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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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