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A voltage across a circuit component at time \$t\$ can be calculated by \$V\cdot\sin(\omega t + \phi)\$.

However, in the section of my book that introduces complex numbers to represent voltages, it is stated that a voltage \$V\cdot\cos(\omega t)\$ can be represented by \$V\cdot e^{j\theta}\$ where \$e^{j\theta} = \cos(\theta) + j\sin(\theta)\$.

My confusion arises from the fact that the actual voltage is its amplitude times the cosine function and not the sine function since it is a sinusoid. Is this an error or I am just confused?

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  • \$\begingroup\$ Be aware that \$\cos(\theta) = \sin(\theta + \frac{\pi}{2})\$ so the distinction you seem to be drawing doesn't exist. A sinusoidal function of time can equally be expressed as \$f(t) = \cos(\omega t + \phi) = \sin(\omega t + \phi + \frac{\pi}{2}) = \sin(\omega t + \phi')\$ \$\endgroup\$ – Alfred Centauri Jan 12 '15 at 3:32
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Both sin and cos are considered sinusoidal waveforms. As a practical matter, sin and cos are essentially the same thing, just offset by 90 degrees.Since "time 0" is arbitrary, the distinction between sin and cos only matters if you are comparing phase against another signal.

Since Euler's equation works out to cos for the real component and sin for the imaginary component, it's just more useful to use the cos than the sin for real voltage.

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  • \$\begingroup\$ But if lets say at time = 0 I start with a voltage = 0, then the correct function that would model the circuit's input voltage would be the sin function, attempting to solve the voltage with the cos function would be 90 out of phase. \$\endgroup\$ – AlanZ2223 Jan 12 '15 at 13:32
  • \$\begingroup\$ yes, correct. and the idea of time 0 is theoretical. the whole point of using phasor notation is to simplify the amplitude and phase relationships for complicated circuits. also note a theoretical sinusoid starts at t "minus infinity" and extends to "positive infinity". real circuits only temporarily approximate sinusoids. so without a phase 0 reference, either sin or cos or -sin or -cos fits. \$\endgroup\$ – MarkU Jan 12 '15 at 20:49

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