arccos(x) - arccos(y) = arccos(xy + \(\sqrt{1 - x^{2}}\)\(\sqrt{1 - y^{2}}\))

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


Proof:  

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

From cos\(^{-1}\) x = α we get,

x = cos α

and from cos\(^{-1}\) y = β we get,

y = cos β

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

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

⇒ cos (α - β) = (xy + \(\sqrt{1 - x^{2}}\)\(\sqrt{1 - y^{2}}\))

⇒ α - β = cos\(^{-1}\)(xy + \(\sqrt{1 - x^{2}}\)\(\sqrt{1 - y^{2}}\))

or, cos\(^{-1}\) x - cos\(^{-1}\) y = cos\(^{-1}\)(xy + \(\sqrt{1 - x^{2}}\)\(\sqrt{1 - y^{2}}\))

Therefore, arccos(x) - arccos(y) = arccos(xy) + \(\sqrt{1 - x^{2}}\)\(\sqrt{1 - y^{2}}\))       Proved.

 

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

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

 Inverse Trigonometric Functions






11 and 12 Grade Math

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