In an auxiliary angle question, the step subsequent to establishing that $$5\sin x+12\cos x=R\sin\left(x+α\right)$$ is $$R=\sqrt{5^{2}+12^{2}}=\ 13$$ How does one prove that the value of $R^2$ equates to $5^{2}+12^{2}$ ? I understand this has something to do with the Pythagorean Therorem, yet I am not exactly sure in what way.
All guidance is much appreciated.
If we expand the right-hand side, we get:
$\begin{eqnarray} R \sin (x + \alpha) & = & R(\sin x \cos \alpha + \cos x \sin \alpha ) \\ & = & R \cos \alpha \sin x + R \sin \alpha \cos x\end{eqnarray}$
then equating that with the left-hand side gives the equations
$R \cos \alpha = 5, R \sin \alpha = 12$
and we can eliminate the $\alpha$ terms using the Pythagorean identity:
$\begin{eqnarray} (R \cos \alpha)^2 + (R \sin \alpha)^2 & = & 5^2 + 12^2 \\ R^2(\cos^2 \alpha + \sin^2 \alpha) & = & 13^2 \\ R^2 & = & 13^2 \end{eqnarray}$