How do you use the sum and difference identity to evaluate #sin(pi/12)#?
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 … Leggi tutto