# 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
• yes sir, that is right! – Zacky Apr 23 '18 at 16:12

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).

• Thank you sir, makes sense! Mind If I ask how the integral arises to evaluate this? – Zacky Apr 23 '18 at 16:17
• @Zacky, it should be easily evaluated using the usual rules for integrals. (i.e., antideriative of cosine is negative sine, and the symmetry will give you two opposite values of the sine at 0 and T). – The Photon Apr 23 '18 at 16:23
• Yes I know to evaluate the integral, I am curious how it appears here. – Zacky Apr 23 '18 at 16:27
• @Zacky, there is an identity $\sin^2 \theta \equiv \frac{1-\cos 2\theta}{2}$. – The Photon Apr 23 '18 at 16:32

Even though its intuitive why it's zero. We can just integrate to find that result.

Solution:

$$\int_0^Tcos(2\omega t+\phi)$$ $$=\int_0^Tcos(2.\frac{2\pi}{T} t+\phi)$$ $$=\int_0^Tcos(\frac{4\pi}{T} t+\phi)$$ $$=\left[\frac{sin(\frac{4\pi}{T}t+\phi)}{4\pi/T}\right]^T_0$$ $$=\frac{sin(4\pi+\phi)-sin\phi}{\frac{4\pi}{T}}$$ $$=\frac{sin4\pi.cos\phi+sin\phi.cos4\pi-sin\phi}{\frac{4\pi}{T}}$$ $$= \frac{0.cos\phi+sin\phi.1-sin\phi}{\frac{4\pi}{T}}$$ $$= \frac{sin\phi-sin\phi}{\frac{4\pi}{T}}$$ $$=0$$

• nice, I didnt thought to use the alternate version for $\omega$. Thank you alot! – Zacky Apr 23 '18 at 16:29