The equations for rms value of voltage, note that voltage with phase angle. Follow the equations easily.
\[{V_{rms}} = {\left[ {\frac{1}{T}\int\limits_0^T {{V^2}_{\max }} {{\cos }^2}(\omega t + {\theta _v})d(\omega t)} \right]^{\frac{1}{2}}}\]
\[\int {{{\cos }^2}} x = \frac{x}{2} + \frac{{\sin 2x}}{4} + c\]
\[{V_{rms}} = {\left[ {\frac{1}{T}{V^2}_{\max }\left[ {\frac{{\omega t + {\theta _v}}}{2} + \frac{{\sin 2(\omega t + {\theta _v})}}{4}} \right]_0^T} \right]^{\frac{1}{2}}}\]
\[\omega = \frac{{2\pi }}{T}\]
\[\sin (x + y) = \sin x.\cos y + \sin y.\cos x\]
\[\sin (2\omega t + 2{\theta _v}) = \sin 2\omega t.\cos 2{\theta _v} + \sin 2{\theta _v}.\cos 2\omega t\]
\[\begin{array}{l} \cos 0 = 1\\ \sin 0 = 0 \end{array}\]
\[\begin{array}{l} \sin 4\pi = 0\\ \cos 4\pi = 1 \end{array}\]
\[{V_{rms}} = {\left[ {\frac{1}{T}{V^2}_{\max }.\pi } \right]^{\frac{1}{2}}} = {V_{\max }}\sqrt {\frac{\pi }{T}}\]
\[T = 2\pi \Rightarrow {V_{rms}} = \frac{{{V_{\max }}}}{{\sqrt 2 }}\]
