在三角形abc中,cosb＝负13分之5.cosc＝5分之4,求sina的值

由(sinA+sinB)/sinC=(a+b)/c=cosA+cosB=(b^2+c^2-a^2)/2bc+(a^2+c^2-b^2)/2ac得：a^3+b^3+a^2b+ab^2-ac^2-bc^2

化为c/a=2cosB又c/a=sinC/sinA所以sinC=2sinAcosB因为A+B+C=180sinC=sin(A+B)=sinAcosB+sinBcosA于是sinAcosB=sinBco

由正弦定理,b/sinB=c/sinC得b=sinB·c/sinC代入原式得cosC·sinB·c/sinC=c·cosBsinB·cosC=sinC·cosBsinB·cosC-sinC·cosB=

a/cosB=b/cosAa/b=cosB/cosA由正弦定理a/sinA=b/sinB所以a/b=sinA/sinB所以cosB/cosA=sinA/sinBsinAcosA=sinBcosB2si

反例:A=120,B=30,则sinA=cosB=sin60,此三角形显然不是直角三角形

正弦定理a/sinA=b/sinB=>a/b=sinA/sinBa*cosA=b*cosB=>a/b=cosB/cosA则cosB/cosA=sinA/sinB即sinAcosA-cosBsinB=0

sinB=√(1-cos^2B)=12/13sinC=√(1-cos^2C)=3/5sinA=sin(B+C)=sinBcosC+cosBsinC=33/65.(1/2)AB×AC×sinA=33/2

cosB/cosC=-b/(3a+c)=-sinB/(3sinA+sinC)(由正弦定理得到此步）之后,等号左右变形-3cosBsinA-cosBsinC=cosCsinB-3cosBsinA=cos

在三角形ABC中sinA=sin(B+C)所以sinA+cosB=根2/2即sin(B+C)+cosB=根2/2由AC=b=2AB=c=3以及正弦定理a/SinA=b/SinB=c/SinC可知3*s

sinA=sin(A+B)所以有2sin(B+C)*(cosB+cosC)=sinB+sinC2(sinB*cosC+csB*sinC)*(cosB+cosC)=sinB+sinC化解得sin（B+2

解由sinC=（sinA+sinB）／（cosA+cosB）即sinA+sinB=sinCcosA+sinCcosB即sin（B+C)+sin(A+C)=sinCcosA+sinCcosB即sinBc

cosB/cosA=b/a=sinB/sinAsinAcosB-cosAsinB=0sin(A-B)=0A=B同理,B=C所以三角形是正三角形☆⌒_⌒☆希望可以帮到you~

三角形ABC中,若COSA＋COSB＝SINC,则三角形ABC是直角三角形证明如下cosA+cosB=sinC=sin[π-(A+B)]∴cosA+cosB=sin(A+B)∴cosA+cosB=si

cosB等于负十三分之五,那么sinB=12/13cosC等于五分之四,那么sinC=3/5cosA=cos[180°-(B+C)]=-cos(B+C)=-cosBcosC+sinBsinC=-(-5

∵cosB/cosA=a/b又：根据正弦定理：a/b=sinA/sinB∴cosB/cosA=sinA/sinB∴cosAsinA=cosBsinB∴2sinAcosA=2sinBcosB∴sin2A

cosA/cosB＝b/aa/b=cosB/cosA由正弦定理a/sinA=b/sinB所以a/b=sinA/sinB所以cosB/cosA=sinA/sinBsinAcosA=sinBcosB2si

A是内角所以0

即sinAsinCcosB/cosC

cosB等于负十三分之五,那么sinB=12/13cosC等于五分之四,那么sinC=3/5cosA=cos[180°-(B+C)]=-cos(B+C)=-cosBcosC+sinBsinC=-(-5

cosA＝3/5.sinA=4/5,cosB＝5/13,sinB=12/13(由于余弦是正,所以角度是锐角）cosC=cos(π-A-B)=-cos(A+B)=-(cosAcosB-sinAsinB)