sin(a-b)=-sinc吗
来源:学生作业帮助网 编辑:作业帮 时间:2024/09/22 00:59:37
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
sinA=sinB在三角形中,这句话等价于A=B或者A+B=π通过诱导公式sin(π-A)=sinA可以证明所以A-B=A+BORA-B+A+B=π显然前者推得B=0舍去,所以只能去后者,所以是个直角
根据正弦及余弦定理可得sin(A-B)/sinC=(sinAcosB-cosAsinB)/sinC=(acosB-bcosA)/c=[(a²+c²-b²)/2c-(b
因为a/sinA=b/sinB=c/sinC=2r(r是三角形外接圆半径)所以a=2rsinA,b=2rsinB,c=2rsinC代入,等式左边=[(sinA)^2-(sinB)^2]/(sinC)^
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
C=180-(A+B)而sin(180-x)=sinx所以sinC=sin[180-(A+B)]=sin(A+B)
【无条件相等】∵A+B+C=π∴C=π-(A+B)∴sinC=sin[π-(A+B)]∵sin[π-(A+B)]=sin(A+B)∴sinC=sin(A+B)
左边=sin(A+B)sin(B-A)+sin²C=sin(180-C)sin(B-A)+sin²C=sinCsin(B-A)+sin²C=sinC[sin(B-A)+s
左边=sin(A+B)sin(B-A)+sin²C=sin(180-C)sin(B-A)+sin²C=sinCsin(B-A)+sin²C=sinC[sin(B-A)+s
∵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(a²+b²)sin(A-B)=(a²-b²)sinC=)=(a²-b²)sin(A+B)可得a²tanB=b²tan
由正弦定理,原式可化为a^2+c^2-ac=b^2即[(a^2+c^2-b^2)/2ac]=0.5即cosB=0.5∴B=π/3
,{sin(A-B)+sinC)/{cos(A-B)+cosC}=,{sin(A-B)+sin(A+B))/{cos(A-B)-cos(A+B)}=2sinAcosB/2sinAsinB=cosB/s
因为正弦定理a/sinA=b/sinB=c/sinC所以sin(A-B)/sinC=(sinAcosB-cosAsinB)/sinC=(acosB-bcosA)/c=[a*(a²+c
sin²A-sin²B-sin²C=sinBsinCa/sinA=b/sinB=c/sinC则由sin²A-sin²B-sin²C=sinB
由sinc+cosc=2sina平方可得1+2sinc*cosc=4sin^2a因sinc*cosc=sin^2b所以1+2sin^2b=4sin^2a2-4sin^2a=1-2sin^2b2cos2
解题思路:利用三角函数的和差角公式和二倍角公式解答解题过程:
星凝冰雨您好在三角形ABC中,sin(A+B)=sinC这种说法是正确的因为在△ABC中,A+B+C=π,所以sin(A+B)=sinC我在百度上搜了一下,有如下可能:1您的书上的sin(A+B)=s