Answer
(a) $f(t)=C(1+cos\alpha)sin(\omega t)+Csin\alpha\cdot cos(\omega t)$
(b) $k=30\sqrt 3$ and $\Phi=\frac{\pi}{6}$
Work Step by Step
(a) $f(t)=Csin(\omega t)+Csin(\omega t+\alpha)=C(sin(\omega t)+sin(\omega t)cos\alpha+cos(\omega t)sin\alpha$
Thus $f(t)=C(1+cos\alpha)sin(\omega t)+Csin\alpha\cdot cos(\omega t)$
(b) With $C=10, \alpha=\frac{\pi}{3}$, we have $f(t)=10(1+cos\frac{\pi}{3})sin(\omega t)+10sin\frac{\pi}{3}\cdot cos(\omega t)=45sin(\omega t)+15\sqrt 3cos(\omega t)=45sin(\omega t)+15\sqrt 3cos(\omega t)=30\sqrt 3(\frac{\sqrt 3}{2}sin(\omega t)+\frac{1}{2}cos(\omega t))=30\sqrt 3(sin(\omega t)cos\frac{\pi}{6}+cos(\omega t)sin\frac{\pi}{6})=30\sqrt 3sin(\omega t+\frac{\pi}{6})$, which gives $k=30\sqrt 3$ and $\Phi=\frac{\pi}{6}$
