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We will learn how to prove the property of the inverse trigonometric function arcsin (x) + arcsin(y) = arcsin (x √1−y2 + y√1−x2)
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 √1−sin2β + √1−sin2α sin β
⇒ sin (α + β) = x ∙ √1−y2 + √1−x2 ∙ y
Therefore, α + β = sin−1 (x √1−y2 + y√1−x2)
or, sin−1 x + sin−1 y = sin−1 (x √1−y2 + y√1−x2). Proved.
Note: If x > 0, y > 0 and x2 + y2 > 1, then the sin−1 x + sin−1 y may be an angle more than π/2 while sin−1 (x √1−y2 + y√1−x2), is an angle between – π/2 and π/2.
Therefore, sin−1 x + sin−1 y = π - sin−1 (x √1−y2 + y√1−x2)
1. Prove that sin−1 35 + sin−1 817 = sin−1 7785
Solution:
L. H. S. = sin−1 35 + sin−1 817
Now, we will apply the formula sin−1 x + sin−1 y = sin−1 (x √1−y2 + y√1−x2)
= sin−1 (35 √1−(817)2 + 817√1−(35)2)
= sin−1 (35 × 1517 + 817 × 45)
= sin−1 7785 = R. H. S. Proved.
2. Show that, sin−1 45 + sin−1 513 + sin−1 1665 = π2.
Solution:
L. H. S. = (sin−145 + sin−1513) + sin−11665
Now, we will apply the formula sin−1 x + sin−1 y = sin−1 (x√1−y2 + y√1−x2)
= sin−1 (45 √1−(513)2 + 513√1−(45)2 + sin−11665
= sin−1 (45 × 1213 + 513 × 35) + sin−11665
= sin−1 6365 + sin−11665
= sin−1 6365 + cos−16365, [Since, sin−1 1665 = cos−1 6365]
= π2, [Since, sin−1 x + cos−1 x = π2] = R. H. S. Proved.
Note: sin−1 = arcsin (x)
● Inverse Trigonometric Functions
11 and 12 Grade Math
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