Root mean square of a translated sine wave The sine wave 
$${\displaystyle y=A_{1}\sin(\omega t)\,}$$
has a root mean square value of 
$$rms = \frac{A_1}{\sqrt{2}}$$
Does this change if the wave is translated up? For example,
$${\displaystyle y=A_{1}\sin(\omega t)\,+20}$$
What would the $rms$ be then?
 A: If you look at the RMS calculation:
\begin{align}
\text{RMS}(y + C) 
&= \sqrt{\frac{1}{T} \int\limits_0^T (y(t) + C)^2 \, dt } \\
&= \sqrt{\frac{1}{T} \int\limits_0^T (A_1 \sin(\omega t) + C)^2 \, dt } \\
&= \sqrt{\frac{1}{T} \left( \int\limits_0^T (A_1 \sin(\omega t))^2 \, dt +
2C A_1 \underbrace{\int\limits_0^T \sin(\omega t) \, dt}_0 + C^2 \int\limits_0^T \, dt  \right) }\\
&= \sqrt{\frac{1}{T} \int\limits_0^T (A_1 \sin(\omega t))^2 \, dt + C^2} \\
&= \sqrt{\text{RMS}(y)^2 + C^2}\\
&\ge \text{RMS}(y) \\
\end{align}
you see it will increase for $C \ne 0$.
For your example it is
$$
\sqrt{\frac{A_1^2}{2} + 400}
$$
A: $$f_{\mathrm{rms}}^2=\frac{1}{T}\int_Tf(t)^2dt$$
Hence, for $f(t)=A_0+A_1\cos(\omega t)$
$$\begin{align}
f_{\mathrm{rms}}^2&=\frac{1}{T}\int_T{(A_0+A_1\cos(\omega t))}^2dt\\
&=\frac{1}{T}\int_T{(A_0^2+2A_0A_1\cos(\omega t)+A_1^2\cos^2(\omega t))}dt\\
&=\cdots\\
&=A_0^2+\frac{A_1^2}{2}
\end{align}$$
Therefore, the rms value is $$\boxed{f_{\mathrm{rms}}=\sqrt{A_0^2+\frac{A_1^2}{2}}}$$ 
Notice that it reduces to $\frac{A_1}{\sqrt{2}}$ for $A_0=0$.
A: When we look at a generalized problem, find the RMS of (where $\text{K}_1$, $\text{K}_2$ and $\varphi$ are constants):
$$f(t)=\text{K}_1\cdot\cos\left(\frac{2\pi t}{\text{T}}+\varphi\right)+\text{K}_2$$
We get:
$$\text{RMS}\left[f(t)\right]=\sqrt{\frac{1}{\text{T}}\int_0^\text{T}\left\{\text{K}_1\cdot\cos\left(\frac{2\pi t}{\text{T}}+\varphi\right)+\text{K}_2\right\}^2\space\text{d}t}=\sqrt{\frac{\text{K}_1^2}{2}+\text{K}_2^2}$$
