How do you use the sum and difference identity to evaluate #sin(pi/12)#?
Risposta:
#(sqrt2/4)(1 - sqrt3)#
Spiegazione:
Usa identità trig:
sin (a - b) = sin a.cos b - sin b.cos a
#sin (pi/12) = sin (pi/3 - pi/4) = sin (pi/3).cos (pi/4) - sin (pi/4).(cos pi/3)#
Trig table gives;
#sin (pi/3) = 1/2# e #cos (pi/3) = sqrt3/2#
#sin (pi/4) = sqrt2/2# e #cos (pi/4) = sqrt2/2#
Otteniamo:
#sin (pi/12) = (1/2)(sqrt2/2) - (sqrt2/2)(sqrt3/2)#
#= (sqrt2/4)(1 - sqrt3)#