# 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.

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$$\sqrt{1 - x^{2}}$$)

1. Prove that sin$$^{-1}$$ $$\frac{3}{5}$$ + sin$$^{-1}$$ $$\frac{8}{17}$$ = sin$$^{-1}$$ $$\frac{77}{85}$$

Solution:

L. H. S. = sin$$^{-1}$$ $$\frac{3}{5}$$ + sin$$^{-1}$$ $$\frac{8}{17}$$

Now, we will apply the formula sin$$^{-1}$$ x + sin$$^{-1}$$ y = sin$$^{-1}$$ (x $$\sqrt{1 - y^{2}}$$ + y$$\sqrt{1 - x^{2}}$$)

= sin$$^{-1}$$ ($$\frac{3}{5}$$ $$\sqrt{1 - (\frac{8}{17})^{2}}$$ + $$\frac{8}{17}$$$$\sqrt{1 - (\frac{3}{5})^{2}}$$)

= sin$$^{-1}$$ ($$\frac{3}{5}$$ ×  $$\frac{15}{17}$$ + $$\frac{8}{17}$$ ×  $$\frac{4}{5}$$)

= sin$$^{-1}$$  $$\frac{77}{85}$$ = R. H. S.                  Proved.

2. Show that, sin$$^{-1}$$ $$\frac{4}{5}$$ + sin$$^{-1}$$ $$\frac{5}{13}$$ + sin$$^{-1}$$ $$\frac{16}{65}$$ = $$\frac{π}{2}$$.

Solution:

L. H. S. = (sin$$^{-1}$$$$\frac{4}{5}$$ + sin$$^{-1}$$$$\frac{5}{13}$$) + sin$$^{-1}$$$$\frac{16}{65}$$

Now, we will apply the formula sin$$^{-1}$$ x + sin$$^{-1}$$ y = sin$$^{-1}$$ (x$$\sqrt{1 - y^{2}}$$ + y$$\sqrt{1 - x^{2}}$$)

= 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$$^{-1}$$$$\frac{16}{65}$$

= sin$$^{-1}$$ $$\frac{63}{65}$$ + cos$$^{-1}$$$$\frac{63}{65}$$, [Since, sin$$^{-1}$$ $$\frac{16}{65}$$ = cos$$^{-1}$$ $$\frac{63}{65}$$]

= $$\frac{π}{2}$$, [Since, sin$$^{-1}$$ x + cos$$^{-1}$$ x = $$\frac{π}{2}$$] = R. H. S.            Proved.

Note: sin$$^{-1}$$ = arcsin (x)

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Inverse Trigonometric Functions