. Ta có:$ A + B +C = \piVT=(sinA+sinB)+sinC=2.\sin \frac {A+B}{2}.\cos \frac {A-B}{2} + 2.\sin \frac{C}{2}.\cos \frac {C}{2}(∗ÁpdụngCTbiếntổngthànhtíchvàCTnhânđôi.)=2.\cos \frac {C}{2}.\cos \frac {A-B}{2} + 2.\sin \frac {C}{2}.\cos \frac {C}{2}(∗\sin \frac {A+B}{2} = \sin (\frac {\pi}{2} - \frac {C}{2}) = \cos \frac {C}{2})=2.\cos \frac {C}{2}.(\cos \frac {A-B}{2} + \sin \frac {C}{2})=2.cosC2.(cosA−B2+cosA+B2)$(∗$cosA+B2=cos(π2−C2)=sinC2$)$=2.cosC2.(2.cosA2cosB2) = 4.\cos \frac {A}{2} \cos \frac {B}{2} \cos \frac {C}{2}$$ = VP $ => Đpcm
. Ta có:
A+B+C=π$VT = (\sin A + \sin B) +
\sin C
=2.sinA+B2.cosA−B2+2.sinC2.cosC2$(∗ÁpdụngCTbiếntổngthànhtíchvàCTnhânđôi.)$=2.cosC2.cosA−B2+2.sinC2.cosC2$(∗$sinA+B2=sin(π2−C2)=cosC2$)$=2.cosC2.(cosA−B2+sinC2)=2.\cos \frac {C}{2}.(\cos \frac {A-B}{2} + \cos \frac {A+B}{2})
(∗\cos \frac {A+B}{2} = \cos (\frac {\pi}{2} - \frac {C}{2}) = \sin \frac {C}{2}
)=2.\cos \frac {C}{2}.(2.\cos \frac {A}{2} \cos \frac {B}{2})
=4.cosA2cosB2cosC2 = VP $ => Đpcm