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

From arcsin(x) - arcsin(y) to HOME PAGE




Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.



New! Comments

Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.