The definition for root-mean-square (rms) voltage is:
$$v_{rms}=\sqrt{\frac{1}{T}\int_{0}^{T}v^{2}dt}$$
If \$v=v_{1}+v_{2}\$ are all functions of time.
Then:
$$v_{rms}=\sqrt{\frac{1}{T}\int_{0}^{T}\left(v_{1}+v_{2}\right)^{2}dt}=\sqrt{\frac{1}{T}\int_{0}^{T}\left(v_{1}^{2}+v_{2}^{2}+2v_{1}v_{2}\right)dt}$$
The cross term \$2v_{1}v_{2}\$ will integrate to zero if \$v_{1}\$ and \$v_{2}\$ are periodic and \$T\$ is an integer multiple of the period.
$$v_{rms}=\sqrt{\frac{1}{T}\int_{0}^{T}\left(v_{1}^{2}+v_{2}^{2}\right)dt}$$
The integral can be distributed as follows:
$$v_{rms}=\sqrt{\frac{1}{T}\int_{0}^{T}v_{1}^{2}dt+\frac{1}{T}\int_{0}^{T}v_{2}^{2}dt}$$
Since your sinusoids have different frequencies, their frequency difference provides a "beat" frequency. \$T\$ can be assigned to the beat frequency period. Then the periods \$T_{1}\$ and \$T_{2}\$ are integer multiples of \$T\$.
Averaging \$v^{2}_{1}\$ over \$T_{1}\$ will give the same result as averaging over \$T\$. Similarly for \$v^{2}_{1}\$.
$$v_{rms}=\sqrt{\frac{1}{T_{1}}\int_{0}^{T_{1}}v_{1}^{2}dt+\frac{1}{T_{2}}\int_{0}^{T_{2}}v_{2}^{2}dt}=\sqrt{v_{1rms}^{2}+v_{2rms}^{2}}$$
A dc voltage of the rms value will produce the same heating in a pure resistance as the ac voltage. Since the separate signals produce heating of \$P_{1}\$ and \$P_{2}\$ watts separately. Then the combined heating is \$P_{1}+P_{2}\$ so adding the square of the voltages makes physical sense.
