This is a very simple problem, but I can't figure out where I am going wrong! Say you have the following:

$a \sin\theta + b \cos\theta = c. \tag{1}$

Now, this for example can be rewritten using: $R \sin (\theta + \alpha) = R \cos\alpha\sin\theta + R\sin\alpha\cos\theta, \tag{2}$ whereby we obtain: $R^2 = a^2 + b^2, \tag{3}$ $\alpha = arc\tan(\frac{b}{a}). \tag{4}$

We can now substitute $R$ and $\alpha$ into Eq. (2). However, which values of $R$ and $\alpha$ should one use? Since $R = \pm \sqrt{a^2 + b^2}$ and $\alpha = P(arc\tan(\frac{b}{a}))$ or $\alpha = P(arc\tan(\frac{b}{a})) + \pi$, where $P(arc\tan(\frac{b}{a}))$ gives the principle value for $\alpha$.


Firstly, you can use $R$ as strictly $+\sqrt{a^2+b^2}$ and still make things work. The right values of $\alpha$ are:

If $a>0$, $\alpha=\arctan{b \over a}$.

If $a<0$, $\alpha=\pi + \arctan{b \over a}$

This works whenever you define $a=R\cos{\alpha}$.

  • $\begingroup$ You can also do it with $R$ negative and $\alpha$ of the opposite signs as Ud779 gives, or that same $\alpha + k\pi$ for some integer $k$ and it will also work, but why make things harder for yourself? $\endgroup$ – Paul Sinclair Sep 2 '15 at 3:53
  • $\begingroup$ So, I understand how one can pick a specific $\alpha$. However, once you have chosen a specific $\alpha$, what allows you pick either $R < 0$ or $R > 0$? Why shouldn't we find solutions corresponding to both negative and positive $R$? Wouldn't the solutions corresponding to the specific $\alpha$ be different for $R < 0$ and $R > 0$? $\endgroup$ – Jonathan Prescott Sep 2 '15 at 5:14
  • $\begingroup$ The reason I am asking is that we can use Eqs. (1) and (2) in conjunction to obtain: $\cos\alpha = \frac{a}{R} \tag{5}$ $\sin\alpha = \frac{b}{R}. \tag{6}$ Now, the quadrant that $\alpha$ lies in depends on the signs of $\cos \alpha$ and $\sin \alpha$. However, given specific $a, b$, the signs of $\cos \alpha$ and $\sin \alpha$ then depend on the whether $R<0$ or $R>0$. What allows us to pick a specific $R$? Please elaborate on this! $\endgroup$ – Jonathan Prescott Sep 2 '15 at 6:11
  • $\begingroup$ Ah, I see. Either choice of $R$, yields the same solutions! Got, it! Thanks! $\endgroup$ – Jonathan Prescott Sep 2 '15 at 7:47

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