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MBAѧĿ֮رʽʮ

2014-01-10 14:55:22| չԱ
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Ǻ

Ǻ͹ʽ

sin(A+B)=sinAcosB+cosAsinB sin(A-B)=sinAcosB-sinBcosA

cos(A+B)=cosAcosB-sinAsinB cos(A-B)=cosAcosB+sinAsinB

tan(A+B)=(tanA+tanB)/(1-tanAtanB) tan(A-B)=(tanA-tanB)/(1+tanAtanB)

ctg(A+B)=(ctgActgB-1)/(ctgB+ctgA) ctg(A-B)=(ctgActgB+1)/(ctgB-ctgA)

ǹʽ

tan2A=2tanA/(1-tan2A) ctg2A=(ctg2A-1)/2ctga

cos2a=cos2a-sin2a=2cos2a-1=1-2sin2a

ǹʽ

sin(A/2)=√((1-cosA)/2) sin(A/2)=-√((1-cosA)/2)

cos(A/2)=√((1+cosA)/2) cos(A/2)=-√((1+cosA)/2)

tan(A/2)=√((1-cosA)/((1+cosA)) tan(A/2)=-√((1-cosA)/((1+cosA))

ctg(A/2)=√((1+cosA)/((1-cosA)) ctg(A/2)=-√((1+cosA)/((1-cosA))

Ͳ

2sinAcosB=sin(A+B)+sin(A-B) 2cosAsinB=sin(A+B)-sin(A-B)

2cosAcosB=cos(A+B)-sin(A-B) -2sinAsinB=cos(A+B)-cos(A-B)

sinA+sinB=2sin((A+B)/2)cos((A-B)/2 cosA+cosB=2cos((A+B)/2)sin((A-B)/2)

tanA+tanB=sin(A+B)/cosAcosB tanA-tanB=sin(A-B)/cosAcosB

ctgA+ctgBsin(A+B)/sinAsinB -ctgA+ctgBsin(A+B)/sinAsinB

ijЩǰn

1+2+3+4+5+6+7+8+9+…+n=n(n+1)/2 1+3+5+7+9+11+13+15+…+(2n-1)=n2

2+4+6+8+10+12+14+…+(2n)=n(n+1) 1^2+2^2+3^2+4^2+5^2+6^2+7^2+8^2+…+n^2=n(n+1)(2n+1)/6

1^3+2^3+3^3+4^3+5^3+6^3+…n^3=(n(n+1)/2)^2 1*2+2*3+3*4+4*5+5*6+6*7+…+n(n+1)=n(n+1)(n+2)/3

Ҷ a/sinA=b/sinB=c/sinC=2R ע R ʾεԲ뾶

Ҷ b2=a2+c2-2accosB עBDZaͱcļн

ʽ ʽʽ

˷ʽ a2-b2=(a+b)(a-b) a3+b3=(a+b)(a2-ab+b2) a3-b3=(a-b(a2+ab+b2)

Dzʽ |a+b|≤|a|+|b| |a-b|≤|a|+|b| |a|≤b<=>-b≤a≤b

|a-b|≥|a|-|b| -|a|≤a≤|a|

һԪη̵Ľ -b+√(b2-4ac)/2a -b-√(b2-4ac)/2a

ϵĹϵ x1+x2=-b/a x1*x2=c/a עΤﶨ

ijЩǰn

1+2+3+4+5+6+7+8+9+…+n=n(n+1)/2 1+3+5+7+9+11+13+15+…+(2n-1)=n2

2+4+6+8+10+12+14+…+(2n)=n(n+1) 12+22+32+42+52+62+72+82+…+n2=n(n+1)(2n+1)/6

13+23+33+43+53+63+…n3=n2(n+1)2/4 1*2+2*3+3*4+4*5+5*6+6*7+…+n(n+1)=n(n+1)(n+2)/3

Ҷ a/sina=b/sinb=c/sinc=2r ע r ʾεԲ뾶

Ҷ b2=a2+c2-2accosb עbDZaͱcļн

Բı׼ (x-a)2+(y-b)2=r2 ע(a,b)Բ

Բһ㷽 x2+y2+dx+ey+f=0 עd2+e2-4f>0

߱׼ y2=2px y2=-2px x2=2py x2=-2py

ֱ s=c*h б s=c'*h

׶ s=1/2c*h' ̨ s=1/2(c+c')h'

Բ̨ s=1/2(c+c')l=pi(r+r)l ı s=4pi*r2

Բ s=c*h=2pi*h Բ׶ s=1/2*c*l=pi*r*l

ʽ l=a*r aԲĽǵĻr >0 ʽ s=1/2*l*r

׶ʽ v=1/3*s*h Բ׶ʽ v=1/3*pi*r2h

б v=s'l ע,s'ֱ lDzⳤ

ʽ v=s*h Բ v=pi*r2h



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