- #1

- 1,462

- 44

## Homework Statement

Show that ##a \sin x + b\cos x = c \sin (x + \theta)##, where ##c = \sqrt{a^2 + b^2}## and ## \displaystyle \theta = \arctan (\frac{b}{a})##

## Homework Equations

## The Attempt at a Solution

We see that ##c \sin (x + \theta) = c \cos \theta (\sin x) + x \sin \theta (\cos x)##. So we compare coefficients. ##a = c \cos \theta## and ##b = c sin \theta##. We can assume that neither ##a## or ##b## are zero, so we find by division that ##c = \sqrt{a^2 + b^2}##. Next, we see that ##a^2 + b^ = c^2 \cos^2 \theta + c^2 \sin^2 \theta = c^2##, so ##c = \pm \sqrt{a^2+b^2}##. So it seems that I am done, since the we can take the positive root. However, what about the negative root? Would that give a valid answer too? Why isn't the negative root the one that is desired by the question?