Let's say I have the composite wave:
$y(t) = \cos(40t) - 0.3 \cos(40t - 16)$
This combines two waves of the same frequency but different phases, and the objective here is to add them and present the result in the simplified form, as a single real amplitude multiplied by a single cosine. I have tried complex analysis, via $\operatorname{Re}[e^{i40t}(1-0.3e^{-i16})]$, but I still can't get a result out of it. May I ask what I might be missing here?
On
$$\cos{(40t)}-0.3\cos{(40t-16)}$$ $$=\cos{(40t)}-0.3\cos{(16)}\cos{(40t)}-0.3\sin{(16)}\sin{(40t)}$$ $$=(1-0.3\cos{(16)})\cos{(40t)}-0.3\sin{(16)}\sin{(40t)}$$ Now we want the above expression in the form $$R \cos{(40t+\alpha)}=R\cos{(\alpha)}\cos{(40t)}-R\sin{(\alpha)}\sin{(40t)}$$ So, equating coefficients of $\cos{(40t)}$ and $\sin{(40t)}$ we have $$R\cos{(\alpha)}=1-0.3\cos{(16)}$$ $$R\sin{(\alpha)}=0.3\sin{(16)}$$ Which we can solve simultaneously to give $$R=\sqrt{(1-0.3\cos{(16)})^2+(0.3\sin{(16)})^2}$$ $$\alpha=\arctan{\bigg(\frac{0.3\sin{(16)}}{1-0.3\cos{(16)}}\bigg)}$$ So, the answer is $$\cos{(40t)}-0.3\cos{(40t-16)}$$ $$=\sqrt{(1-0.3\cos{(16)})^2+(0.3\sin{(16)})^2} \times\cos{\bigg(40t+\arctan{\bigg(\frac{0.3\sin{(16)}}{1-0.3\cos{(16)}}\bigg)}\bigg)}$$ $$\approx 1.290192113\cos{(40t-0.06699439337)}$$
your working is fine until
$\Re[e^{j40t}[1-0.3e^{-j16}]]$ now, remove time varying part ( harmonic part) and focus on it's phasor and write it in Euler form as i.e, $[1-0.3e^{-j16}]=1.28-0.0863j=1.282\ e^{-j0.0673}$
since wave is nothing but $=$ Re part of $[\ \ \text{phasor}$$\times$ $e^{j\omega t}]$
so, your wave is $1.282 \cos(\omega t-0.0673)$ in your case phasor is rotating with frequency 40 radian per second in Anti clock wise direction i.e,put $\omega=40$