# Reexpress Side of Triangle with Angle - Rodrigues' Rotation Formula

I can't get this step in the geometric proof of Rodrigues' Rotation Formula --- It's claimed $\color{green}{NV = (\mathbf{r}-\mathbf{n}(\mathbf{n} \cdot \mathbf{r}))\cos \phi}$ and $\color{red}{VQ = (\mathbf{n} \times \mathbf{r})\sin\phi}$. The picture is below.

But $\cos\phi = \frac{\color{green}{NV}}{\color{blue}{NQ}}$ and $\sin\phi = \frac{\color{red}{VQ}}{{\color{blue}{NQ}}}$ because $\color{blue}{NQ}$ is the hypothesis. How do they make their claim from this? Thanks a lot.

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$NQ$ is (the length of) the hypotenuse (not the hypothesis). –  joriki Jan 28 '13 at 12:30

You've got several notational confusions there. First, what you denote by $\mathbf p$ in the first paragraph seems to be denoted by $\mathbf r$ in the rest of the post. More importantly, they're using expressions like $NV$ to refer to vectors, whereas in your expressions for $\cos\phi$ and $\sin\phi$ you're using them as the magnitudes of these vectors. If you multiply your equations through by $NQ$, you get the magnitudes of their equations (noting that both $\mathbf r-\mathbf n(\mathbf n\cdot\mathbf r)$ and $\mathbf n\times\mathbf r$ have length $NQ$ if $\mathbf n$ is a unit vector), and then you just have to check that the directions work out, too.