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

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

Proof:  

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

From sin\(^{-1}\) x = α we get,

x = sin α

and from sin\(^{-1}\) y = β we get,

y = sin β

Now, sin (α - β) = sin α cos β - cos α sin β

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

⇒ sin (α - β) = x ∙ \(\sqrt{1 - y^{2}}\) - \(\sqrt{1 - x^{2}}\) ∙ y

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

or, sin\(^{-1}\) x - sin\(^{-1}\) y = sin\(^{-1}\) (x \(\sqrt{1 - y^{2}}\) - y\(\sqrt{1 - x^{2}}\)).       Proved.

 

Note: If x > 0, y > 0 and x\(^{2}\) + y\(^{2}\) > 1, then the sin\(^{-1}\) x + sin\(^{-1}\) y may be an angle more than π/2 while sin\(^{-1}\) (x \(\sqrt{1 - y^{2}}\) + y\(\sqrt{1 - x^{2}}\)), is an angle between – π/2 and π/2.

Therefore, sin\(^{-1}\) x - sin\(^{-1}\) y = π - sin\(^{-1}\) (x \(\sqrt{1 - y^{2}}\) + y\(\sqrt{1 - x^{2}}\))







11 and 12 Grade Math

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