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

Solved examples on property of inverse circular function arccos (x) + arccos(y) = arccos(xy - $$\sqrt{1 - x^{2}}$$$$\sqrt{1 - y^{2}}$$)

1. If cos$$^{-1}$$$$\frac{x}{a}$$ + cos$$^{-1}$$$$\frac{y}{b}$$ = α prove that,

$$\frac{x^{2}}{a^{2}}$$ - $$\frac{2xy}{ab}$$ cos α + $$\frac{y^{2}}{b^{2}}$$ = sin$$^{2}$$ α.

Solution:

L. H. S. = cos$$^{-1}$$$$\frac{x}{a}$$ + cos$$^{-1}$$$$\frac{y}{b}$$ = α

We have, cos$$^{-1}$$ x - cos$$^{-1}$$ y = cos$$^{-1}$$(xy - $$\sqrt{1 - x^{2}}$$$$\sqrt{1 - y^{2}}$$)

⇒ cos$$^{-1}$$ [$$\frac{x}{a}$$ · $$\frac{y}{b}$$ - $$\sqrt{1 - \frac{x^{2}}{a^{2}}}$$ $$\sqrt{1 - \frac{y^{2}}{b^{2}}}$$] = α

⇒ $$\frac{xy}{ab}$$ - $$\sqrt{(1 - \frac{x^{2}}{a^{2}})(1 - \frac{y^{2}}{b^{2}})}$$ = cos α

⇒ $$\frac{xy}{ab}$$ - cos α = $$\sqrt{(1 - \frac{x^{2}}{a^{2}})(1 - \frac{y^{2}}{b^{2}})}$$

⇒ ($$\frac{xy}{ab}$$ - cos α)$$^{2}$$ = $$(1 - \frac{x^{2}}{a^{2}})(1 - \frac{y^{2}}{b^{2}})$$, (squaring both the sides)

⇒ $$\frac{x^{2}y^{2}}{a^{2}b^{2}}$$  - 2$$\frac{xy}{ab}$$cos α + cos$$^{2}$$ α = 1 - $$\frac{x^{2}}{a^{2}}$$ - $$\frac{y^{2}}{b^{2}}$$ +  $$\frac{x^{2}y^{2}}{a^{2}b^{2}}$$

⇒ $$\frac{x^{2}}{a^{2}}$$ - - 2$$\frac{xy}{ab}$$cos α + cos$$^{2}$$ α + $$\frac{y^{2}}{b^{2}}$$ = 1 - cos$$^{2}$$ α

⇒ $$\frac{x^{2}}{a^{2}}$$ - - 2$$\frac{xy}{ab}$$cos α + cos$$^{2}$$ α + $$\frac{y^{2}}{b^{2}}$$ = sin$$^{2}$$ α.            Proved.

2. If cos$$^{-1}$$ x + cos$$^{-1}$$ y + cos$$^{-1}$$ z = π, prove that x$$^{2}$$ + y$$^{2}$$ + z$$^{2}$$ + 2xyz = 1.

Solution:

cos$$^{-1}$$ x + cos$$^{-1}$$ y + cos$$^{-1}$$ z = π

⇒ cos$$^{-1}$$ x + cos$$^{-1}$$ y = π - cos$$^{-1}$$ z

⇒ cos$$^{-1}$$ x + cos$$^{-1}$$ y = cos$$^{-1}$$ (-z), [Since, cos$$^{-1}$$ (-θ) = π - cos$$^{-1}$$ θ]

⇒ cos$$^{-1}$$(xy - $$\sqrt{1 - x^{2}}$$$$\sqrt{1 - y^{2}}$$) = cos$$^{-1}$$ (-z)

⇒ xy - $$\sqrt{1 - x^{2}}$$$$\sqrt{1 - y^{2}}$$ = -z

⇒ xy + z = $$\sqrt{1 - x^{2}}$$$$\sqrt{1 - y^{2}}$$

Now squaring both sides

⇒ (xy + z)$$^{2}$$ = (1 - x$$^{2}$$)(1 - y$$^{2}$$)

⇒ x$$^{2}$$y$$^{2}$$ + z$$^{2}$$ + 2xyz = 1 - x$$^{2}$$ - y$$^{2}$$ + x$$^{2}$$y$$^{2}$$

⇒ x$$^{2}$$ + y$$^{2}$$ + z$$^{2}$$ + 2xyz = 1                Proved.

Inverse Trigonometric Functions