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
● Inverse Trigonometric Functions
11 and 12 Grade Math
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