# About Root mean square voltage

Hello there I am trying to understand this proof: foo. I dont understand why $$\cos{(2 \omega t +\phi)}$$ vanishes when it's integrated between 0 and T. When I try to do this alone I get: $$V_{rms}=\sqrt{\frac{V_{pk}^2}{2}-\frac{V_{pk}^2}{(2T)(2\omega)}(\sin{(2\omega T+ 2\phi)-\sin{(2\phi))}}}$$ Can I get some help to understand this please?

• Can we assume $T=2\pi/\omega$? – The Photon Apr 23 '18 at 16:09

I dont understand why $$\cos{(2 \omega t +\phi)}$$ vanishes when it's integrated between 0 and T.
Because regardless of what $\phi$ is, if $T$ is the period of the sine wave (or an integer multiple of the period), then within the time $T$ the sine wave spends half it's time below 0 and half the time above 0 (and the shape is indeed symmetric).
• @Zacky, there is an identity $\sin^2 \theta \equiv \frac{1-\cos 2\theta}{2}$. – The Photon Apr 23 '18 at 16:32
$$\int_0^Tcos(2\omega t+\phi)$$ $$=\int_0^Tcos(2.\frac{2\pi}{T} t+\phi)$$ $$=\int_0^Tcos(\frac{4\pi}{T} t+\phi)$$ $$=\left[\frac{sin(\frac{4\pi}{T}t+\phi)}{4\pi/T}\right]^T_0$$ $$=\frac{sin(4\pi+\phi)-sin\phi}{\frac{4\pi}{T}}$$ $$=\frac{sin4\pi.cos\phi+sin\phi.cos4\pi-sin\phi}{\frac{4\pi}{T}}$$ $$= \frac{0.cos\phi+sin\phi.1-sin\phi}{\frac{4\pi}{T}}$$ $$= \frac{sin\phi-sin\phi}{\frac{4\pi}{T}}$$ $$=0$$