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Let's solve:

$\sqrt{3}\sin x - \cos x=2$

The left hand side may be expressed as $R\sin(x+ \phi)$

We know that $R=\sqrt{3+1}=2$

We also know that $\tan \phi= \frac{-1}{\sqrt{3}}$

The solution to $\tan \phi=\frac{-1}{\sqrt{3}}$ has many solutions, for example, -30, 150, 330 degrees etc.

Which of these solutions do we accept? Or is it irrelevant which we will accept? Which of these solutions are acceptable?

Thanks!

Asaf Karagila
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mathos
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2 Answers2

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$$\frac{\sqrt{3}}{2}\sin x - \frac{1}{2}\cos x=1$$ $$\sin\left(x -\frac{\pi}{6}\right) = 1$$ $$\sin\left(x -\frac{\pi}{6}\right) = \sin\left(\frac{\pi}{2}\right)$$ $$ x = n\pi + (-1)^n \frac{\pi}{2} +\frac{\pi}{6}$$

Aakash Kumar
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It does not matter which solution you choose; the algebra works out in the end. If you choose to let $0 \leq \phi \leq \pi$, then R would actually be negative, not positive, which you can see from equating coefficients. It still gives you the same function however.

Let $\sqrt 3 \sin x - \cos x=R\sin(x+ \phi)$ Expanding the RHS, $\sqrt 3 \sin x - \cos x = R \sin x \cos \phi + R\cos x \sin \phi $

Equating coefficients, then $R\cos \phi= \sqrt 3 (1),\\ R \sin \phi= -1 (2)$

So $\tan \phi =\frac{-1}{\sqrt 3}$

Arbitrarily, we choose the solutions $\phi=\frac{-\pi}{6},\frac{5\pi}{6}$. If $\phi=\frac{-\pi}{6}$, $\sin \phi =\frac{-1}{2}$

From substitution into $(2)$, $\frac{R}{-2}=-1 \implies R=2$

However, if $\phi=\frac{5\pi}{6}$, $\sin \phi=\frac{1}{2}$.

Substitution into (2) gives $\frac{R}{2}=-1 =-2\implies R=-2$

If you substitute these values in and expand the entire RHS out you will get it to match the LHS regardless of which value you initially chose.

Sherlock Holmes
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  • How can R be negative? That's impossible! – mathos Oct 16 '14 at 07:44
  • @mathos: why would that be impossible ? The relations are $R\cos\phi=\sqrt3$ and $R\sin\phi=-1$. That tells you that $R^2=4$. –  Oct 16 '14 at 07:53
  • we know that R= square root of (a^2 plus b^2) and by definition we don't accept the negative of this square root! – mathos Oct 16 '14 at 08:00
  • @mathos please refer to my edited answer. – Sherlock Holmes Oct 16 '14 at 08:01
  • I got more confused. I thought that R can only be negative (something like the radius of the trigonometric circle) and only phee can be either negative or positive. Now you say that R can also be negative. So, how do I actually solve equations like this? – mathos Oct 16 '14 at 08:05
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    @mathos: you only know that $R^2=a^2+b^2$, which means $R=\pm\sqrt{a^2+b^2}$. The exact sign depends on the choice of the angle quadrant. –  Oct 16 '14 at 08:07
  • @mathos Yves Daoust has hit it on the nail head precisely. For solving equations like this, pick any values you want - the algebra will always work out in the end so long as you are consistent and accurate. – Sherlock Holmes Oct 16 '14 at 08:09
  • so, R is not a radius and thus an always positive number? – mathos Oct 16 '14 at 08:12
  • so, R is not a radius and thus an always positive number? R is just a coefficient? And how we ended up with R^2=a^2+b^2? To solve an equation of asinx+bcosx=c, what steps do we take? Until now, I thought we could calculate the R (and take only its positive result) and then calculate the phee, by its tan. Now you are telling me to find first the phee, pick ANY value I want for phee, then calculate the R (not by the square root of a^2 plus b^2, but by Rsin(phee)=b) and then move on? – mathos Oct 16 '14 at 08:19
  • @mathos Yes, that's right. – Sherlock Holmes Oct 16 '14 at 08:47
  • Can you point me please to a textbook or online resource that has a tutorial on how to solve these equations with this new to me way? – mathos Oct 16 '14 at 13:18
  • @mathos I suggest googling auxiliary angle. – Sherlock Holmes Oct 17 '14 at 02:54