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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