Multi-Sine RMS

We will consider the function on the interval $[-\pi,\pi]$ , since the function has a period of $2\pi$ . Then, the mean square of the function is given by $\begin{aligned}f_{\text{MS}}&=\frac1{2\pi} \int_{-\pi}^{\pi} \left(\sin(t)+\frac12\sin(2t)+\frac13\sin(3t)\right)^2dt\\ &=\frac1{2\pi} \int_{-\pi}^{\pi} \left(\sin^2(t)+\frac14\sin^2(2t)+\frac19\sin^2(3t)+\sin(t)\sin(2t)+\frac23\sin(t)\sin(3t)+\frac13\sin(2t)\sin(3t)\right)dt \end{aligned}$ We make use of the following identities: $\begin{aligned}\sin^2(x)&=\frac12-\frac12\cos(2x)&\qquad&(1)\\ \sin(x)\sin(y)&=\frac12\cos(x-y)-\frac12\cos(x+y)&\qquad&(2)\end{aligned}$ After applying (1) to the first 3 terms and (2) to the others, we will be left with a sum of cosines and constants. The cosines integrate to become sines, which evaluate to zero because of the terminals $-\pi$ and $\pi$ , leaving only the constants from identity (1). This leaves $f_{\text{MS}}=\frac1{2\pi} \int_{-\pi}^{\pi} \left(\frac12+\frac18+\frac1{18}\right)dt=\frac1{2\pi} \int_{-\pi}^{\pi} \left(\frac{49}{72}\right)dt =\frac1{2\pi}\frac{49}{72}(\pi+\pi)=\frac{49}{72}$ Thus, the root mean square value is simply the square root of this result, $f_{\text{RMS}}=\sqrt{\frac{49}{72}}=\frac7{6\sqrt2}\approx0.82496$