# Addition Properties of Inverse Trigonometric Functions

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)

${\tan ^{ - 1}}x - {\tan ^{ - 1}}y = \left\{ \begin{gathered} {\tan ^{ - 1}}\left( {\frac{{x - y}}{{1 + xy}}} \right), xy > - 1 \\ \pi + {\tan ^{ - 1}}\left( {\frac{{x - y}}{{1 + xy}}} \right), x > 0,\;y < 0,\;xy < - 1 \\ - \pi + {\tan ^{ - 1}}\left( {\frac{{x - y}}{{1 + xy}}} \right), x < 0,\;y > 0,\;xy < - 1 \\ \end{gathered} \right\}$