Combining two sinusoids with differing amplitudes and similar frequencies This seems likes a super simple problem, but I cannot figure out how to combine these two sinusoids.
$$x(t)=-3*\cos(2\pi f_0 t)+4\sin(2 \pi f_0 t)$$
I tried using the first two relationships mentioned here on wikipedia, but I ended up back where I started.
What identity or property can I use here to combine this two sinusoids into one?
 A: I ended up using the two relationships I found on wikipedia.
$$\cos x=\frac {e^{ix}+e^{-ix}}{2},\space
\sin x= \frac{e^{ix}-e^{-ix}}{2i}$$
$$-3*\cos(2\pi f_0 t)=-3(\frac {e^{i2 \pi f_0 t}+e^{-i2 \pi f_0 t}}{2})$$
$$4\sin(2 \pi f_0 t)=4(\frac {e^{i2 \pi f_0 t}-e^{-i2 \pi f_0 t}}{2})$$
$$x(t)=-3(\frac {e^{i2 \pi f_0 t}+e^{-i2 \pi f_0 t}}{2})+4(\frac {e^{i2 \pi f_0 t}-e^{-i2 \pi f_0 t}}{2})$$
$$=-\frac{3}{2}e^{i2 \pi f_0 t}-\frac{3}{2}e^{-i2 \pi f_0 t}-2ie^{i2 \pi f_0 t}+2ie^{-i2 \pi f_0 t}$$
$$=e^{i2 \pi f_0 t}(-\frac{3}{2}-2i)+e^{-i2 \pi f_0 t}(-\frac{3}{2}+2i)$$
$$a+ib=\sqrt{a^2+b^2}e^{i \arctan{(\frac{b}{a})}}$$
$$|x(t)|=\sqrt{({-\frac{3}{2}})^2+(-2)^2}=\frac{5}{2}$$
$$\angle x(t)= \arctan{\big(-\frac{2}{-\frac{3}{2}}\big)}=\arctan{\big(\frac{4}{3}\big)}$$
$$x(t)=\frac{5}{2}\large(e^{i2 \pi f_0 t+\arctan{(\frac{4}{3})}}+e^{-i2 \pi f_0 t+\arctan{(\frac{-4}{3})}}\large)$$
$$x(t)=\frac{5}{2}[ \cos({2 \pi f_0 t+\arctan{(\frac{4}{3})}}) +j\sin({2 \pi f_0 t+\arctan{(\frac{4}{3})}}) + \cos({2 \pi f_0 t+\arctan{(\frac{4}{3})}})-  j\sin({2 \pi f_0 t+\arctan{(\frac{4}{3})}})]$$
$$\boxed{x(t)=5\cos({2 \pi f_0 t+\arctan{(\frac{4}{3})}})}$$
