Switch from $a\cdot \sin(t) + b \cdot \cos(t)$ to $c \cdot \cos(t+f)$ How could I switch from $a\cdot \sin(t) + b \cdot \cos(t)$ to $c \cdot \cos(t+f)$?
Thank you for your time.
 A: We see that by the addition formula,
$$c \cdot \cos(t + f) = c \cdot \left( \cos(t)\cos(f) - \sin(t)\sin(f)\right).$$
If you set $a = -\sin(f), c = 1,$ and $b = \cos(f),$ You can make a reasonable comparison between the expressions $a \sin(t) + b \cos(t)$ and $c \cdot \cos(t + f).$
A: $$
a\sin t+b\cos t=\sqrt{a^2+b^2}\left[\frac{b}{\sqrt{a^2+b^2}}\cos t +\frac{a}{\sqrt{a^2+b^2}}\sin t\right]$$
$$
=\sqrt{a^2+b^2}\left[\cos \alpha \cos t +\sin\alpha \sin t\right]$$
$$ =\sqrt{a^2+b^2} \cos(t-\alpha). $$
$ (c,f) $ have different symbols.
Useful in combining two waves of same frequency but different amplitude.
A: Set $t=0$, giving
$$a\sin(0)+b\cos(0)=b=c\cos(f).$$
Then set $t=\frac\pi2$, and
$$a\sin\left(\frac\pi2\right)+b\cos\left(\frac\pi2\right)=a=c\cos\left(\frac\pi2+f\right)=-c\sin(f).$$
This gives you the transfrom from $c,f$ to $a,b$.
You can invert it by noting that
$$a^2+b^2=c^2\cos^2(f)+c^2\sin^2(f)=c^2,$$ and
$$\frac ab=-\frac{c\sin(f)}{c\cos(f)}=-\tan(f).$$

Alternatively, the left expression cancels when
$$a\sin(t)+b\cos(t)=0,$$ i.e.
$$\tan(t)=-\frac ba.$$
This must coincide with 
$$\cos(t+f)=0,$$ or
$$f=\frac\pi2-t,$$
$$\tan(f)=\cot(t)=-\frac ab.$$
Then, with $t=-f$,
$$-a\sin(f)+b\cos(f)=c,$$
$$c=-a\sin\left(\arctan\left(-\frac ab\right)\right)+b\cos\left(\arctan\left(-\frac ab\right)\right).$$
