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.
 A: 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?
A: Here's a trigonograph (with $a$ and $b$ reversed, just because):

$$\begin{align}
2\;\angle X &\;=\; \stackrel{\frown}{PQ} - \stackrel{\frown}{RS} \\[6pt]
\implies \qquad\qquad\qquad 2\;\gamma &\;=\; \alpha - \beta \\[6pt]
\implies \qquad 2\,\arcsin \frac{a-b}{2\sqrt{c}} &\;=\; \arcsin\frac{a}{c} - \arcsin\frac{b}{c}
\end{align}$$
A: *

*$\arcsin \dfrac bc -\arcsin \dfrac ac =2\arcsin \dfrac{b-a}{c \sqrt 2}$

*$a,b,c \gt 0$

*$a^2 + b^2 = c^2$
Let's rewrite this as
$\dfrac 12 \left(\arcsin \dfrac bc -\arcsin \dfrac ac \right) 
 =\arcsin \dfrac{b-a}{c \sqrt 2}$
Since you said that sides $a, b, $ and $c$ are sides of a right triangle, then we know that $a,b,c \gt 0$.
Let $\theta$ be the angle that corresponds to the point $(a,b)$ in the first quadrant.
Then $\sin \theta= \dfrac bc$ and $\cos \theta = \dfrac ac$
So $\dfrac 12 \left( \arcsin \dfrac bc -\arcsin \dfrac ac \right) = 
    \dfrac 12 \left( \theta - \left(\dfrac{\pi}{2} - \theta\right) \right) =
    \theta - \dfrac{\pi}{4}$
If you can show that $\sin \left( \theta - \dfrac{\pi}{4} \right)
                     = \dfrac{b-a}{c \sqrt 2}$,
then you are done.
