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

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