已知sinB sinA(sinC-cosC)=0,a=2,c=根号2,则C=

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

设三边长分别为a、b、c,已知a+b+c=√2+1sinA+sinB=√2sinCS=(SinC)/6正弦定理：sinA/a=sinB/b=sinC/c变形得sinA/a=sinB/b=sinC/c=

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

（Ⅰ）△ABC中，由已知条件可得sin2A-sin2B=2sinAsinC-sin2C，再由正弦定理可得a2+c2-b2=2ac，∴cosB=a2+c2−b22ac=22，∴B=π4．（Ⅱ）∵B=π4

应当是sin^2A+sin^2B【+】sin^2C=sinB*sinC+sinC*sinA+sinA*sinB吧括号中是要改的.两边同乘以22sin²A+2sin²B+2sin&s

120°利用前两个比例：5（sinB+sinC）=4(sinC+sinA)化简得到sinC=4sinA-5sinB利用后两个比例：6（sinC+sinA)=5(sinA+sinB)化简得到sinA=5

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

1）.对于第一个等式,两边除以sinA·sinB,并利用正弦定理得到：1=(c^2-a^2)/ab,c^2=a^2+ab,再用余弦定理：c^2=a^2+b^2-2ab·cosC,联立得到：b^2=ab

（1）由a−cb−c＝sinBsinA+sinC，得a−cb−c＝ba+c，即a2=b2+c2-bc，由余弦定理，得cosA＝12，∴A＝π3．（2）f（x）=cos2（x+A）-sin2（x-A）=

(b-a)(sinA+sinB)=bsinA,(b-a)/b=sinA/(sinA+sinB),(1)根据正弦定理,sinA/a=sinB/b,(sinA+sinB)/(a+b)=sinA/a,(等比

由sinA/a=sinB/b=sinC/c(其中a,b,c为角A,B,C对应的三条边)设sinA/a=sinB/b=sinC/c=k则a=sinA/k,b=sinB/k,c=sinC/k带入(sinB

tan(A-B)=(tanA-tanB)/(1+tanA*tanB)tan(A-B)/tanA+sin²C/sin²A=1左右移项得1-[(tanA-tanB)/(1+tanA*t

sinA+sinB=根号2sinCa+b=根号2ca+b+c=根号2+1c=1a+b=根号2a*b=1/3a^2+b^2+2ab=24ab=4/3a^2+b^2-2ab=2/3a-b=正负根号6/32

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

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

90度

（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

∵acosA+bcosB=ccosC∴sinAcosA+sinBcosB=sinCcosC∴sin2A+sin2B=sin2C=sin(2π-2A-2B)=-sin(2A+2B)∴0=sin2A+si