1}x+\tan ^{-1}y=\tan ^{-1}\frac{x+y}{1-xy}$

    Here, we replace y to -y, we get the following identity and by replacing y to x we get the next identity.

  14. $\tan ^{-1}x-\tan ^{-1}y=\tan ^{-1}\frac{x-y}{1-xy}$ ; xy$>$ -1
  15. $2tan^{-1}x=\tan ^{-1}\frac{2x}{1-x^{2}}$ ; $\vert x\vert <1$
  16. $2tan^{-1}x=\sin ^{-1}\frac{2x}{1+x^{2}}, \vert x\vert \le 1$

    Let $\tan ^{-1}x=y$, then x = tan y

    Therefore,

    $\sin ^{-1}\frac{2x}{1+x^{2}}=\sin ^{-1}\frac{2\tan y}{1+\tan ^{2}y}$

    $=\sin ^{-1}(\sin 2y)=2y=2\tan ^{-1}x$

  17. $2tan^{-1}x=\cos ^{-1}\frac{1-x^{2}}{1+x^{2}}, x\ge 0$

    $\cos ^{-1}\frac{1-x^{2}}{1+x^{2}}=\cos ^{-1}\frac{1-\tan ^{2}y}{1+\tan ^{2}y}=\cos ^{-1}(\cos 2y)=2y=2\tan ^{-1}x$

  18. $2tan^{-1}x=\tan ^{-1}\frac{2x}{1-x^{2}}, -1< x<1$

Example: Express $\tan ^{-1}\frac{\cos x}{1-\sin x}, -\frac{\pi }{2}

Solution: $\tan ^{-1}\frac{\cos x}{1-\sin x}=\tan ^{-1}\lbrack \frac{\cos ^{2}x/2 -\sin ^{2}x/2}{\cos ^{2}x/2 +\sin ^{2}x/2-2\sin \frac{x}{2}\cos \frac{x}{2}}\rbrack $

$=\tan ^{-1}\lbrack \frac{(cos\frac{x}{2}+\sin \frac{x}{2})(\cos \frac{x}{2}-\sin \frac{x}{2})}{(\cos \frac{x}{2}-\sin \frac{x}{2} )^{2}}\rbrack $

$=\tan ^{-1}\lbrack \frac{(cos\frac{x}{2}+\sin \frac{x}{2})}{(\cos \frac{x}{2}-\sin \frac{x}{2})}\rbrack =\tan ^{-1}\lbrack \frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}}\rbrack $

$=\tan ^{-1}\lbrack \tan (\frac{\pi }{4}+\frac{x}{2})\rbrack =\frac{\pi }{4}+\frac{x}{2}$

Alternatively,

$\tan ^{-1}(\frac{\cos x}{1-\sin x})=\tan ^{-1}\lbrack \frac{\sin (\frac{\pi }{2}-x)}{1- cos (\frac{\pi }{2}-x)}\rbrack =\tan ^{-1}\lbrack \frac{\sin (\frac{\pi -2x}{2})}{1- cos (\frac{\pi -2x}{2})}\rbrack $

$=\tan ^{-1}\lbrack \frac{2\sin (\frac{\pi -2x}{4})\cos (\frac{\pi -2x}{4})}{2 \sin ^{2}( \frac{\pi -2x}{4})}\rbrack $

$=tan^{-1}\lbrack \cot (\frac{\pi -2x}{4})\rbrack =\tan ^{-1}\lbrack \tan (\frac{\pi }{2}-\frac{\pi -2x}{4})\rbrack $

$=\tan ^{-1}\lbrack \tan (\frac{\pi }{4}+\frac{x}{2})\rbrack =\frac{\pi }{4}+\frac{x}{2}$

Example: What is the value of $\cos (\sec ^{-1}x+\csc ^{-1}x), \vert x\vert \ge 1$

Solution: Recall that $\cos (\sec ^{-1}x+\csc ^{-1}x)=\cos \frac{\pi }{2}=0$