How can I Show that, $\arcsin\left(\frac{b}{c}\right)-\arcsin\left(\frac{a}{c}\right)=2\arcsin\left(\frac{b-a}{c\sqrt2}\right)$ Where $c^2=a^2+b^2$ is Pythagoras theorem. Sides a,b and c are of a right angle triangle.
Show that,
$$\arcsin\left(\frac{b}{c}\right)-\arcsin\left(\frac{a}{c}\right)=2\arcsin\left(\frac{b-a}{c\sqrt2}\right)$$ 
How do I go about proving this identity? Can anybody help? I know of the arctan.

HINT:
$$\dfrac a{\sin A}=\dfrac b{\sin B}=\dfrac c1$$
$\implies\arcsin\dfrac ac=\arcsin(\sin A)=A$  as $0<A<\dfrac\pi2$
Now $A=\dfrac\pi2-B$
and  $\dfrac{b-a}{\sqrt2c}=\dfrac{\sin B-\sin A}{\sqrt2}=\dfrac{\sin B-\cos B}{\sqrt2}=\sin\left(B-\dfrac\pi4\right)$
Can you take it from here?
