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

From arccos(x) + arccos(y) to HOME PAGE




Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.



New! Comments

Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.