Trigonometry: Solve $(1-\cos\alpha)^2 + \sin^2\alpha = d^2$ for $\alpha$ My next step in implementing my algorithm in Java is following.
It is quite difficult to explain, but I know what I need. I have this equation:

Given: d
Asked: $\alpha$
$$(1-\cos\alpha)^2 + \sin^2\alpha = d^2 $$

Which I "simplified" to this, using this formula's:
$$2+2\cos\alpha+\frac{\cos2\alpha}{2}-\frac{\sin\alpha}{2} = d^2$$
But now, I'm stuck. This is probably pretty easy, but I'm 15 years old at the moment. I didn't see that much trigonometry in school yet.
Can you help me?
 A: Make the following manipulations:
$$(1-\cos(\alpha))^2+\sin^2(\alpha)=d^2$$
$$1-2\cos(\alpha)+\cos^2(\alpha)+\sin^2(\alpha)=d^2$$
$$1-2\cos(\alpha)+1=d^2$$
$$2-2\cos(\alpha)=d^2$$
$$\cos(\alpha)=1-\frac{d^2}{2}$$
$$\alpha=\cos^{-1}\left(1-\frac{d^2}{2}\right)$$
A: A slightly more general tip that I sometimes tell my students:
If an equation contains more than one trig function, it might be nice if there were a way to rewrite it using only one trig function.
Sometimes that's easy and sometimes that's hard. But if you see precisely $\sin^2 \alpha$ or $\cos^2 \alpha$, those are very easy to rewrite using another trig function.
(The identity $\sin^2 \alpha + \cos^2 \alpha = 1$ is far and away the most important trig identity of them all.)
A: You could just expand the square... Hint : $\left( 1-\cos \alpha \right)^2 + \sin ^2 \alpha = 1 - 2 \cos \alpha + \cos^2 \alpha + \sin^2 \alpha$, then recall that $\cos^2 \alpha + \sin^2 \alpha = 1$.
A: Write $\sin^{2}x = 1-\cos^{2}{x}$. Then you have 
\begin{align*}
(1-\cos{x})^{2} + \sin^{2}{x} &= (1-\cos{x})^{2} + (1+\cos{x})\cdot (1-\cos{x}) \\ &= (1-\cos{x}) \cdot \Bigl[ 1- \cos{x} + 1 + \cos{x} \Bigr] \\ &= 2\: \sin^{2}\frac{x}{2} \cdot 2 = 4\:\sin^{2} \frac{x}{2}
\end{align*}
Now $$ \frac{d^{2}}{4} =\sin^{2}\frac{x}{2} \Longrightarrow \sin\frac{x}{2} = \pm\frac{d}{2}$$
