arcsin (x) - arcsin(y) = arcsin (x $\sqrt{1 - y^{2}}$ - y $\sqrt{1 - x^{2}}$ )

We will learn how to prove the property of the inverse trigonometric function arcsin (x) - arcsin(y) = arcsin (x $\sqrt{1 - y^{2}}$ - y $\sqrt{1 - x^{2}}$ )

Proof:

Let, sin $^{-1}$ x = α and sin $^{-1}$ y = β

From sin $^{-1}$ x = α we get,

x = sin α

and from sin $^{-1}$ y = β we get,

y = sin β

Now, sin (α - β) = sin α cos β - cos α sin β

⇒ sin (α - β) = sin α $\sqrt{1 - sin^{2} β}$ - $\sqrt{1 - sin^{2} α}$ sin β

⇒ sin (α - β) = x ∙ $\sqrt{1 - y^{2}}$ - $\sqrt{1 - x^{2}}$ ∙ y

Therefore, α - β = sin $^{-1}$ (x $\sqrt{1 - y^{2}}$ - y $\sqrt{1 - x^{2}}$ )

or, sin $^{-1}$ x - sin $^{-1}$ y = sin $^{-1}$ (x $\sqrt{1 - y^{2}}$ - y $\sqrt{1 - x^{2}}$ ). Proved.

If x > 0, y > 0 and x $^{2}$ + y $^{2}$ > 1, then the sin $^{-1}$ x + sin $^{-1}$ y may be an angle more than π/2 while sin $^{-1}$ (x $\sqrt{1 - y^{2}}$ + y $\sqrt{1 - x^{2}}$ ), is an angle between – π/2 and π/2.

Therefore, sin $^{-1}$ x - sin $^{-1}$ y = π - sin $^{-1}$ (x $\sqrt{1 - y^{2}}$ + y $\sqrt{1 - x^{2}}$ )

● Inverse Trigonometric Functions

11 and 12 Grade Math

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arcsin (x) - arcsin(y) = arcsin (x √1−y2\sqrt{1 - y^{2}} - y√1−x2\sqrt{1 - x^{2}})

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arcsin (x) - arcsin(y) = arcsin (x $\sqrt{1 - y^{2}}$ - y $\sqrt{1 - x^{2}}$ )