Time domain of a Sine wave?
No, you are incorrect:

Time domain equation for a sin wave is:

$$v(t)= V_M \sin(2\pi f\:t+\theta) = V_M \sin\left(\frac{2\pi}{T}t+\theta\right)$$

Where:


\$V_M\$ is the amplitude of the sin wave, (relate to the peak to peak and RMS: \$V_M=\frac{V_{pk-pk}}{2}= \sqrt{2} V_{RMS}\$ )
\$f\$ is the frequency, in Hz
\$T\$ is the period in seconds (\$T=\frac{1}{f}\$) 
\$\theta\$ is a phase shift in radians.
\$t\$ is the time in seconds.




So for example, if we pass Australia's mains grid power:
50Hz, \$V_{RMS}=240\text{V} \implies V_M=339.4\$V though a suitable RLC circuit that causes a phase shift  of \$60^\circ=\frac{\pi}{3}\$ radians, the output would be:
$$v(t)=339.4\sin(2\pi\times 50 \times t + \frac{\pi}{3})$$
This is a expression in terms of time -- the voltage is not constant, it goes up and down with time.
You need to evaluate it at some fixed point in time to find some voltage.

Which if we wanted to know the voltage after 3 seconds: \$v(3)=339.4\sin(2\pi\times 50 \times 3 + \frac{\pi}{3})=293.9\$V.

The thing you had wrong was you were using the frequency form of the equation, but instead of putting in the time you  wished to evaluate at, you put in the period.