ok so i'm starting 12th year and took extra maths class to prepair for belgium's polytechnic exam so basically what we do are exercices from those exams and we're reviewing trigonometry and this exercice has been messing with me for like 2 hours I can't do it so festo or someone a lil bit smart help me out pls
if a+b+c+d=2pi
show that:
sin a + sin b + sin c + sin d = 4sin[(a+b)/2]sin[(b+c)/2]sin[(a+c)/2]
i'll be wery thanksful if smone help me out with this <.<
all i know is you have to get rid of sin d since it's not back on the right side so what I did was
d = 2pi - (a+b+c)
sin d = sin (2pi-[a+b+c])
sin d = -sin (a+b+c)
good luck guys gonna go study french <.<
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