Cho tam giác ∆ABC chứng minh rằng:
a) \(\sin \left( {\frac{{\widehat A + \widehat B}}{2}} \right) = \cos \frac{{\widehat C}}{2}\);
b) \(\tan \left( {2\widehat A + \widehat B + \widehat C} \right) = \tan \widehat A\);
c) \(\sin \left( {\frac{{\widehat A + \widehat B + 3\widehat C}}{2}} \right) = \cos \widehat C\).
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: \(\widehat A + \widehat B + \widehat C = {180^o}\)
\( \Leftrightarrow \frac{{\widehat A + \widehat B + 3\widehat C}}{2} = \frac{{{{180}^o} + 2\widehat C}}{2} = {90^o} - \widehat C\)
\( \Rightarrow \sin \left( {\frac{{\widehat A + \widehat B + 3\widehat C}}{2}} \right) = \sin \left( {{{90}^o} - \widehat C} \right) = \cos \widehat C\)