18.10 - Property VI

In the previous section, we saw the fourth and fifth properties of inverse trigonometric functions. In this section, we will see the sixth property.

Property VI
• This has 3 parts:

$\begin{array}{cc}{}    &{\text{(i)}}    &{2 \tan^{-1}x}    &{}={}    &{\sin^{-1}\frac{2x}{1+x^2}},    &{|x| \le 1}\\ {}    &{\text{(ii)}}    &{2\tan^{-1}x}    &{}={}    &{\cos^{-1}\frac{1-x^2}{1+x^2}},    &{x \ge 0}\\ {} &{\text{(iii)}}    &{2\tan^{-1}x}    &{}={}    &{\tan^{-1}\frac{2x}{1-x^2}},    &{-1 < x < 1}\\ \end{array}$

• Let us prove part (i):

◼ Remarks:
• In line A, we assume that $\tan^{-1}(x)$ = y
• In line B, we use the identity 15.
   ♦ List of identities can be seen here.
• In line C, we substitute for y from A.

• Let us prove part (ii):

◼ Remarks:
• In line A, we assume that $\tan^{-1}(x)$ = y
• In line B, we use the identity 14.
   ♦ List of identities can be seen here.
• In line C, we substitute for y from A.

• Let us prove part (iii). We already saw the proof in property V. Here we will see an alternate method:

◼ Remarks:
• In line A, we assume that $\tan^{-1}(x)$ = y
• In line B, we use the identity 16.
   ♦ List of identities can be seen here.
• In line C, we substitute for y from A.

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Now we will see some solved examples.

Solved example 18.3
Show that
$\begin{array}{cc}{}    &{\text{(i)}}    &{\sin^{-1} \left(2x{\sqrt{1-x^2}} \right)}    &{}={}    &{2\sin^{-1}x},    &{-{\frac{1}{\sqrt2}} \le x \le \frac{1}{\sqrt2}}\\ {}    &{\text{(ii)}}    &{\sin^{-1} \left(2x{\sqrt{1-x^2}} \right)}    &{}={}    &{2\cos^{-1}x},    &{-{\frac{1}{\sqrt2}} \le x \le 1}\\ {}     \end{array}$
Solution:

Part (i):

◼ Remarks:
• Line 1: We assume that $\sin^{-1}(x)$ = y
• Line 3: We use the identity sin²𝜃 + cos²𝜃 = 1.
• Line 6: We use the identity 15.
   ♦ List of identities can be seen here.
• Line 8: We take the inverse.
• Line 9: We substitute for y based on the assumption in (1).

Part (ii):

 

◼ Remarks:
• Line 1: We assume that $\cos^{-1}(x)$ = y
• Line 3: We use the identity sin²𝜃 + cos²𝜃 = 1.
• Line 6: We use the identity 15.
   ♦ List of identities can be seen here.
• Line 8: We take the inverse.
• Line 9: We substitute for y based on the assumption in (1).

Solved example 18.4
Show that $\tan^{-1}\frac{1}{2} + \tan^{-1}\frac{2}{11} = \tan^{-1}\frac{3}{4}$
Solution:

◼ Remarks:
• Line 1: We use part 2 of property 5.
• Line 2: We substitute the known values.

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In the next section, we will see a few more solved examples.

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