(1)过B点作BD垂直于AC于D点,所以BD=a·sinC=2·根号3/3 所以sinA=BD/c=(2·根号3/3)÷根号3=2/3
(2)因为sin2C=2sinCcosC 所以sinCcosC=(sin2C)/2 因为cos2C=2cosC^2一1 所以cosC^2=(cos2C+1)/2 所以f(C)=根号3sinCcosC一(cosC)^2=3·sin2C/2一(cos2C+1)/2=3·sin2C/2一cos2C/2一1/2
=(根号3/2)·2sinCcosC
(1)过B点作BD垂直于AC于D点,所以BD=a·sinC=2·根号3/3 所以sinA=BD/c=(2·根号3/3)÷根号3=2/3
(2)因为sin2C=2sinCcosC 所以sinCcosC=(sin2C)/2 因为cos2C=2cosC^2一1 所以cosC^2=(cos2C+1)/2 所以f(C)=根号3sinCcosC一(cosC)^2=3·sin2C/2一(cos2C+1)/2=3·sin2C/2一cos2C/2一1/2
=(根号3/2)·2sinCcosC