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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