# Multiple Angle Formulae of Inverse Trigonometric Functions

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## Multiple-Angle Relations

(1)  $$2{\sin ^{ - 1}}x$$

Let  \begin{align}{\sin ^{ - 1}}x = \theta \Rightarrow x = \sin \theta \;\;{\text{and}}\;\;\theta \in \left[ {\frac{{ - \pi }}{2},\;\frac{\pi }{2}} \right]\end{align}

Now,

\begin{align} \sin 2\theta = 2\sin \theta \cos \theta\\ \,\,\,\,\,\,\,\,\,\,\,\, = 2x\sqrt {1 - {x^2}}\\\end{align}

To invert this, we have to check the interval in which  $$2\theta$$  lies.

If \begin{align}2\theta \in \left[ {\frac{{ - \pi }}{2},\;\frac{\pi }{2}} \right]\end{align}                    That is, if \begin{align}\theta \in \left[ {\frac{{ - \pi }}{4},\;\frac{\pi }{4}} \right]\;{\text{or}}\;x \in \left[ {\frac{{ - 1}}{{\sqrt 2 }},\;\frac{1}{{\sqrt 2 }}} \right]\end{align},

then we can simply invert the relation above to obtain

$2\theta = {\sin ^{ - 1}}\left( {2x\sqrt {1 - {x^2}} } \right)$

If  \begin{align}{\mathbf{2}}\theta {\mathbf{ > }}\frac{{\mathbf{\pi }}}{{\mathbf{2}}}\end{align}  That is,\begin{align}\theta > \frac{\pi }{4}\;{\text{or}}\;x > \frac{1}{{\sqrt 2 }},\end{align}  if then

$$\sin 2\theta = \sin (\pi - 2\theta ) = 2x\sqrt {1 - {x^2}}$$, so that

\begin{align}&\pi- 2\theta = 2x\sqrt {1 - {x^2}} \\ \Rightarrow\qquad &2\theta = \pi - {\sin ^{ - 1}}\left( {2x\sqrt {1 - {x^2}} } \right) \\ \end{align}

If \begin{align}{\mathbf{2}}\theta {\mathbf{ < --}}\frac{{\mathbf{\pi }}}{{\mathbf{2}}}\end{align}    That is, if      \begin{align}\theta < \frac{{ - \pi }}{4}\;{\text{or}}\;x < - \frac{1}{{\sqrt 2 }}\end{align}, then

Thus, we can summarize this as

2{\sin ^{ - 1}}x = \left\{ \begin{align} {\sin ^{ - 1}}\left( {2x\sqrt {1 - {x^2}} } \right), x \in \left[ {\frac{{ - 1}}{{\sqrt 2 }},\;\frac{1}{{\sqrt 2 }}} \right] \\ \pi - {\sin ^{ - 1}}\left( {2x\sqrt {1 - {x^2}} } \right), x \in \left[ {\frac{1}{{\sqrt 2 }},\;1} \right] \\ - \pi - \sin \left( {2x\sqrt {1 - {x^2}} } \right), x \in \left[ { - 1,\;\frac{{ - 1}}{{\sqrt 2 }}} \right]\\ \end{align} \right\}

For more clarity, try drawing the graph of just $${\sin ^{ - 1}}\left( {2x\sqrt {1 - {x^2}} } \right)$$

(2)  $$3{\sin ^{ - 1}}x$$

Let  $${\sin ^{ - 1}}x = \theta ,$$ so that \begin{align}x = \sin \theta ,\;\;\theta \in \;\left[ {\frac{{ - \pi }}{2},\;\frac{\pi }{2}} \right]\end{align}

Now,

$\sin 3\theta = 3\sin \theta - 4{\sin ^3}\theta = 3x - 4{x^3}$

We now need to check the interval in which $$3\theta$$ lies. We have to consider the following three cases:

Therefore, we can write  $$3{\sin ^{ - 1}}x$$ as

\;\left. \begin{gathered} \,\,\,\,\,\,\,\,\,\,\,\,3\theta \in \left[ {\frac{{ - \pi }}{2},\;\frac{\pi }{2}} \right]\; \\ \,\, \,\,\,\,\,\,\,\,\,\,\,\,\,\,\theta \in \left[ {\frac{{ - \pi }}{6},\;\frac{\pi }{6}} \right] \\ \Rightarrow \; x = \sin \theta \in \left[ {\frac{{ - 1}}{2},\;\frac{1}{2}} \right]\;\;\\ \end{gathered} \right|\;\;\left. \begin{gathered} \,\,\,\,\,\,\,\,\,\,\,3\theta > \frac{\pi }{2}\;\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\theta > \frac{\pi }{6} \\ x = \sin \theta > \frac{1}{2}\;\; \\ \end{gathered} \right|\;\;\;\begin{align} {\,\,\,\,\,\,\,\,\,\,3\theta < - \frac{\pi }{2}} \\ {\,\,\,\,\,\,\,\,\,\,\,\,\,\theta < - \frac{\pi }{6}} \\ {x = \sin \theta < - \frac{1}{2}} \end{align}

(3)  $$2{\cos ^{ - 1}}x$$

Let  $${\cos ^{ - 1}}x = \theta \;{\text{so}}\;x = \cos \theta ,\;\theta \in [0,\;\pi ]$$

Now,

$\cos 2\theta = 2{\cos ^2}\theta - 1 = 2{x^2} - 1$

We now check the interval in which  $$2\theta$$  lies.

If  \begin{align}2\theta \in \left[ {0,\;\pi } \right]\end{align}  That is  \begin{align}\theta \in \left[ {0,\;\frac{\pi }{2}} \right]\end{align}, if so that $$x = \cos \theta \in \left[ {0,\;1} \right]$$

$2\theta = {\cos ^{ - 1}}\left( {2{x^2} - 1} \right)$

If $$2\theta > \pi$$                                             \begin{align} \Rightarrow \;\theta > \frac{\pi }{2} \Rightarrow x = \cos \theta < 0,\end{align}    then

$2\theta = 2\pi - {\cos ^{ - 1}}(2{x^2} - 1)$

Thus,

$2{\cos ^{ - 1}}x = \left\{ \begin{gathered} {\cos ^{ - 1}}(2{x^2} - 1), x \in \left[ {0,\;1} \right] \\ 2\pi - {\cos ^{ - 1}}(2{x^2} - 1), x \in \left[ { - 1,\;0} \right) \\ \end{gathered} \right\}$

(4)  $$3{\cos ^{ - 1}}x$$

Similar to the reasoning followed for $$3{\sin ^{ - 1}}x$$, we have

3{\cos ^{ - 1}}x = \left\{ \begin{align} {\cos ^{ - 1}}(4{x^3} - 3x), x \in \left[ {\frac{1}{2},\;1} \right] \\ 2\pi - {\cos ^{ - 1}}(4{x^3} - 3x), x \in \left[ {\frac{{ - 1}}{2},\;\frac{1}{2}} \right] \\ 2\pi + {\cos ^{ - 1}}(4{x^3} - 3x), x \in \left[ { - 1,\;\frac{{ - 1}}{2}} \right] \\ \end{align} \right\}

Make sure you prove this on your own.

(5)  $$2{\tan ^{ - 1}}x$$

Let $${\tan ^{ - 1}}x = \theta \Rightarrow x = \tan \theta$$ such that  \begin{align}\theta \in \left( {\frac{{ - \pi }}{2},\;\frac{\pi }{2}} \right)\end{align}

Now,

$\tan 2\theta = \frac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }} = \frac{{2x}}{{1 - {x^2}}}$

To invert this relation, we find the interval in which lies.

 If  \begin{align}2\theta \in \left( {\frac{{ - \pi }}{2},\;\frac{\pi }{2}} \right)\end{align} : \begin{align}&\Rightarrow \theta \in \left( {\frac{{ - \pi }}{4},\;\frac{\pi }{4}} \right) \Rightarrow x = \tan \theta \in ( - 1,\;1)\\&\text{In this case,}\\&2\theta = {\tan ^{ - 1}}\left( {\frac{{2x}}{{1 - {x^2}}}} \right)\end{align} If  \begin{align}2\theta > \frac{\pi }{2} \end{align} : \begin{align}&\Rightarrow\quad \theta > \frac{\pi }{4} \Rightarrow x > 1\\\ &\Rightarrow\quad \tan 2\theta = \tan ( - \pi + 2\theta ) = \frac{{2x}}{{1 - {x^2}}}\\\ &\Rightarrow\quad 2\theta = \pi + {\tan ^{ - 1}}\left( {\frac{{2x}}{{1 - {x^2}}}} \right)\end{align} If  \begin{align}2\theta < \frac{\pi }{2}\end{align} : \begin{align}&\Rightarrow\quad \theta < \frac{\pi }{4} \Rightarrow\quad x < 1\\\ &\Rightarrow \tan 2\theta = \tan (\pi + 2\theta ) = \frac{{2x}}{{1 - {x^2}}}\\\ &\Rightarrow\quad 2\theta = - \pi + {\tan ^{ - 1}}\left( {\frac{{2x}}{{1 - {x^2}}}} \right)\end{align}

Thus,

2{\tan ^{ - 1}}x = \left\{ \begin{align} \;\;\;\;\;\;\;{\tan ^{ - 1}}\left( {\frac{{2x}}{{1 - {x^2}}}} \right), x \in ( - 1,\;1) \\ \;\;\pi + {\tan ^{ - 1}}\left( {\frac{{2x}}{{1 - {x^2}}}} \right), x > 1 \\ - \pi + {\tan ^{ - 1}}\left( {\frac{{2x}}{{1 - {x^2}}}} \right), x < 1 \\ \end{align} \right\}

Also, at $$x = 1$$, $$2{\tan ^{ - 1}}x = \pi /2$$

(6)  $$3{\tan ^{ - 1}}x$$

Proving the following is similar to the earlier parts and is left to the reader as an exercise.

3{\tan ^{ - 1}}x = \left\{ \begin{align} {\tan ^{ - 1}}\left( {\frac{{3x - {x^3}}}{{1 - 3\;{x^2}}}} \right), x \in \left( {\frac{{ - 1}}{{\sqrt 3 }},\;\frac{1}{{\sqrt 3 }}} \right) \\ \pi + {\tan ^{ - 1}}\left( {\frac{{3x - {x^3}}}{{1 - 3{x^2}}}} \right), x > \frac{1}{3} \\ - \pi + {\tan ^{ - 1}}\left( {\frac{{3x - {x^3}}}{{1 - 3{x^2}}}} \right), x < \frac{{ - 1}}{{\sqrt 3 }} \\ \end{align} \right\}

Note that  $$3{\tan ^{ - 1}}x$$  is also defined for $x = \pm \frac{1}{{\sqrt 3 }} \quad \Rightarrow \;\;\;\;3{\tan ^{ - 1}}x = \pm \frac{\pi }{2}$

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