2 arcsin(x) = arcsin(2x\(\sqrt{1 - x^{2}}\))

We will learn how to prove the property of the inverse trigonometric function 2 arcsin(x) = arcsin(2x\(\sqrt{1 - x^{2}}\)) or, 2 sin\(^{-1}\) x = sin\(^{-1}\) (2x\(\sqrt{1 - x^{2}}\)).

Proof:

Let, sin\(^{-1}\) x = α                 

Therefore, sin α = x

Now, sin 2α = 2 sin α cos α

sin 2α = 2 sin α \(\sqrt{1 - sin^{2}α}\)

sin 2α = 2x . \(\sqrt{1 - x^{2}}\)

sin 2α =  2x\(\sqrt{1 - x^{2}}\)

Therefore, 2α = sin\(^{-1}\) (2x\(\sqrt{1 - x^{2}}\))

2 sin\(^{-1}\) x = sin\(^{-1}\) (2x\(\sqrt{1 - x^{2}}\)).     

or, 2 arcsin(x) = arcsin(2x\(\sqrt{1 - x^{2}}\))             Proved






















11 and 12 Grade Math

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