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arcsin (x) - arcsin(y) = arcsin (x √1βˆ’y2 - y√1βˆ’x2)

We will learn how to prove the property of the inverse trigonometric function arcsin (x) - arcsin(y) = arcsin (x √1βˆ’y2 - y√1βˆ’x2)

Proof:  

Let, sinβˆ’1 x = Ξ± and sinβˆ’1 y = Ξ²

From sinβˆ’1 x = Ξ± we get,

x = sin Ξ±

and from sinβˆ’1 y = Ξ² we get,

y = sin Ξ²

Now, sin (Ξ± - Ξ²) = sin Ξ± cos Ξ² - cos Ξ± sin Ξ²

β‡’ sin (Ξ± - Ξ²) = sin Ξ± √1βˆ’sin2Ξ² - √1βˆ’sin2Ξ± sin Ξ²

β‡’ sin (Ξ± - Ξ²) = x βˆ™ √1βˆ’y2 - √1βˆ’x2 βˆ™ y

Therefore, Ξ± - Ξ² = sinβˆ’1 (x √1βˆ’y2 - y√1βˆ’x2

or, sinβˆ’1 x - sinβˆ’1 y = sinβˆ’1 (x √1βˆ’y2 - y√1βˆ’x2).       Proved.

 

Note: If x > 0, y > 0 and x2 + y2 > 1, then the sinβˆ’1 x + sinβˆ’1 y may be an angle more than Ο€/2 while sinβˆ’1 (x √1βˆ’y2 + y√1βˆ’x2), is an angle between – Ο€/2 and Ο€/2.

Therefore, sinβˆ’1 x - sinβˆ’1 y = Ο€ - sinβˆ’1 (x √1βˆ’y2 + y√1βˆ’x2)

● Inverse Trigonometric Functions







11 and 12 Grade Math

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