arcsin (x) + arcsin(y) = arcsin (x \(\sqrt{1 - y^{2}}\) + y\(\sqrt{1 - x^{2}}\))

We will learn how to prove the property of the inverse trigonometric function arcsin (x) + arcsin(y) = arcsin (x \(\sqrt{1 - y^{2}}\) + y\(\sqrt{1 - x^{2}}\))

Proof:  

Let, sin\(^{-1}\) x = α and sin\(^{-1}\) y = β

From sin\(^{-1}\) x = α we get,

x = sin α

and from sin\(^{-1}\) y = β we get,

y = sin β

Now, sin (α + β) = sin α cos β + cos α sin β

sin (α + β) = sin α \(\sqrt{1 - sin^{2} β}\) + \(\sqrt{1 - sin^{2} α}\) sin β

sin (α + β) = x ∙ \(\sqrt{1 - y^{2}}\) + \(\sqrt{1 - x^{2}}\) ∙ y

Therefore, α + β = sin\(^{-1}\) (x \(\sqrt{1 - y^{2}}\) + y\(\sqrt{1 - x^{2}}\)) 

or, sin\(^{-1}\) x + sin\(^{-1}\) y = sin\(^{-1}\) (x \(\sqrt{1 - y^{2}}\) + y\(\sqrt{1 - x^{2}}\)).       Proved.


Note: If x > 0, y > 0 and x\(^{2}\) + y\(^{2}\) > 1, then the sin\(^{-1}\) x + sin\(^{-1}\) y may be an angle more than π/2 while sin\(^{-1}\) (x \(\sqrt{1 - y^{2}}\) + y\(\sqrt{1 - x^{2}}\)), is an angle between – π/2 and π/2.

Therefore, sin\(^{-1}\) x + sin\(^{-1}\) y = π - sin\(^{-1}\) (x \(\sqrt{1 - y^{2}}\) + y\(\sqrt{1 - x^{2}}\))



1. Prove that sin\(^{-1}\) \(\frac{3}{5}\) + sin\(^{-1}\) \(\frac{8}{17}\) = \(^{-1}\) \(\frac{77}{85}\)

Solution:

L. H. S. = sin\(^{-1}\) \(\frac{3}{5}\) + sin\(^{-1}\) \(\frac{8}{17}\)

Now, we will apply the formula sin\(^{-1}\) x + sin\(^{-1}\) y = sin\(^{-1}\) (x \(\sqrt{1 - y^{2}}\) + y\(\sqrt{1 - x^{2}}\))

= sin\(^{-1}\) (\(\frac{3}{5}\) \(\sqrt{1 - (\frac{8}{17})^{2}}\) + \(\frac{8}{17}\)\(\sqrt{1 - (\frac{3}{5})^{2}}\))

= sin\(^{-1}\) (\(\frac{3}{5}\) ×  \(\frac{15}{17}\) + \(\frac{8}{17}\) ×  \(\frac{4}{5}\))

= sin\(^{-1}\)  \(\frac{77}{85}\) = R. H. S.                  Proved.


2. Show that, sin\(^{-1}\) \(\frac{4}{5}\) + sin\(^{-1}\) \(\frac{5}{13}\) + sin\(^{-1}\) \(\frac{16}{65}\) = \(\frac{π}{2}\).  

Solution:     

L. H. S. = (sin\(^{-1}\)\(\frac{4}{5}\) + sin\(^{-1}\)\(\frac{5}{13}\)) + sin\(^{-1}\)\(\frac{16}{65}\)

Now, we will apply the formula sin\(^{-1}\) x + sin\(^{-1}\) y = sin\(^{-1}\) (x\(\sqrt{1 - y^{2}}\) + y\(\sqrt{1 - x^{2}}\))

= sin\(^{-1}\) (\(\frac{4}{5}\) \(\sqrt{1 - (\frac{5}{13})^{2}}\) + \(\frac{5}{13}\)\(\sqrt{1 - (\frac{4}{5})^{2}}\) + sin\(^{-1}\)\(\frac{16}{65}\)

= sin\(^{-1}\) (\(\frac{4}{5}\) ×  \(\frac{12}{13}\) + \(\frac{5}{13}\) ×  \(\frac{3}{5}\)) + sin\(^{-1}\)\(\frac{16}{65}\)

= sin\(^{-1}\) \(\frac{63}{65}\) + sin\(^{-1}\)\(\frac{16}{65}\)

= sin\(^{-1}\) \(\frac{63}{65}\) + cos\(^{-1}\)\(\frac{63}{65}\), [Since, sin\(^{-1}\) \(\frac{16}{65}\) = cos\(^{-1}\) \(\frac{63}{65}\)]

= \(\frac{π}{2}\), [Since, sin\(^{-1}\) x + cos\(^{-1}\) x = \(\frac{π}{2}\)] = R. H. S.            Proved.






11 and 12 Grade Math

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