Solving an Equation with Sin -1 and Sin: Confusion and Help Needed

[DGriffiths]

this example was in a book I bought ( maths methods for physics, Mathews and Walker)

dy/dx + sqrt( (1-y^2) / (1-x^2) ) = 0

dy/(1-y^2) + dx/(1-x^2) = 0

sin -1 y + sin -1 x = c [1]

or, taking the sine of both sides

x (1-y^2)^1/2 + y (1-x^2)^1/2 = sin c [2]



My response/confusion with this...

[1] Ok so I'm fine with this, standard integral

[2] ?Where on Earth did this come from
just by taking the sin of the equation
above.

Please help if you can.

 

[lurflurf]

use from elementary trigonometry
$$\sin(x+y)=\sin(x)\cos(y)+\cos(x)\sin(y)$$
and
$$\sin( \cos^{-1}(x))=\sqrt{1-x^2}$$

 

[HallsofIvy]

Because $sin^2(y)+ cos^2(y)= 1$, $sin(y)= \sqrt{1- cos^2(y)}$ and so if x= cos(y) (so that $y= cos^{-1}(x))$, then $sin(cos^{-1}(x))= sin(y)= \sqrt{1- cos^2(y)}= \sqrt{1- x^2}$.

Another way of seeing this is to think of $cos^{-1}(x)$ as representing the angle in a right triangle that has cosine equal to x. Since "cosine= near side over hypotenuse" and x= x/1, that right triangle can have "near side" of length x and hypotenuse 1. By the Pythagorean theorem, the third side, the "opposite side" to the angle has length $\sqrt{1- x^2}$ and so the sine of that angle, $sin(cos^{-1}(x)$, is $\sqrt{1- x^2}/1= \sqrt{1- x^2}$.

 

[DGriffiths]

Thanks folks for your insight really appreciate it.