Addition Properties of Inverse Trigonometric Functions
Addition Properties
The most frequently used properties in inverse trigonometry problems are the addition properties.
(1) \({\mathbf{si}}{{\mathbf{n}}^{-{\mathbf{1}}}}x + {\mathbf{si}}{{\mathbf{n}}^{-{\mathbf{1}}}}y\)
Note that \(x \in [ - 1,\;1]\;{\text{and}}\;y \in [ - 1,\;1]\) for this expression to be defined.
Let \({\sin ^{ - 1}}x = \theta \;{\text{and}}\;{\sin ^{ - 1}}y = \phi \) so that
\[x = \sin \theta ,\;\;y = \sin \phi ,\;\;\theta ,\,\,\phi \in \left[ {\frac{{ - \pi }}{2},\;\frac{\pi }{2}} \right]\]
Now,
\[\begin{align} \sin (\theta + \phi ) &= \sin \theta \cos \phi + \cos \theta \sin \phi \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &= x\sqrt {1 - {y^2}} + y\sqrt {1 - {x^2}} = z\;({\text{say}}) \\ \end{align} \]
which hints us to the fact that
\[\theta + \phi \; = {\sin ^{ - 1}}x + {\sin ^{ - 1}}y = {\sin ^{ - 1}}\left( {x\sqrt {1 - {y^2}} + y\sqrt {1 - {x^2}} } \right)\]
But wait! This is not entirely correct! Take \(x = y = 1\), and
we have
\[{\sin ^{ - 1}}x + {\sin ^{ - 1}}y = \frac{\pi }{2} + \frac{\pi }{2} = \pi \]
while
\[{\sin ^{ - 1}}\left( {x\sqrt {1 - {y^2}} + y\sqrt {1 - {x^2}} } \right) = {\sin ^{ - 1}}(0) = 0\]
There seems to be some problem in this equality, and we have to modify it some what. To do that, we note that
\[\cos \left( {\theta + \phi } \right) = \cos \theta \cos \phi - \sin \theta \sin \phi = \sqrt {1 - {x^2}} \sqrt {1 - {y^2}} - xy\]
The problem is originating from the fact that \(\theta + \phi \) may not be in the restricted domain of sin, i.e. in \(\begin{align}\left[ {\frac{{ - \pi }}{2},\;\frac{\pi }{2}} \right]\end{align}\). Therefore, we have to determine, based on x and y, the value of \(\theta + \phi \). We consider the following cases:
(a) \({x^2} + {y^2} \leqslant 1\)
\[\begin{align}\Rightarrow\quad &\sqrt {1 - {x^2}} \geqslant \;|y|\;{\text{and}}\;\sqrt {1 - {y^2}} \geqslant \left| x \right| \\\Rightarrow\quad &\sqrt {1 - {x^2}} \;\sqrt {1 - {y^2}} - xy \geqslant 0 \\\Rightarrow\quad &\cos (\theta + \phi ) \geqslant 0\;\; \Rightarrow \;\;\theta + \phi \in \left[ {\frac{{ - \pi }}{2},\frac{\pi }{2}} \right] \\ \Rightarrow \quad &\theta + \phi = {\sin ^{ - 1}}x + {\sin ^{ - 1}}y = {\sin ^{ - 1}}\left\{ {x\sqrt {1 - {y^2}} + y\sqrt {1 - {x^2}} } \right\}\\ \end{align} \]
(b) \({x^2} + {y^2} > 1,\;x\,y < 0\)
\[\begin{align}\Rightarrow \quad &\sqrt {1 - {x^2}} > \;|y|,\sqrt {1 - {y^2}} < \;|x| \\ \Rightarrow \quad&\sqrt {1 - {x^2}} \,\sqrt {1 - {y^2}} < \;|xy| \\ \end{align} \]
Since \(xy < 0,\)
\[\sqrt {1 - {x^2}} \;\sqrt {1 - {y^2}} - xy > 0 \Rightarrow \cos (\theta + \phi ) > 0\]
Once again
\[\theta + \phi = {\sin ^{ - 1}}x + {\sin ^{ - 1}}y = {\sin ^{ - 1}}\left\{ {x\sqrt {1 - {y^2}} + y\sqrt {1 - {x^2}} } \right\}\]
(c) \({x^2} + {y^2} > 1,\;\;x,\;y > 0\)
\(\begin{align}{\text{Since}}\;x > 0,\;\theta \in \left( {0,\;\frac{\pi }{2}} \right)\end{align}\). Similarly, \(\begin{align}\phi \in \left( {0,\;\frac{\pi }{2}} \right)\end{align}\).
Thus \(\theta + \phi \; \in \left( {0,\;\pi } \right]\), and so \(\theta + \phi \) may not be in the restricted domain of sin.
Also,
\[\begin{align}&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\sqrt {1 - {x^2}} < \;|y|,\,\sqrt {1 - {y^2}} < \;|x|\\ \Rightarrow\quad &\sqrt {1 - {x^2}} \,\sqrt {1 - {y^2}} < \;xy,\; {\text{since}}\;xy > 0 \\ \Rightarrow\quad & \cos (\theta + \phi ) = \sqrt {1 - {x^2}} \,\sqrt {1 - {y^2}} \; - xy < 0 \\ \Rightarrow\quad &\theta + \phi \in \left( {\frac{\pi }{2},\;\pi } \right] \\ \end{align}\]
Therefore,
\[\theta + \phi = \pi - {\sin ^{ - 1}}\left\{ {x\sqrt {1 - {y^2}} + y\sqrt {1 - {x^2}} } \right\}\]
to adjust for the fact that \(\begin{align}\theta + \phi \end{align}\) is in \(\begin{align}\left( {\frac{\pi }{2},\;\pi } \right)\end{align}\)
(d) \({x^2} + {y^2} > 1,\;x,\;y < 0\)
We can show that \(\theta + \phi \, \in \left[ { - \pi ,\;0} \right)\) and \(\cos (\theta + \phi ) < 0\), which means that
\[\theta + \phi \; \in \left[ { - \pi ,\;\frac{{ - \pi }}{2}} \right)\]
Thus,
\[\theta + \phi \; = {\sin ^{ - 1}}x + {\sin ^{ - 1}}y = - \pi - {\sin ^{ - 1}}\left\{ {x\sqrt {1 - {y^2}} + y\sqrt {1 - {x^2}} } \right\}\]
Graphically, the four cases can be correlated with four distinct. regions in a unit square:
One way to remember these relations is by making the following observations
(i) In region (c), both \({\sin ^{ - 1}}x\) and \({\sin ^{ - 1}}y\) will be large, so \({\sin ^{ - 1}}x + {\sin ^{ - 1}}y\) will be large. Precisely speaking, it will be above \(\begin{align}\frac{\pi }{2},\end{align}\) while \({\sin ^{ - 1}}\left( {x\sqrt {1 - {y^2}} + y\sqrt {1 - {x^2}} } \right)\) will not be. Thus, an extra term of \(\pi \) will be needed to compensate.
(ii) In region (d), both \({\sin ^{ - 1}}x\) and \({\sin ^{ - 1}}y\) will be large, but negative in sign, so \({\sin ^{ - 1}}x + {\sin ^{ - 1}}y\) will be large and negative. Therefore, we need an extra term of \( - \pi \) to compensate.
(iii) In the region (b) in the second quadrant \({\sin ^{ - 1}}x\), will be negative while \({\sin ^{ - 1}}y\) is positive, so \({\sin ^{ - 1}}x + {\sin ^{ - 1}}y\) falls in the restricted domain of i.e. in \(\begin{align}\left[ {\frac{{ - \pi }}{2},\;\frac{\pi }{2}} \right]\end{align}\). Similarly for region (b) in the fourth quadrant.
(iv) In region (a), both \({\sin ^{ - 1}}x\) and \({\sin ^{ - 1}}y\) are such that their sum always falls in \(\begin{align}\left[ {\frac{{ - \pi }}{2},\;\frac{\pi }{2}} \right]\end{align}\).
(2) \({\cos ^{ - 1}}x + {\cos ^{ - 1}}y\)
Let \({\cos ^{ - 1}}x = \theta \;{\text{and}}\;{\cos ^{ - 1}}y = \phi \), so that
\[x = \cos \theta ,\; y = \cos \phi , \theta ,\phi \in [0,\;\pi ]\]
Now,
\[\begin{align}\cos (\theta + \phi ) = \cos \theta \cos \phi - \sin \theta \sin \phi \\ = xy - \sqrt {1 - {x^2}} \;\sqrt {1 - {y^2}} \\ \end{align} \]
Once again, we can’t straightaway take the \({\cos ^{ - 1}}\) on both sides. We have to determine the interval in which \(\theta + \phi \) lies to make the appropriate adjustments. For example,
If \(\theta + \phi \; \in \;[0,\;\pi ]\)
\[\begin{align}&\theta \leqslant \pi - \phi \\ \Rightarrow\quad &\cos \theta \geqslant \cos \left( {\pi - \phi } \right) = - \cos \phi \\ \Rightarrow\quad & \cos \theta + \cos \phi \; \geqslant \;0 \\ \Rightarrow\quad & x + y\; \geqslant \;0 \\ \end{align} \]
Then, \(\theta + \phi = {\cos ^{ - 1}}\left\{ {xy - \sqrt {1 - {x^2}} \;\sqrt {1 - {y^2}} } \right\}\)
If \(\theta + \phi \; \in [\pi ,\;2\pi ]\)
\[\begin{align}&\theta \geqslant \pi - \phi \\ &\cos \theta \leqslant \cos (\pi - \phi ) = - \cos \phi \\\Rightarrow\quad &\cos \theta + \cos \phi \leqslant 0\\ \Rightarrow\quad &x + y \leqslant 0 \\ \end{align} \]
Then, \(\theta + \phi = 2\pi - {\cos ^{ - 1}}\left\{ {xy - \sqrt {1 - {x^2}} \,\sqrt {1 - {y^2}} } \right\}\)
Thus, when, \(x,\;y\; \in [ - 1,\;1]\), we have
\[{\cos ^{ - 1}}x + {\cos ^{ - 1}}y = \left\{ \begin{align} {\cos ^{ - 1}}\left\{ {xy - \sqrt {1 - {x^2}} \;\sqrt {1 - {y^2}} } \right\},\qquad x + y \geqslant 0 \\ 2\pi - {\cos ^{ - 1}}\left\{ {xy - \sqrt {1 - {x^2}} \;\sqrt {1 - {y^2}} } \right\}, \qquad x + y \leqslant 0 \\ \end{align} \right\}\]
(3) \({\tan ^{ - 1}}x + {\tan ^{ - 1}}y\)
If \({\tan ^{ - 1}}x = \theta \) and \({\tan ^{ - 1}}y = \phi \), then
\[x = \tan \theta , \qquad y = \tan \phi , \qquad \theta , \qquad \phi \in \left( {\frac{{ - \pi }}{2},\;\frac{\pi }{2}} \right)\]
Thus, \(\theta + \phi \; \in ( - \pi ,\;\pi )\)
Also,
\[\tan (\theta + \phi ) = \frac{{\tan \theta + \tan \phi }}{{1 - \tan \theta \tan \phi }} = \frac{{x + y}}{{1 - xy}}\]
Once again, we have to determine whether \(\theta + \phi \) lies in the (restricted) interval \(\begin{align}\left( {\frac{{ - \pi }}{2},\;\frac{\pi }{2}} \right)\end{align}\) or not, because otherwise, we’ll have to make the necessary adjustments.
If \(\begin{align}\theta + \phi > \frac{\pi }{2}\end{align}\) \(\begin{align}\theta > \frac{\pi }{2} - \phi \end{align}\)
\[ \Rightarrow \tan \theta > \tan \left( {\frac{\pi }{2} - \phi } \right) = \cot \phi = \frac{1}{{\tan \phi }}\]
Since both \(\theta ,\;\phi \; > 0\) in this case, we have
\[\tan \theta \tan \phi = xy > 1\]
Also, since \(\begin{align}\theta + \phi \in \left( {\frac{\pi }{2},\pi } \right),\end{align}\)
\[\begin{align}&\pi - (\theta + \phi ) \in \left( {0,\;\frac{\pi }{2}} \right) \hfill \\ \Rightarrow\quad &\tan \left( {\pi - (\theta + \phi )} \right) = - \frac{{x + y}}{{1 - xy}} \\\Rightarrow\quad &\theta + \phi = \pi + {\tan ^{ - 1}}\left( {\frac{{x + y}}{{1 - xy}}} \right) \\ \end{align} \]
If \(\begin{align}\theta + \phi < \frac{{ - \pi }}{2}\end{align}\) \(\begin{align}\theta < \frac{{ - \pi }}{2} - \phi \end{align}\)
\[\tan \theta < \tan \left( {\frac{{ - \pi }}{2} - \phi } \right) = \cot \phi = \frac{1}{{\tan \phi }}\]
Since both \(\theta ,\;\phi < 0,\;\;\tan \phi < 0\) so that
\[\tan \theta \tan \phi = xy > 1\]
Also,
\[\begin{align}&\pi + (\theta + \phi ) \in \left( {0,\;\frac{\pi }{2}} \right) \\ \Rightarrow\quad &\tan (\pi + (\theta + \phi )) = \frac{{x + y}}{{1 - xy}} \\ \Rightarrow \quad &\theta + \phi = - \pi + {\tan ^{ - 1}}\left( {\frac{{x + y}}{{1 - xy}}} \right) \\ \end{align} \]
If \(\begin{align}\theta + \phi \in \left( {\frac{{ - \pi }}{2},\;\frac{\pi }{2}} \right)\end{align}\) In this case,
\(\tan \theta \tan \phi = xy < 1\) (Verify !)
and
\[\theta + \phi = {\tan ^{ - 1}}\left( {\frac{{x + y}}{{1 - xy}}} \right)\]
We can summarize this as
\[{\tan ^{ - 1}}x + {\tan ^{ - 1}}y = \left\{ \begin{align} {\tan ^{ - 1}}\left( {\frac{{x + y}}{{1 - xy}}} \right), \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{\text{if}}\;xy < 1 \\ \;\;\pi + {\tan ^{ - 1}}\left( {\frac{{x + y}}{{1 - xy}}} \right), {\text{if}}\;x > 0,\;y > 0,\;xy > 1 \\ - \pi + {\tan ^{ - 1}}\left( {\frac{{x + y}}{{1 - xy}}} \right), {\text{if}}\;x < 0,\;y < 0,\;xy > 1 \\ \end{align} \right\}\]
As an exercise, prove the following relations.
(a) For \(x,\;y \in [ - 1,\;1]\)
\[{\sin ^{ - 1}}x - {\sin ^{ - 1}}y ={\left\{ \begin{array} {\sin ^{ - 1}}\left\{ {x\sqrt {1 - {y^2}} - y\sqrt {1 - {x^2}} } \right\}, {x^2} + {y^2} \leqslant 1\\ {\text{or}} \\ xy > 0\;{\text{and}}\;{x^2} + {y^2} > 1 \\ \pi - {\sin ^{ - 1}}\left\{ {x\sqrt {1 - {y^2}} - y\sqrt {1 - {x^2}} } \right\}, & {x^2} + {y^2} > 1,\;\;x > 0,\;\;y < 0 \\ - \pi - {\sin ^{ - 1}}\left\{ {x\sqrt {1 - {y^2}} - y\sqrt {1 - {x^2}} } \right\}, & {x^2} + {y^2} > 1,\;\,x < 0,\;y > 0 \\ \end{array} \right\}}\]
(b) For \(x,\;y \in [ - 1,\;1],\)
\[{\cos ^{ - 1}}x - {\cos ^{ - 1}}y = \left\{ \begin{align} {\cos ^{ - 1}}\left\{ {xy + \sqrt {1 - {x^2}} \,\sqrt {1 - {y^2}} } \right\}\;, x \leqslant y \\ - {\cos ^{ - 1}}\left\{ {xy + \sqrt {1 - {x^2}} \,\sqrt {1 - {y^2}} } \right\}, \;x \geqslant y \\ \end{align} \right\}\]
(c)
\[{\tan ^{ - 1}}x - {\tan ^{ - 1}}y = \left\{ \begin{gathered} {\tan ^{ - 1}}\left( {\frac{{x - y}}{{1 + xy}}} \right), xy > - 1 \\ \pi + {\tan ^{ - 1}}\left( {\frac{{x - y}}{{1 + xy}}} \right), x > 0,\;y < 0,\;xy < - 1 \\ - \pi + {\tan ^{ - 1}}\left( {\frac{{x - y}}{{1 + xy}}} \right), x < 0,\;y > 0,\;xy < - 1 \\ \end{gathered} \right\}\]
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