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.

● Inverse Trigonometric Functions

• General and Principal Values of sin\(^{-1}\) x
• General and Principal Values of cos\(^{-1}\) x
• General and Principal Values of tan\(^{-1}\) x
• General and Principal Values of csc\(^{-1}\) x
• General and Principal Values of sec\(^{-1}\) x
• General and Principal Values of cot\(^{-1}\) x
• Principal Values of Inverse Trigonometric Functions
• General Values of Inverse Trigonometric Functions
  
• arcsin(x) + arccos(x) = \(\frac{π}{2}\)
• arctan(x) + arccot(x) = \(\frac{π}{2}\)
• arctan(x) + arctan(y) = arctan(\(\frac{x + y}{1 - xy}\))
• arctan(x) - arctan(y) = arctan(\(\frac{x - y}{1 + xy}\))
• arctan(x) + arctan(y) + arctan(z)= arctan\(\frac{x + y + z – xyz}{1 – xy – yz – zx}\)
• arccot(x) + arccot(y) = arccot(\(\frac{xy - 1}{y + x}\))
• arccot(x) - arccot(y) = arccot(\(\frac{xy + 1}{y - x}\))
• arcsin(x) + arcsin(y) = arcsin(x \(\sqrt{1 - y^{2}}\) + y\(\sqrt{1 - x^{2}}\))
• arcsin (x) - arcsin(y) = arcsin (x \(\sqrt{1 - y^{2}}\) - y\(\sqrt{1 - x^{2}}\))
• arccos (x) + arccos(y) = arccos(xy - \(\sqrt{1 - x^{2}}\)\(\sqrt{1 - y^{2}}\))
• arccos(x) - arccos(y) = arccos(xy + \(\sqrt{1 - x^{2}}\)\(\sqrt{1 - y^{2}}\))
• 2 arcsin(x) = arcsin(2x\(\sqrt{1 - x^{2}}\)) 
• 2 arccos(x) = arccos(2x\(^{2}\) - 1)
• 2 arctan(x) = arctan(\(\frac{2x}{1 - x^{2}}\)) = arcsin(\(\frac{2x}{1 + x^{2}}\)) = arccos(\(\frac{1 - x^{2}}{1 + x^{2}}\))
• 3 arcsin(x) = arcsin(3x - 4x\(^{3}\))
• 3 arccos(x) = arccos(4x\(^{3}\) - 3x)
• 3 arctan(x) = arctan(\(\frac{3x - x^{3}}{1 - 3 x^{2}}\))
• Inverse Trigonometric Function Formula
• Principal Values of Inverse Trigonometric Functions
• Problems on Inverse Trigonometric Function

11 and 12 Grade Math

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