Trigonometric manipulation

Question

From $$\frac{R\sin(\omega t)-\omega L\cos(\omega t)}{\omega^{2}L^{2}+R^{2}}$$ I have to get $$\frac{\sin(\omega t-\alpha)}{\sqrt{R^{2}+\omega^{2}L^{2}}}$$ where $\alpha$ is a constant. How do I do this?

Answer 1

Use the well-known identity $\sin(A\pm B) \equiv \sin A \cos B \pm \sin B \cos A$.

In your case: $\sin(\omega t -\alpha) \equiv \sin\omega t\cos\alpha-\sin\alpha\cos\omega t.$

If you multiply by a constant, say $r$, then: $r\sin(\omega t -\alpha) \equiv (r\cos\alpha)\sin\omega t-(r\sin\alpha)\cos\omega t.$

Can you find the values of $r$ and $\alpha$ for which

$$\begin{eqnarray*} r\cos\alpha &=& \frac{R}{\omega^2L^2+R^2} \\ \\ r\sin\alpha &=& \frac{\omega L}{\omega^2L^2+R^2} \end{eqnarray*}$$

Dividing one equation by the other gives:

$$\tan\alpha = \frac{r\sin\alpha}{r\cos\alpha} = \frac{\omega L}{R} \implies \alpha = \arctan\left(\frac{\omega L}{R}\right)$$

Squaring both equations and summing them gives:

$$\begin{eqnarray*} (r\cos\alpha)^2 + (r\sin\alpha)^2 &=& \left(\frac{R}{\omega^2L^2+R^2}\right)^{\!2}+\left(\frac{\omega L}{\omega^2L^2+R^2}\right)^{\!2} \\ \\ r^2 &=& \frac{R^2 + \omega^2L^2}{(R^2 + \omega^2L^2)^2} \\ \\ &\implies& r = \frac{1}{\sqrt{R^2 + \omega^2L^2}} \end{eqnarray*} $$

Answer 2

$\sin(\omega t - \alpha) = \sin(\omega t)\cos(\alpha) - \cos(\omega t)\sin(\alpha)$, so we need $$\cos(\alpha) = \frac{R}{\sqrt{\omega^2L^2 + R^2}}$$and$$\sin(\alpha) = \frac{\omega L}{\sqrt{\omega^2L^2 + R^2}}$$

These equations are consistent with each other (since $\sin^2(\alpha) + \cos^2(\alpha) = 1$), so $$\alpha =\sin^{-1}\left(\frac{\omega L}{\sqrt{\omega^2L^2 + R^2}}\right)$$