Numarum l de submiltimi de 10 luate cate 3
C^3=10 !/3 !*7!=8*9*10/6=120
numarul ede elemente ale multrimii A care nu contin pe 1=combinari 9 luate cate 3=84
Numarul de elemente care contin cifra 1=120-84=36
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AB=√(2-m)²+(m+2)²=4 =>
2m²+8=16
m=+/-2
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3.Notezi Expresia cu E(π/12)
cos23π/12=cos(24π/12-π/12)=cos(2π-π/12)=c0s2π*cosπ/12+sinπ/12*sin2π=
cosπ/12
E(π/12)= ctg π/12)
ctg x=(1-t²)/2t unde t=tg x/2
ctg π/6
Exprima,c tg π/6 in functie de t unde t=tg π/12
tgπ/6=√3=(1-t²)/2t
t²+2t√3-1=0
t1=-√3-2=>ctgπ/12=1/t1=-1/(√3+2)= a,plifici cu conjugata (√3-2)
t2=-√3+2=> ctg π/12=1/t2=-1/(√3-2)=amplifici cu conjugata=√3-2
E(π/12)=√3-2
