An alternating current is given by the equation $$i = i_1 \sin\omega t + i_2 \cos\omega t$$. The rms current will be:

The instantaneous current is $$i = i_1 \sin \omega t + i_2 \cos \omega t$$.

The rms (root-mean-square) value of any periodic current is defined as $$I_{rms} = \sqrt{\frac{1}{T}\int_{0}^{T} i^2 \, dt}$$ where the time period $$T = \frac{2\pi}{\omega}$$ for angular frequency $$\omega$$.

First square the given current:
$$i^2 = \bigl(i_1 \sin \omega t + i_2 \cos \omega t\bigr)^2$$
$$\quad = i_1^2 \sin^2 \omega t + i_2^2 \cos^2 \omega t + 2 i_1 i_2 \sin \omega t \cos \omega t$$ $$-(1)$$

Now find the time average of each term over one full period.

For any sinusoid, the following standard results hold over a complete cycle:
$$\frac{1}{T}\int_{0}^{T} \sin^2 \omega t \, dt = \frac{1}{2}\,, \qquad \frac{1}{T}\int_{0}^{T} \cos^2 \omega t \, dt = \frac{1}{2}$$
$$\frac{1}{T}\int_{0}^{T} \sin \omega t \cos \omega t \, dt = 0$$

Applying these averages to expression $$(1)$$ gives
$$\frac{1}{T}\int_{0}^{T} i^2 \, dt = i_1^2 \cdot \frac{1}{2} + i_2^2 \cdot \frac{1}{2} + 2 i_1 i_2 \cdot 0$$
$$\quad = \frac{i_1^2 + i_2^2}{2}$$ $$-(2)$$

Finally, take the square root to obtain the rms current:
$$I_{rms} = \sqrt{\frac{i_1^2 + i_2^2}{2}} = \frac{1}{\sqrt{2}}\sqrt{i_1^2 + i_2^2}$$

Therefore, the correct option is Option D.