# 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?

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Make the following manipulations: $$(1-\cos(\alpha))^2+\sin^2(\alpha)=d^2$$ $$1-2\cos(\alpha)+\cos^2(\alpha)+\sin^2(\alpha)=d^2$$ $$1-2\cos(\alpha)+1=d^2$$ $$2-2\cos(\alpha)=d^2$$ $$\cos(\alpha)=1-\frac{d^2}{2}$$ $$\alpha=\cos^{-1}\left(1-\frac{d^2}{2}\right)$$

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Yes, indeed. Many thanks. It was indeed not that difficult. But you saw this: $\cos^2\alpha + \sin^2\alpha = 1$. I didn't see it. – Martijn Courteaux Jul 10 '11 at 19:14
@Martijn: No problem, glad to help :) – Zev Chonoles Jul 10 '11 at 19:20
Alternatively you can state the above as $\alpha=2 \sin^{-1} \left( \frac{d}{2} \right)$ using a tan-half-angle formula (link en.wikipedia.org/wiki/Tangent_half-angle_formula) – ja72 Jul 11 '11 at 2:59
Nitpick: $\alpha = \cos^{-1} (...)$ is not entirely correct as there are infinite solutions... – Aryabhata Jul 11 '11 at 4:23
Is it safe to mix upper index meanings like that? – Alexei Averchenko Jul 11 '11 at 5:03

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.)

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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 $\cos^2 \alpha + \sin^2 \alpha = 1$.

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Guess I was nearly one minute late ^_^ – BS. Jul 10 '11 at 19:15

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 $$\frac{d^{2}}{4} =\sin^{2}\frac{x}{2} \Longrightarrow \sin\frac{x}{2} = \pm\frac{d}{2}$$

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