General Format for the Sinusoidal Voltage or Current

The basic mathematical format for the sinusoidal waveform is $$\bbox[10px,border:1px solid grey]{A_m \sin \alpha}$$ where $A_m$ is the peak value of the waveform and a is the unit of measure for the horizontal axis, as shown in Fig. 1.
Fig. 1: Basic sinusoidal function.
The equation a $\alpha =\omega t$ states that the angle a through which the rotating vector of Fig. 1 will pass is determined by the angular velocity of the rotating vector and the length of time the vector rotates.
For example, for a particular angular velocity (fixed $\omega$), the longer the radius vector is permitted to rotate (that is, the greater the value of t), the greater will be the number of degrees or radians through which the vector will pass. Relating this statement to the sinusoidal waveform, for a particular angular velocity, the longer the time, the greater the number of cycles shown. For a fixed time interval, the greater the angularvelocity, the greater the number of cycles generated.
The general format of a sine wave can also be written $$\bbox[10px,border:1px solid grey]{A_m \sin (\omega t)}$$ with $\omega t$ as the horizontal unit of measure. For electrical quantities such as current and voltage, the general format is $$i = I_m \sin (\omega t)$$ $$v = V_m \sin (\omega t)$$ where the capital letters with the subscript m represent the amplitude, and the lowercase letters $i$ and $e$ represent the instantaneous value of current or voltage, respectively, at any time $t$. This format is particularly important since it presents the sinusoidal voltage or current as a function of time, which is the horizontal scale for the oscilloscope. Recall that the horizontal sensitivity of a scope is in time per division and not degrees per centimeter.
Example 1: Given $e = 5 \sin \alpha$, determine e at $\alpha = 40^\circ$ and $\alpha = 0.8\pi$.
Solution:
For $\alpha = 40^\circ$, $$e = 5 \sin \alpha =5 \sin(40^\circ)$$ $$e = 5(0.6428) = 3.214 V$$ For $\alpha = 0.8\pi$, $$\theta = {180^\circ \over \pi} (0.8\pi) =144^\circ$$ and $$e = 5 \sin 144^\circ = 5(0.5878) = 2.939 V$$
The angle at which a particular voltage level is attained can be determined by rearranging the equation $$e = E_m \sin \alpha$$ in the following manner: $$\sin \alpha = {e \over E_m}$$ which can be written $$\bbox[10px,border:1px solid grey]{\alpha = \sin^{-1}({e \over E_m})} \tag{1}$$ Similarly, for a particular current level, $$\bbox[10px,border:1px solid grey]{\alpha = \sin^{-1}({i \over I_m})} \tag{2}$$ The function $sin^{-1}$ is available on all scientific calculators.
Example 2:
a. Determine the angle at which the magnitude of the sinusoidal function $v = 10 \sin 377t$ is $4 V$.
b. Determine the time at which the magnitude is attained.
Fig. 2: For example 2.
Solution: a. Eq. (1): $$\alpha _1 = \sin^{-1}({v \over V_m}) = \sin^{-1}({4 \over 10}) = 23.578^\circ$$ However, Figure 2 reveals that the magnitude of 4 V (positive) will be attained at two points between $0^\circ$ and $180^\circ$. The second intersection is determined by $$\alpha _2= 180^\circ - 23.578^\circ= 156.422^\circ$$ In general, therefore, keep in mind that Equations (1) and (2) will provide an angle with a magnitude between $0^\circ$ and $90^\circ$.
b. $\alpha = \omega t$, and so $t = \alpha/\omega$. However, $\alpha$ must be in radians. Thus, $$\alpha (rad) = {\pi \over 180^\circ} 23.578^\circ = 0.411 rad$$ and $$t_1 = {\alpha \over \omega} = {0.411 rad \over 377} = 1.09ms$$ For the second intersection, $$\alpha (rad) = {\pi \over 180^\circ} 156.422^\circ = 2.73 rad$$ and $$t_2 = {\alpha \over \omega} = {2.73 rad \over 377} = 2.24ms$$