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

• DGriffiths

#### 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 

or, taking the sine of both sides

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

My response/confusion with this...

 Ok so I'm fine with this, standard integral

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

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

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 $\sqrt{1- x^2}/1= \sqrt{1- x^2}$.

Thanks folks for your insight really appreciate it.