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\begin{document}

{\bf Question}

Find the mean and the rms values of $f(t)=\cos \omega t$ over the
half period $0 \leq y \leq \frac{\pi}{\omega}$ and over the full
period $0 \leq t \leq \frac{2\pi}{\omega}$.
\medskip

{\bf Answer}

\un{Mean over $\frac{1}{2}$ period}:

$\begin{array} {rcl} \overline{f_{\frac{1}{2}}} &amp; = &amp;
\ds\frac{1}{(\frac{\pi}{\omega}-0)}\ds\int_0^{\frac{\pi}{\omega}}\,dt
\cos \omega t\\ &amp; &amp; \\ &amp; = &amp;
\ds\frac{\omega}{\pi}\left(\ds\frac{\sin
\omega2}{\omega}\right]_0^{\frac{\pi}{\omega}}=\un{0}\end{array}$

Since we have $\cos\un{\omega t}$ a full period is
$\ds\frac{2\pi}{\omega}$

\un{Mean over full period}:

$\begin{array} {rcl} \overline{f_{full}} &amp; = &amp;
\ds\frac{1}{(\frac{2\pi}{\omega}-0)}\ds\int_0^{\frac{2\pi}{\omega}}\,dt
\cos \omega t\\ &amp; &amp; \\ &amp; = &amp;
\ds\frac{\omega}{2\pi}\left(\ds\frac{\sin \omega
t}{\omega}\right]_0 ^{\frac{2\pi}{\omega}}=\un{0}\end{array}$

\un{RMS value over $\frac{1}{2}$ period}:

$\begin{array} {rcl} f_{rms,\frac{1}{2}} &amp; = &amp;
\sqrt{\ds\frac{1}{\frac{\pi}{\omega}-0}}\ds\int_0^{\frac{\pi}{\omega}}
\,dt \cos^2 \omega t\\ &amp; &amp; \\ &amp; = &amp;
\sqrt{\ds\frac{\omega}{\pi}\ds\frac{1}{2}\ds\int_0^{\frac{\pi}{\omega}}\,dt(1+\cos
2\omega t)}\\ &amp; &amp; \\ &amp; = &amp;
\left[\ds\frac{\omega}{2\pi}\left[t+\ds\frac{\sin 2\omega
t}{2\omega}\right]_0^{\frac{\pi}{\omega}}\right]^{\frac{1}{2}}\\ &amp;
&amp; \\ &amp; = &amp; \left(\ds\frac{\omega}{2\pi} \cdot
\ds\frac{\pi}{\omega}\right)^{\frac{1}{2}}\\ &amp; &amp; \\ &amp; = &amp;
\un{\ds\frac{1}{\sqrt 2}} \end{array}$

\un{RMS value over full period}:

$\begin{array} {rcl} f_{rms,full} &amp; = &amp;
\sqrt{\ds\frac{1}{\frac{2\pi}{\omega}-0}}\ds\int_0^{\frac{\pi}{\omega}}
\,dt \cos^2 \omega t\\ &amp; &amp; \\ &amp; = &amp; \sqrt{\ds\frac{\omega}{2\pi}
\times \ds\frac{1}{2}\left[t+\ds\frac{\sin 2\omega
t}{2\omega}\right]_0^{\frac{2\pi}{\omega}}}\\ &amp; &amp; \\ &amp; = &amp;
\un{\ds\frac{1}{\sqrt{2}}} \end{array}$

\end{document}
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