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.






11 and 12 Grade Math

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