sinC=0.6的角

sinC+cosC=1-sin（C/2）sinC=1-cosC-sin（C/2）2sin(C/2)cos(C/2)=2sin²(C/2)-sin（C/2)∵sin(C/2)≠0∴2cos(C

sinC+cosC=1-sinC/2sinC=1-sinC/2-cosC2sinC/2cosC/2=1-sinC/2-1+2sin^2C/22sinC/2cosC/2=sinC/2(2sinC/2-1

在△ABC中A+B=180°-C(A+B)/2=90°-C/2tan(A+B)/2=tan(90°-C/2)=1/tan(C/2)=cos(C/2)/sin(C/2)tan(A+B)/2=sincco

由正余弦课得.a/sinA=b/sinB=c/sinC=2RsinA=a/2RsinB=b/2RsinC=c/2R将sinA=a/2RsinB=b/2RsinC=c/2R带入sinA^2=sinB^2

和差化积公式sinA+sinC=2sin[(A+C)/2]cos[(A-C)/2]因为A+C固定120所以=2sin60cos[(A-C)/2]=（根三）cos[(A-C)/2]容易知道cos[(A-

m垂直n,则有m*n=1-sinC/2-(sinC+cosC)=0,sinC+cosC=1-sinC/2（1）sinC+cosC=1-sinC/2,移项得sinC-sinC/2=1-cosC由二倍角公

（1）∵在△ABC中,b2=a2+c2-2accosB,∴b2-a2-c2=-2accosB,同理可得c2-a2-b2=-2abcosC∵sinC/(2sinA−sinC)＝(b2W

tan(A+B)=2因为C=180º-(A+B)所以,tanC=-tan(A+B)tanC=-2sinC=-2cosC=-2√(1-sin²C)sin²C=4-4sin&

∵sinC=2sin0.5C×cos0.5C,cosC=cos0.5C×cos0.5C-sin0.5C×sin0.5C∴2sin0.5C×cos0.5C+cos0.5C×cos0.5C-sin0.5C

因为a/sinA=b/sinB=c/sinC所以sinA：sinB：sinC=a：b：c所以a：b：c=3：2：4设a=3,b=2,c=4cosC=(a^2+b^2-c^2)/2ab=(9+4-16)

m⊥n=>m.n=0(2cos(C/2),-sinC).(cos(C/2),2sinC)=02(cosC/2)^2-2(sinC)^2=0(2(cosC/2)^2-1)-2(sinC)^2+1=0co

把它变化为正玄定理(a+b+c)(a+b-c)=aba^2+b^2+2ab-c^2=ab(a^2+b^2-c^2)/ab=-1由余弦定理(a^2+b^2-c^2)/2ab=-1/2=cosCc=120

1.sinC+cosC化成半角,2sinc/2cosc/2+1-2sinc/2sinc/2原式化为cosC/2-sinC/2=0两边平方,得到1-sinC=0即sinC=12.条件不足,看看题是否写错

∵sinC=2sin0.5C×cos0.5C,cosC=cos0.5C×cos0.5C-sin0.5C×sin0.5C∴2sin0.5C×cos0.5C+cos0.5C×cos0.5C-sin0.5C

由已知表达出sinA^2-sinC^2+sinB^2-sinAsinB=0由正弦定理c^2=a^2+b^2-ab所以C是60°再表达s+t=(COSA,-1+2COS(B/2)^2)=(COSA,CO

（1）sinC+cosC=1-sinC/2,移项得sinC-sinC/2=1-cosC由二倍角公式得2sinC/2cosC/2-sinC/2=2(sinC/2)^2因为sinC/2≠0,所以两边消去s

（1）sinC+cosC=1-sinC/2,移项得sinC-sinC/2=1-cosC由二倍角公式得2sinC/2cosC/2-sinC/2=2(sinC/2)^2因为sinC/2≠0,所以两边消去s

由sinC=4/5得cosC=±3/5sinB=sin(A+C)=sinAcosC+cosAsinC=1/2*(±3/5)+√3/2*4/5sinB=(4√3±3)/10再由正弦定理,AB/sinC=

B=90则A+C=90sinA+sinC=sinA+sin(90-A)=sinA+cosA=√2*sin(A+45)B=90,所以0

由已知得：(sinB+cosB)sinC+（sinB-cosB)cosC=-1/5即-cos(B+C)+sin(B+C)=-1/5即cosA+sinA=-1/5联立cosA^2+sin^2=1得sin