Question: Write $2\cos(100t + \frac{1}{3}) - \sin(100t-1)$ in the form $A\cos(\omega t + \phi)$. Find A, $\omega$, $\phi$ . (Hint: Phasor approach may simplify your task.) (Remark: Leave your answers exact as real numbers maybe in the form of mathematical expressions, and do not attempt to approximate them using decimal representations.)
My attempt:
$$-\sin(100t-1)=\cos\left(100t - 1 + \frac{\pi}{2}\right)$$
Rewrite using this identity:
$$\underbrace{2\cos\left(100t + \frac{1}{3}\right)}_{=z_1} + \underbrace{\cos\left(100t - 1 + \frac{\pi}{2}\right)}_{=z_2} $$
$z_1$ in phasor is $z_1 = 2 \angle{\frac{1}{3}rad}$ which is $z_1 = 2 \angle{19.0986^{\circ}}$
$z_2$ in phasor is $z_2 = 1 \angle{-1 + \frac{\pi}{2}rad}$ which is $z_2 = 1 \angle{32.7042^{\circ}}$
This is where I am struggling because in order to proceed I have to take the approximations otherwise how can I convert it to rectangular form in order to perform addition over these two complex numbers?
See that $$A\cos(\omega t+\phi)=\big(A\cos(\phi)\big)\cos(\omega t)-\big(A\sin(\phi)\big)\sin(\omega t)$$ And $$2\cos(100t+1/3)-\sin(100t-1)\\=2\left(\cos(100t)\cos(1/3)-\sin(100t)\sin(1/3)\right)-\big(\sin(100t)\cos(1)-\cos(100t)\sin(1)\big) \\ =\big(2\cos(1/3)+\sin(1)\big)\cos(100t)-\big(2\sin(1/3)+\cos(1)\big)\sin(100t)$$ So to write $2\cos(100t+1/3)-\sin(100t-1)$ as $A\cos(\omega t+\phi)$ we clearly need $\omega=100$ and we need to find $A,\phi$ such that
$$A\begin{bmatrix}\cos\phi \\ \sin \phi\end{bmatrix}=\begin{bmatrix}2\cos(1/3)+\sin(1) \\ 2\sin(1/3)+\cos(1)\end{bmatrix}$$
You can see that $$\frac{\sin \phi}{\cos \phi}=\tan\phi=\frac{2\sin(1/3)+\cos(1)}{2\cos(1/3)+\sin (1)}$$ Can you finish?