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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