# Geometric identity, cannot show equivalence using trigonometric identities

clearly

$$(x+a \cos\theta)^2+(y-a \sin\theta)^2=b^2$$

expanding and using the Weierstrass substitution we find that

$$\theta= 2 \arctan \frac{\left( 2ay- \sqrt{ 4a^2y^2 - ( (x-a)^2+y^2-b^2)( (x+a)^2+y^2-b^2) }\right)}{(x-a)^2+y^2-b^2}$$

if we use the law of cosines, with $c^2=x^2+y^2$

$$\theta = \arctan_2(y,x) - \arccos( (c^2+a^2-b^2) / (2ac) )$$

Is there a way to pass from one expression to the other using trigonometric identities?

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What do you mean by arctan2(y,x) – Amr Nov 13 '12 at 16:44
– Henry Nov 13 '12 at 16:45
an extended version of the tangent that returns the correct angle (between -pi and pi instead of -pi/2.. pi/2) considering the signs of y and x en.wikipedia.org/wiki/Atan2 – Fabio Dalla Libera Nov 13 '12 at 16:47

Try this: $$(x+a \cos\theta)^2+(y-a \sin\theta)^2=b^2$$ Working through we get $$x\cos\theta-y\sin\theta=\frac {(b^2-a^2-y^2-x^2)}{2a}$$
Now use $x^2+y^2=c^2$ on both sides and set $\phi=\arctan \frac y x$. Divide through by $c$ to obtain: $$\cos \phi \cos \theta - \sin\phi\sin\theta =\cos(\theta+\phi)=\frac {(b^2-a^2-c^2)}{2ac}$$ and you can do it from there.
You could look at the angles at the vertex between sides $c$ and $a$. The arctan comes from the $x,y$ triangle and the arccos from the $abc$ one.