Question

What is the solution of Cos ( 2 sin^-1X) ?

Answer 1

AnswEr :

Given Expression,

$\sf \: cos(2 {sin}^{ - 1} x)$

Let us assume that,

$\sf \theta = {sin}^{ - 1} x \\ \\ \implies \: \sf \: x = sin \theta \: , \theta \in [ - \dfrac{\pi}{2} , \dfrac{\pi}{2} ]$

The expression can be re-written as :

$\sf \longrightarrow \: cos(2 \theta) \\ \\ \longrightarrow \: \sf \: cos {}^{2} \theta - sin {}^{2} \theta \\ \\ \longrightarrow \: \sf \: 1 - 2 {sin}^{2} \theta \\ \\ \longrightarrow \: \sf \:1 - 2sin {}^{2} (\theta) \\ \\ \large{\longrightarrow \: \underline{ \boxed{ \sf \: 1 - 2x^2} }}$

The solution of cos(arcsin(x)) would be 1 - 2x².

Answer 2

Question:

$What \: is \: the \: solution \: of \: the \: expression \: cos(2 {sin}^{ - 1} x)?$

To find:

$To \: find \: the \: solution \: of \: the \: expression.$

Answer:

$The \: solution \: is \: \underline\bold{1 - 2 {x}^{2} }$

Step-by-step explanation:

$cos(2 {sin}^{ - 1} x)$

Now Assuming the value of θ,

$\longrightarrow θ = {sin}^{ - 1} x$

$\longrightarrow\underline\bold{x = sinθ}$

$θ∈( - \frac{\pi}{2} , \frac{\pi}{2} )$

The expression can also be written as,

$\longrightarrow cos2θ$

$= {cos}^{2}θ - 2{sin}^{2}θ$

$= 1 - 2 {sin}^{2} θ$

$\boxed{= 1 - 2 {x}^{2}}$

$Solution \: of \: the \: expression \: is \: \underline\bold{1 - 2 {x}^{2} }$