# inverse trig identity

#### rww88

Is there an identity for the expression: [arccos (x/z) - arccos (y/z)] ? or can the expression be otherwise simplified?

#### gelatine1

Let's write $$\displaystyle x/z=u$$ and $$\displaystyle y/z=w$$ for simplicity.

Suppose $$\displaystyle \arccos(u)-\arccos(w)=t$$
then $$\displaystyle \cos(\arccos(u)-\arccos(w))=\cos(t)$$
and thus by sum formula for cosinus we have $$\displaystyle uw+\sin(\arccos(u))\sin(\arccos(w))=\cos(t)$$
and now because $$\displaystyle \sin(\arccos(u))=\sqrt{1-u^2}$$
we have $$\displaystyle uw+\sqrt{1-u^2}\sqrt{1-w^2}=\cos(t)$$
so $$\displaystyle t = \arccos(uw+\sqrt{1-u^2}\sqrt{1-w^2})$$

Now replace u and w by their original values again...

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