Friday, January 19, 2024

18.8 - Properties II and III

In the previous section, we saw the first property of inverse trigonometric functions. In this section, we will see the second and third properties.

Property II
• This has 3 parts:
$\begin{array}{cc}{}    &{\text{(i)}}    &{\sin^{-1}\left(-x \right)}    {}={}    &{-\sin^{-1}x},    &{x \in [-1,1]}\\
{}    &{\text{(ii)}}    &{\tan^{-1}\left(-x \right)}    {}={}    &{-\tan^{-1}x},    &{x \in R}\\
{}    &{\text{(iii)}}    &{\csc^{-1}\left(-x \right)}    {}={}    &{-\csc^{-1}x},    &{|x| \ge 1}\\
\end{array}$

• Let us prove part (i):
$\begin{array}{ll}{}    &{\text{Let}~\sin^{-1} \left(-x \right)}    & {~=~}    &{y~\color{magenta}{\text{- - - (A)}}}    &{} \\
{\implies}    &{\sin y}    & {~=~}    &{-x}    &{} \\
{\implies}    &{- \sin y}    & {~=~}    &{x}    &{} \\
{\implies}    &{\sin (-y)}    & {~=~}    &{x~\color{magenta}{\text{- - - (B)}}}    &{} \\
{\implies}    &{\sin^{-1} x}    & {~=~}    &{-y}    &{} \\
{\implies}    &{-\sin^{-1} x}    & {~=~}    &{y}    &{} \\
{\implies}    &{-\sin^{-1} x}    & {~=~}    &{\sin^{-1} \left(-x \right)~\color{magenta}{\text{- - - (C)}}}    &{} \\
\end{array}$               

◼ Remarks:
• In line (A), we assume that $\sin^{-1}(-x)$ = y
• In line B, we use the identity 1: sin (-𝜃) = - sin 𝜃
    ♦ List of identities can be seen here.
• In line C, we substitute for y, using the assumption in (A).

• Let us prove part (ii):
$\begin{array}{ll}{}    &{\text{Let}~\tan^{-1} \left(-x \right)}    & {~=~}    &{y~\color{magenta}{\text{- - - (A)}}}    &{} \\
{\implies}    &{\tan y}    & {~=~}    &{-x}    &{} \\
{\implies}    &{- \tan y}    & {~=~}    &{x}    &{} \\
{\implies}    &{\tan (-y)}    & {~=~}    &{x~\color{magenta}{\text{- - - (B)}}}    &{} \\
{\implies}    &{\tan^{-1} x}    & {~=~}    &{-y}    &{} \\
{\implies}    &{-\tan^{-1} x}    & {~=~}    &{y}    &{} \\
{\implies}    &{-\tan^{-1} x}    & {~=~}    &{\tan^{-1} \left(-x \right)~\color{magenta}{\text{- - - (C)}}}    &{} \\
\end{array}$               

◼ Remarks:
• In line (A), we assume that $\tan^{-1}(-x)$ = y
• In line B, we use the identities 1 and 2 to get: tan (-𝜃) = - tan 𝜃
    ♦ List of identities can be seen here.
• In line C, we substitute for y, using the assumption in (A).

• Let us prove part (iii):
$\begin{array}{ll}{}    &{\text{Let}~\csc^{-1} \left(-x \right)}    & {~=~}    &{y~\color{magenta}{\text{- - - (A)}}}    &{} \\
{\implies}    &{\csc y}    & {~=~}    &{-x}    &{} \\
{\implies}    &{- \csc y}    & {~=~}    &{x}    &{} \\
{\implies}    &{\frac{-1}{\sin y}}    & {~=~}    &{x}    &{} \\
{\implies}    &{-\sin y}    & {~=~}    &{\frac{1}{x}}    &{} \\
{\implies}    &{\sin (-y)}    & {~=~}    &{\frac{1}{x}~\color{magenta}{\text{- - - (B)}}}    &{} \\
{\implies}    &{\sin^{-1} \left(\frac{1}{x} \right)}    & {~=~}    &{-y}    &{} \\
{\implies}    &{\csc^{-1} x }    & {~=~}    &{-y~\color{magenta}{\text{- - - (C)}}}    &{} \\
{\implies}    &{\csc^{-1} x}    & {~=~}    &{-\csc^{-1} \left(-x \right)~\color{magenta}{\text{- - - (D)}}}    &{} \\
{\implies}    &{-\csc^{-1} x}    & {~=~}    &{\csc^{-1} \left(-x \right)}    &{} \\
\end{array}$

◼ Remarks:
• In line (A), we assume that $\csc^{-1}(-x)$ = y
• In line B, we use the identity 1: sin (-𝜃) = - sin 𝜃
    ♦ List of identities can be seen here.
• In line C, we use part (i) of property I
• In line D, we substitute for y, using the assumption in (A).


Property III
• This has 3 parts:
$\begin{array}{cc}{}    &{\text{(i)}}    &{\cos^{-1}\left(-x \right)}    {}={}    &{\pi-\cos^{-1}x},    &{x \in [-1,1]}\\
{}    &{\text{(ii)}}    &{\sec^{-1}\left(-x \right)}    {}={}    &{\pi-\sec^{-1}x},    &{|x| \ge 1}\\
{}    &{\text{(iii)}}    &{\cot^{-1}\left(-x \right)}    {}={}    &{\pi-\cot^{-1}x},    &{x \in R}\\
\end{array}$

• Let us prove part (i):


◼ Remarks:
• In line (A), we assume that $\cos^{-1}(-x)$ = y
• In line B, we use the identity 1: cos (π-𝜃) = - cos 𝜃
    ♦ List of identities can be seen here.
• In line C, we substitute for y, using the assumption in (A).

• Let us prove part (ii):


◼ Remarks:
• In line (A), we assume that $\sec^{-1}(-x)$ = y
• In line B, we use the identity 1: cos (π-𝜃) = - cos 𝜃
    ♦ List of identities can be seen here.
• In line C, we use part (ii) of property I
• In line D, we substitute for y, using the assumption in (A).

• Let us prove part (iii):


◼ Remarks:
• In line (A), we assume that $\cot^{-1}(-x)$ = y
• In line B, we use the identities 9(c) and 9(d) to get: cot (π-𝜃) = - cot 𝜃
    ♦ List of identities can be seen here.
• In line C, we substitute for y, using the assumption in (A).


In the next section, we will see fourth and fifth properties.

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