arcsin(x) + arccos(x) = \(\frac{π}{2}\)

We will learn how to prove the property of the inverse trigonometric function arcsin(x) + arccos(x) = \(\frac{π}{2}\).

Proof: Let, sin\(^{-1}\) x = θ

Therefore, x = sin θ

x = cos (\(\frac{π}{2}\) - θ), [Since, cos (\(\frac{π}{2}\) - θ) = sin θ]

⇒ cos\(^{-1}\) x = \(\frac{π}{2}\) - θ

⇒ cos\(^{-1}\) x= \(\frac{π}{2}\) - sin\(^{-1}\) x, [Since, θ = sin\(^{-1}\) x]

⇒ sin\(^{-1}\) x + cos\(^{-1}\) x = \(\frac{π}{2}\)

Therefore, sin\(^{-1}\) x + cos\(^{-1}\) x = \(\frac{π}{2}\).             Proved.


Solved examples on property of inverse circular function sin\(^{-1}\) x + cos\(^{-1}\) x = \(\frac{π}{2}\).

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

Solution:

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

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

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

= \(cos^{-1}\sqrt{1 - (\frac{63}{65})^{2}})\) + sin\(^{-1}\) \(\frac{16}{65}\)

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

= π/2, since \(sin^{-1} x + cos^{-1} x = \frac{π}{2}\)

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

 

2. Solve the trigonometric equation: sin\(^{-1}\) \(\frac{5}{x}\) + sin\(^{-1}\) \(\frac{12}{x}\) = \(\frac{π}{2}\)

Solution:

sin\(^{-1}\) \(\frac{5}{x}\) + sin\(^{-1}\) \(\frac{12}{x}\) = \(\frac{π}{2}\)

⇒ sin\(^{-1}\) \(\frac{12}{x}\) = \(\frac{π}{2}\) - sin\(^{-1}\) \(\frac{5}{x}\)

⇒ sin\(^{-1}\) \(\frac{12}{x}\) = cos\(^{-1}\) \(\frac{5}{x}\), [Since, we know that, sin\(^{-1}\) \(\frac{5}{x}\) + cos\(^{-1}\) \(\frac{5}{x}\) = \(\frac{π}{2}\)]

⇒ sin\(^{-1}\) \(\frac{12}{x}\) = sin\(^{-1}\) \(\frac{\sqrt{x^{2} - 25}}{x}\)

⇒ \(\frac{12}{x}\) = \(\frac{\sqrt{x^{2} - 25}}{x}\)

⇒ \(\sqrt{x^{2} - 25}\) = 12, [Since, x ≠ 0]

⇒ x\(^{2}\) - 25 = 144       

⇒ x\(^{2}\) = 144 + 25   

⇒ x\(^{2}\) = 169        

⇒ x = ± 13

The solution x = - 13 does not satisfy the given equation. 

Therefore the required solution is x = 13.








11 and 12 Grade Math

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