trigonometry expression simplification with inverse cosine While working on a problem, I ended up with this expression for y:
$$
y=x\sin\left(\arccos\left(\frac{\sqrt{x^2-y^2}}x\right)\right)
$$
Is there any way to express $y$ in terms of $x$ only, with no $y$ on RHS?
Thanks.
 A: Start by applying $\sin(\arccos(t)) = \sqrt{1-t^2}$. Then it should be clear that this is a tautology $y=y$, perhaps up to sign and up to choice of branch of $\arccos$.
A: With $$ \cos^{-1} x = \sin ^{-1} \sqrt { 1- x^2}  $$
you have $$ y^2 = x ( x - x^2 + y^2) $$ which you can simplify.
EDIT1:
After Latex correction indicated by RobJohn we have the identity $ y= y $
So it is true for any or all values of $x,y$.
A: $$y=x\sin\left(\cos^{-1}\left(\frac{\sqrt{x^2-y^2}}x\right)\right)=$$
$$y=x\left(\sin\left(\cos^{-1}\left(\frac{\sqrt{x^2-y^2}}x\right)\right)\right)=$$
$$y=x\left(\sqrt{1-\left(\frac{\sqrt{x^2-y^2}}x\right)^2}\right)=$$
$$y=x\sqrt{1-\left(\frac{\sqrt{x^2-y^2}}x\right)^2}=$$
$$y=x\sqrt{1-\frac{x^2-y^2}{x^2}}$$

$$y=x\sqrt{1-\frac{x^2-y^2}{x^2}}\Longleftrightarrow$$
$$y^2=x^2\left(1-\frac{x^2-y^2}{x^2}\right)\Longleftrightarrow$$
$$y^2=x^2-\frac{x^2\left(x^2-y^2\right)}{x^2}\Longleftrightarrow$$
$$y^2=x^2-\frac{x^4-x^2y^2}{x^2}\Longleftrightarrow$$
$$y^2=x^2-\left(x^2-y^2\right)\Longleftrightarrow$$
$$y^2=x^2-x^2+y^2\Longleftrightarrow$$
$$y^2=0+y^2\Longleftrightarrow$$
$$y^2=y^2\Longleftrightarrow$$
$$y=y$$
So it's true foe every value of $x$ and $y$
A: Your equation is
$$y=x\sin\left(\arccos\left(\frac{\sqrt{x^2-y^2}}x\right)\right)$$
So we must have $0 \le |y| \le |x|$ and $x \ne 0$. We can rewrite your equation as
$$ \frac y x =\sin \arccos {\sqrt{1-\left(\frac y x \right)^2}} $$
Since the range of $\arccos$ is $[0, \pi]$, $\dfrac y x$ must be non-negative:
$\require{AMScd}$
\begin{CD}
    [-1,1] @>\arccos>> [0, \pi]@>\sin>>[0,1]\\
    \end{CD}
When $\dfrac y x \ge 0$, the equation is an identity.
When $\dfrac y x \lt 0$, the equation is false.
So the answer is no, there is no way to express $y$ as a function of $x$.
