3.4 Property 4

  Inverse Trigonometric Function

(i) ${\sin ^{ - 1}}(\sin x) = $

\[\left\{ {\begin{array}{c} x&{0 \leqslant x \leqslant \frac{\pi }{2}} \\ {\pi - x}&{\frac{\pi }{2} \leqslant x \leqslant \pi } \\ {\pi - x}&{\pi \leqslant x \leqslant \frac{{3\pi }}{2}} \\ { - 2\pi + x}&{\frac{{3\pi }}{2} \leqslant x \leqslant 2\pi } \end{array}} \right.\]

Proof: We need to plot the graph of this function

$$y = {\sin ^{ - 1}}(\sin x)...(i)$$

Let us break the intervals to plot the graph i.e, $x \in \left[ {2n\pi - \frac{\pi }{2},2n\pi + \frac{\pi }{2}} \right]$

so $0 \leqslant x \leqslant \frac{\pi }{2},\frac{\pi }{2} \leqslant x \leqslant \pi ,.....$

As we know that after completing one rotation i.e $2\pi $, the graph will repeat. So we plot the graph for $ - 2\pi \leqslant x \leqslant 2\pi $

Take the interval $0 \leqslant x \leqslant \frac{\pi }{2}$

Put $\sin x = t$ where $(t \in [ - 1,1])$

$$x = {\sin ^{ - 1}}t...(ii)$$

From $(i)$, as we know that the range of ${\sin ^{ - 1}}$ is $\left[ { - \frac{\pi }{2},\frac{\pi }{2}} \right]$, $x$ should be in that range which is satisfied. From $(i)$ and $(ii)$, we have $$x = {\sin ^{ - 1}}t = y$$

Now, take the interval $\left[ {\frac{\pi }{2},\pi } \right]$,

Put $\sin x = t$ where $(t \in [ - 1,1])$

$$x = {\sin ^{ - 1}}t...(ii)$$

From $(ii)$, $x$ should be in the range of $\left[ { - \frac{\pi }{2},\frac{\pi }{2}} \right]$ which is not satisfied because we assume $\frac{\pi }{2} \leqslant x \leqslant \pi $.

To bring it in the range,

$$\begin{equation} \begin{aligned} \pi - \frac{\pi }{2} \geqslant \pi - x \geqslant \pi - \pi \\ \frac{\pi }{2} \geqslant \pi - x \geqslant 0 \\\end{aligned} \end{equation} $$

$\pi -x$ is in the range so we can replace $x$ by $\pi -x$ , we get

$$\begin{equation} \begin{aligned} \pi - x = {\sin ^{ - 1}}t \\ y = {\sin ^{ - 1}}(\sin x) = {\sin ^{ - 1}}t \\ y = \pi - x \\\end{aligned} \end{equation} $$

Now take the interval $\pi \leqslant x \leqslant \frac{{3\pi }}{2}$

Put $$\begin{equation} \begin{aligned} \sin x = t \\ t \in [ - 1,1] \\ x = {\sin ^{ - 1}}t...(ii) \\\end{aligned} \end{equation} $$

Since $\pi \leqslant x \leqslant \frac{{3\pi }}{2}$ is not in the range so to bring it in the range,

$$\begin{equation} \begin{aligned} \pi - \pi \geqslant \pi - x \geqslant \pi - \frac{{3\pi }}{2} \\ 0 \geqslant \pi - x \geqslant \frac{{ - \pi }}{2} \\\end{aligned} \end{equation} $$

$\pi -x$ is in the range so we can replace $x$ by $\pi -x$ , we get $\pi - x = {\sin ^{ - 1}}t$.

From $(i)$, we get $$\begin{equation} \begin{aligned} y = {\sin ^{ - 1}}(\sin x) = {\sin ^{ - 1}}(t) \\ y = \pi - x \\\end{aligned} \end{equation} $$

Take the interval $\frac{{3\pi }}{2} \leqslant x \leqslant 2\pi $

Put $$\begin{equation} \begin{aligned} \sin x = t \\ t \in [ - 1,1] \\ x = {\sin ^{ - 1}}t...(ii) \\\end{aligned} \end{equation} $$

Since it is not in the range so to bring it in the range,

$$\begin{equation} \begin{aligned} - 2\pi + \frac{{3\pi }}{2} \leqslant - 2\pi + x \leqslant - 2\pi + 2\pi \\ \frac{{ - \pi }}{2} \leqslant - 2\pi + x \leqslant 0 \\\end{aligned} \end{equation} $$

Since $ - 2\pi + x$ is in the range. Replace $x$ by $ - 2\pi + x$, we get $$ - 2\pi + x = {\sin ^{ - 1}}t$$ From $(i)$, $$\begin{equation} \begin{aligned} y = {\sin ^{ - 1}}(\sin x) = {\sin ^{ - 1}}(t) \\ y = - 2\pi + x \\\end{aligned} \end{equation} $$

Similarly we can prove for other intervals. Since ${\sin ^{ - 1}}(\sin x)$ is an odd function, it is symmetric about origin as shown in figure.

(ii) ${\cos ^{ - 1}}(\cos \,\,x) = $

\[\left\{ {\begin{array}{c} x&{0 \leqslant x \leqslant \frac{\pi }{2}} \\ x&{\frac{\pi }{2} \leqslant x \leqslant \pi } \\ {2\pi - x}&{\pi \leqslant x \leqslant \frac{{3\pi }}{2}} \\ {2\pi - x}&{\frac{{3\pi }}{2} \leqslant x \leqslant 2\pi } \end{array}} \right.\]

Proof: We need to plot the graph of this function

$$y = {\cos ^{ - 1}}(\cos x)...(i)$$

Let us break the intervals to plot the graph i.e, $x \in \left[ {2n\pi - \frac{\pi }{2},2n\pi + \frac{\pi }{2}} \right]$

so $0 \leqslant x \leqslant \frac{\pi }{2},\frac{\pi }{2} \leqslant x \leqslant \pi ,.....$

As we know that after completing one rotation i.e $2\pi $, the graph will repeat. So we plot the graph for $ - 2\pi \leqslant x \leqslant 2\pi $

Take the interval $0 \leqslant x \leqslant \frac{\pi }{2}$

Put $\cos x = t$ where $(t \in [ - 1,1])$

$$x = {\cos ^{ - 1}}t...(ii)$$

From $(i)$, as we know that the range of ${\cos ^{ - 1}}$ is $[0,\pi]$, $x$ should be in that range which is satisfied. From $(i)$ and $(ii)$, we have $$x = {\cos ^{ - 1}}t = y$$

Now, take the interval $\left[ {\frac{\pi }{2},\pi } \right]$,

Put $\cos x = t$ where $(t \in [ - 1,1])$

$$x = {\cos ^{ - 1}}t...(ii)$$

From $(ii)$, $x$ should be in the range of $[0,\pi]$ which is satisfied. From $(i)$ and $(ii)$, we have $$x = {\cos ^{ - 1}}t = y$$

Now take the interval $\pi \leqslant x \leqslant \frac{{3\pi }}{2}$

Put $$\begin{equation} \begin{aligned} \cos x = t \\ t \in [ - 1,1] \\ x = {\cos ^{ - 1}}t...(ii) \\\end{aligned} \end{equation} $$

Since $\pi \leqslant x \leqslant \frac{{3\pi }}{2}$ is not in the range so to bring it in the range,

$$\begin{equation} \begin{aligned} - \pi \geqslant - x \geqslant - \frac{{3\pi }}{2} \\ 2\pi - \pi \geqslant 2\pi - x \geqslant 2\pi - \frac{{3\pi }}{2} \\ \pi \geqslant 2\pi - x \geqslant \frac{\pi }{2} \\\end{aligned} \end{equation} $$

$2\pi -x$ is in the range so we can replace $x$ by $2\pi -x$ , we get $2\pi - x = {\cos ^{ - 1}}t$.

From $(i)$, we get $$\begin{equation} \begin{aligned} y = {\cos ^{ - 1}}(\cos x) = {\cos ^{ - 1}}(t) \\ y = 2\pi - x \\\end{aligned} \end{equation} $$

Take the interval $\frac{{3\pi }}{2} \leqslant x \leqslant 2\pi $

Put $$\begin{equation} \begin{aligned} \cos x = t \\ t \in [ - 1,1] \\ x = {\cos ^{ - 1}}t...(ii) \\\end{aligned} \end{equation} $$

Since $\frac{{3\pi }}{2} \leqslant x \leqslant 2\pi $ is not in the range so to bring it in the range,

$$\begin{equation} \begin{aligned} - \frac{{3\pi }}{2} \geqslant - x \geqslant - 2\pi \\ 2\pi - \frac{{3\pi }}{2} \geqslant 2\pi - x \geqslant 2\pi - 2\pi \\ \frac{\pi }{2} \geqslant 2\pi - x \geqslant 0 \\\end{aligned} \end{equation} $$

$2\pi -x$ is in the range so we can replace $x$ by $2\pi -x$ , we get $2\pi - x = {\cos ^{ - 1}}t$.

From $(i)$, we get $$\begin{equation} \begin{aligned} y = {\cos ^{ - 1}}(\cos x) = {\cos ^{ - 1}}(t) \\ y = 2\pi - x \\\end{aligned} \end{equation} $$

Similarly we can prove for other intervals. Since ${\cos ^{ - 1}}(\cos x)$ is an even function, it is symmetric about $y$ axis as shown in figure.

(iii) ${\tan ^{ - 1}}(\tan x)=$

\[\left\{ {\begin{array}{c} x&{0 \leqslant x \leqslant \frac{\pi }{2}} \\ {x - \pi }&{\frac{\pi }{2} \leqslant x \leqslant \pi } \\ {x - \pi }&{\pi \leqslant x \leqslant \frac{{3\pi }}{2}} \\ {x - 2\pi }&{\frac{{3\pi }}{2} \leqslant x \leqslant 2\pi } \end{array}} \right.\]

Proof: Since ${\tan ^{ - 1}}(\tan x)$ is an odd function, it is symmetric about origin as shown in figure.

(iv) ${\text{cose}}{{\text{c}}^{ - 1}}({\text{cosec}}\;x)$. The graph of this function is similar to ${\sin ^{ - 1}}(\sin x)$

(v) $se{c^{ - 1}}(secx)$. The graph of this function is similar to ${\cos ^{ - 1}}(\cos x)$

(vi) ${\cot ^{ - 1}}(\cot x)$. The graph of this function is not similar to ${\tan ^{ - 1}}(\tan x)$. It consists only the positive part of ${\tan ^{ - 1}}(\tan x)$ as shown in figure.

Note: All the functions discussed above are periodic functions.