## R-alpha Method

The $R-\alpha$ method is an application of the composite angle formulae which allows functions of the form $f(x) = a\sin(x) + b\cos(x)$ to be rewritten in the form $f(x) = R\sin(x \pm\alpha_1)$ or $f(x) = R\cos(x \pm \alpha_2)$ The key idea here is to assert that $$ a\sin(x) + b\cos(x) = R\sin(x + \alpha_1) $$ and then apply the compound angle formulae $\sin(\theta + \phi) = \sin\theta\cos\phi +\cos\theta\sin\phi$ $$ a\sin(x) + b\cos(x) = R\sin(x)\cos(\alpha_1)+R\cos(x)\sin(\alpha_1) $$ Treating $\sin(x)$ and $\cos(x)$ independently yields $$a = R\cos(\alpha_1),\;b = R\sin(\alpha_1)$$ This simultaneous equation now gives $R = \sqrt{a^2 + b^2}$. Note we have chosen the positive square root here, so a little bit of care needs to be taken solving for $\alpha_1$. My personal preference is to solve both equations and choose the solution satisfying both. (You can assert that $\frac{b}{a} = \tan\alpha_1$, but a little care needs to be taken to ensure no ambiguity has arisen from your choice of $R$ being positive! )

## R-alpha Method 1

Write the following expression in the form $R\cos(x-\alpha)$, giving $\alpha$ to 4 decimal places. $$ 12\cos(x) + 5\sin(x) $$

solution - press button to display

$$ \begin{array}{rcl} R\cos(x-\alpha) &=& 12 \cos(x) + 5\sin(x) \\ R\left(\cos(x)\cos(\alpha) +\sin(x)\sin(\alpha)\right) &=& 12\cos(x) + 5\sin(x) \end{array} $$ Comparing the coefficients of $\cos(x)$ and $\sin(x)$ here yields the following two equations $$ R\cos(\alpha) = 12,\;R\sin(\alpha) = 5 $$ It follows from the identity $\cos^2(x) + \sin^2(x) \equiv 1$ then that $R^2 = R^2\left(\cos^2(\alpha) + \sin^2(\alpha)\right) = 12^2 + 5^2 = 13^2$ We take $R$ to be positive giving $R=13$ If we now solve $\cos(\alpha) = \frac{12}{13}$ and $\sin(\alpha) = \frac{5}{13}$, we get $\alpha = 0.3948 (\mbox{4dp})$ Therefore $$ 12\cos(x) + 5\sin(x) = 13\cos(x - 0.3948) $$