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.
Note: 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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