# 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 + $$\sqrt{1 - x^{2}}$$$$\sqrt{1 - y^{2}}$$)

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Inverse Trigonometric Functions

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