Elementary properties of inverse trigonometric functions

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

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