Question

Sin^-1 (x+√(1-x²))/√2
​

Answer 1

GIVEN :-

$\\ \bigstar \displaystyle \sf \: \sin ^{ - 1} \bigg( \frac{x + \sqrt{1 - x ^{2} } }{ \sqrt{2} } \bigg)$

TO FIND :-

$\\ \bigstar \: \displaystyle \sf value \: of\: \sin ^{ - 1} \bigg( \frac{x + \sqrt{1 - x ^{2} } }{ \sqrt{2} } \bigg)$

SOLUTION :-

$\\ \dashrightarrow\displaystyle \sf \sin ^{ - 1} \bigg( \frac{x + \sqrt{1 - x ^{2} } }{ \sqrt{2} } \bigg)$

Now let the function,

$\\ \\ \dashrightarrow\displaystyle \sf y = \sin ^{ - 1} \bigg( \frac{x + \sqrt{1 - x ^{2} } }{ \sqrt{2} } \bigg)$

$\bullet\displaystyle \sf \: x = \sin \beta$

$\bullet\displaystyle \sf \: \beta = \dfrac{1}{ \sin} x$

$\bullet\displaystyle \sf \: \beta = \sin ^{ - 1} x$

$\dashrightarrow\displaystyle \sf \: y = \sin ^{ - 1} \bigg( \frac{ \sin \beta + \sqrt{1 - \sin ^{2} \beta } }{ \sqrt{2} } \bigg)$

$\dashrightarrow\displaystyle \sf \: y = \sin ^{ - 1} \bigg( \frac{ \sin \beta + \sqrt{ \cos ^{2} \beta } }{ \sqrt{2} } \bigg)$

$\dashrightarrow\displaystyle \sf \: y = \sin ^{ - 1} \bigg( \frac{ \sin \beta + \cos\beta }{ \sqrt{2} } \bigg)$

$\dashrightarrow\displaystyle \sf \: y = \sin ^{ - 1} \Bigg[ \bigg( \sin \beta + \cos\beta \bigg)\Bigg] \dfrac{1}{ \sqrt{2} }$

$\dashrightarrow\displaystyle \sf \: y = \sin ^{ - 1} \bigg( \sin \beta \frac{1}{ \sqrt{2} } + \cos\beta \frac{1}{ \sqrt{2} } \bigg)$

$\dashrightarrow\displaystyle \sf \: y = \sin ^{ - 1} \bigg( \sin \beta \frac{\pi}{4} + \cos\beta \frac{\pi}{4} \bigg) \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bigg \lgroup \frac{1}{ \sqrt{2} } =\cos\frac{\pi}{4} \bigg \rgroup$

$\dashrightarrow\displaystyle \sf \: y = \sin ^{ - 1} \Bigg[ \sin \beta . \cos \bigg( \frac{\pi}{4} \bigg) + \cos \beta . \sin \bigg( \frac{\pi}{4} \bigg) \Bigg]$

$\dashrightarrow\displaystyle \sf \: y = \sin ^{ - 1} \Bigg[ \sin \bigg( \beta + \frac{\pi}{4} \bigg)\Bigg]$

$\dashrightarrow\displaystyle \sf \: y = \dfrac{1}{\sin } \Bigg[\sin \bigg( \beta + \frac{\pi}{4} \bigg)\Bigg]$

$\dashrightarrow\displaystyle \sf \: y = \beta + \frac{\pi}{4}$

$\dashrightarrow\displaystyle \sf \: y = \sin ^{ - 1} x + \frac{\pi}{4}$

$\dashrightarrow\underline{ \boxed{\displaystyle \sf\sin ^{ - 1} \bigg( \frac{x + \sqrt{1 - x ^{2} } }{ \sqrt{2} } \bigg) = sin ^{ -1}x + \frac{\pi}{4} }}$

Answer 2

Answer :

⇝$\bold{dy\:/\:dy\:=\:1\:/\:√1\:-\:x^2}$

Explanation :

According to the question :

↦Let y = $sin^{-1}\:[{x\:+\:√1\:-\:x^2\:/\:√2\:}]$

↦Put x = sin θ , so

⟹ $sin^{-1}\:[{sin\:θ\:+\:√1\:-\:sin^2\:θ\:/\:√2\:}]$

⟹ $sin^{-1}\:[{sin\:θ\:+\:cos\:θ\:/\:√2\:}]$

⟹ $sin^{-1}\:{[sin\:θ\:(\:1\:/\:√2\:)\:+\:cos\:θ\:(\:1\:/\:√2\:)\:}]$

⟹ $sin^{-1}\:[{sin\:θ\:cos\:π/4\:+\:cos\:θ\:sin\:π/4\:}]$

⟹ $y\:=\:sin^{-1}\:[{\:sin\:(\:θ\:+\:π/4\:)\:}]$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

Here,⇛$\:-1\:<\:x\:<\:1$

⇛$\:-1\:<\:sin\:θ\:<\:1$

⇛$-π\:/\:2\:<\:θ\:<\:π\:/\:2$

⇛$( -π\:/\:2\:+\:π\:/\:4 )\:<\:( π\:/\:4\:+\:θ )\:<\:3π\:/\:4$

From Equation 1 :

➳ $y\:=\:θ\:+\:π\:/\:4$ $[sin^{-1}\:(\:sin\:θ\:)\:=\:θ\:)]$

➳ $y\:=\:sin^{-1}\:x\:+\:π\:/\:4$ $[sin\:θ\:=\:x]$

Differentiating it with respect to x :

⇝$dx\:/\:dy\:=\:1\:/\:√1\:-\:x^2\:+\:0$

⇝$\bold{dy\:/\:dy\:=\:1\:/\:√1\:-\:x^2}$

So, It's Done !!