Can I simplity $A(Bcos(\omega t) - sin(\omega t))$?

Question

I have this ugly solution to a circuits problem.

I have a solution that is pretty much:

$$ A(Bcos(\omega t) - sin(\omega t)) $$ Where $A,B, \omega$ are constants, and $t$ is a variable. Can I simplify this to a single trig function.

Answer 1

We have $$A (B\cos \omega t - \sin \omega t) $$ $$=A\sqrt {B^2+1} (\frac {B}{\sqrt {B^2+1}}\cos \omega t - \frac {1}{\sqrt {B^2+1}} \sin \omega t) $$ $$=A\sqrt {B^2+1} (\cos \alpha \cos \omega t - \sin \alpha \sin \omega t) $$ $$=A\sqrt {B^2+1}(\cos (\alpha + \omega t)) $$

Take a right-angled triangle whose leg measures $B$ and the other side of length $1$. It is then obvious that the hypotenuse will measure $\sqrt {B^2+1} $, thus $\cos \alpha = \frac {B}{\sqrt {B^2+1}} $ and similarly $\sin \alpha $. Hope it helps.

Answer 2

Recall that $$ \sin(\alpha-\beta)=\sin\alpha\cos\beta-\sin\beta\cos\alpha $$ Let's use $\beta=\omega t$. Note that $\sin^2\alpha+\cos^2\alpha=1$; so, let's rewrite your expression a little.

If we write $$ B\cos(\omega t)-\sin(\omega t)=\sqrt{B^2+1}\left(\frac{B}{\sqrt{B^2+1}}\cos(\omega t)-\frac{1}{\sqrt{B^2+1}}\sin(\omega t)\right), $$ then it is clear that we can simplify things if we can find $\alpha$ such that $$ \sin\alpha=\frac{B}{\sqrt{B^2+1}}\qquad \cos\alpha=\frac{1}{\sqrt{B^2+1}}. $$ Find such an $\alpha$, and your entire expression can be simplified as $$ A\sqrt{B^2+1}\sin(\alpha-\omega t). $$