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就 一视频 一网友 一评论 之 具体过程 飨以诸君

2022-04-26 22:12 作者:Mynasty  | 我要投稿


BV1ZA4y1X71H


A/2+B+(C-B)/2=π/2

sin∠ABP=sin(A/2+B)=cos((C-B)/2)

2Rcos((B-C)/2)=AP

AO

=

AP-OP

=

AP-BP

=

2Rcos((B-C)/2)-2Rcos((B+C)/2)

=

4Rsin(B/2)sin(C/2)



PC=PO=PB

a/OP

=

a/PC

=

2cos∠PCB

=

2cos∠PAB

=

(b+c-a)/AO

AO/OP=(b+c-a)/a



AO/OP=(sinB+sinC-sinA)/sinA

OP

=

4Rsin(B/2)sin(C/2)sinA

/

(sinB+sinC-sinA)

=

2Rsin(A/2)(4sin(B/2)sin(C/2)cos(A/2)

/

(sinB+sinC-sinA)

=

2Rsin(A/2)

(2cos((B-C)/2)-2cos((B+C)/2))

cos(A/2)

/

(sinB+sinC-sinA)

=

2Rsin(A/2)

(2cos((B-C)/2)-2cos((B+C)/2))

sin((B+C)/2)

/

(sinB+sinC-sinA)

=

2Rsin(A/2)

(sinB+sinC-sin(B+C))

/

(sinB+sinC-sinA)

=

2Rsin(A/2)

(sinB+sinC-sinA)

/

(sinB+sinC-sinA)

=

2Rsin(A/2)




cosA+cosB+cosC-1

=

ab²+ac²-a³+a²b+bc²-b³+a²c+b²c-c³-2abc

/

2abc

=

(4a²b²-(a²+b²-c²)²)/(a+b+c)

/

2abc

=

2ab/(c(a+b+c))

(4a²b²-(a²+b²-c²)²)/(4a²b²)

=

absinC/(a+b+c)

/

c/(2sinC)

=

r/R


AO·OP

=

8R²sin(A/2)sin(B/2)sin(C/2)

=

2R

4Rsin(A/2)sin(B/2)sin(C/2)

=

2R

2R

(cos((A-B)/2)-cos((A+B)/2))

sin(C/2)

=

2R

2R

(cos((A-B)/2)-cos((A+B)/2))

cos((A+B)/2)

=

2R

R

(cosA+cosB-cos(A+B)-1)

=

2R

R

(cosA+cosB+cosC-1)

=

2Rr

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