a) Áp dụng tổng các góc trong ∆ABC có:

\(\widehat A + \widehat B + \widehat C = {180^o}\)

\( \Rightarrow \widehat A + \widehat B = {180^o} - \widehat C\)

\( \Leftrightarrow \frac{{\widehat A + \widehat B}}{2} = \frac{{{{180}^o} - \widehat C}}{2} = {90^o} - \frac{{\widehat C}}{2}\)

\( \Rightarrow \sin \left( {\frac{{\widehat A + \widehat B}}{2}} \right) = \sin \left( {{{90}^o} - \frac{{\widehat C}}{2}} \right) = \cos \frac{{\widehat C}}{2}\);

b) Ta có: \(\widehat A + \widehat B + \widehat C = {180^o}\)

\(\tan \left( {2\widehat A + \widehat B + \widehat C} \right) = \tan \left( {\widehat A + \widehat A + \widehat B + \widehat C} \right)\)

\[ = \tan \left( {\widehat A + {{180}^o}} \right) = \tan \widehat A\]

\( \Rightarrow \tan \left( {2\widehat A + \widehat B + \widehat C} \right) = \tan \widehat A\)

c) Ta có: \(\widehat A + \widehat B + \widehat C = {180^o}\)

\( \Leftrightarrow \frac{{\widehat A + \widehat B + 3\widehat C}}{2} = \frac{{{{180}^o} + 2\widehat C}}{2} = {90^o} - \widehat C\)

\( \Rightarrow \sin \left( {\frac{{\widehat A + \widehat B + 3\widehat C}}{2}} \right) = \sin \left( {{{90}^o} - \widehat C} \right) = \cos \widehat C\)