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)

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