1 2a·sinC
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/13 17:58:47
sinA+sinC=2sinBa+c=2bb=(a+c)/2,代人a^2-c^2=ac-bca^2-c^2=ac-((a+c)/2)c2a^2-ac-3c^2=0(2a-3c)(a+c)=02a-3c
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
1.假设a/sinA=b/sinB=c/sinC=2R那么sinA=a/2RsinB=b/2RsinC=c/2R因为(sinA)平方=(sinB)平方+sinC(sinB+sinC)所以(a/2R)^
C=180-(A+B)而sin(180-x)=sinx所以sinC=sin[180-(A+B)]=sin(A+B)
解由题意可得三角形ABC面积为18则有,S=1/2bcsina=1/2*c*12*1/2=18解得c=6又有余弦定理可得cosa=再求出sina有正弦定理可得sinbsinc代入即可求出
作CD垂直于AB交AB于点D,算了CD,再算BC根据BC/sinA=AB/sinC算出sinC从而,sinA+sinC=(√3)/2+(√21)/7
(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
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
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
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
A+B=180-C所以原式=cos(180-C)=-cosC选B再问:为什么A+B=180-C呢?再答:三角形内角和=?度
假设ABC外接圆半径R,有a=2RsinAb=2RsinBc=2RsinCc=2S/(b*sinA)=6a^2=b^2+c^2-2bc*cosA=180-72√3a=6√(5-2√3)(sinA+si
1.1)a×sinA+c×sinC-√2a×sinC=b×sinB2)sin(5π/6)=2sin(5π/12)cos(5π/12)a^2+c^2-√2a×c=b^2=sin(π/6)=1/2a^2+
(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
sinC=sin(180°-A-B)=sin(A+B)=2sin[(A+B)/2]cos[(A+B)/2]tan[(A+B)/2]=sin[(A+B)/2]/cos[(A+B)/2]即sin[(A+B
∵acosA+bcosB=ccosC∴sinAcosA+sinBcosB=sinCcosC∴sin2A+sin2B=sin2C=sin(2π-2A-2B)=-sin(2A+2B)∴0=sin2A+si
a(sinB-sinC)+b(sinC-sinA)+c(sinA-sinB)=2RsinAsinB-2RsinAsinC+2RsinBsinC-2RsinBsinA+2RsinCsinA-2RsinC
∵A+B+C=180°,2B=A+C,∴B=60°sinAsinC=cos²BsinAsinC=1/4sinAsin(A+π/3)=1/41/2sin²A+根号3/2sinAcos
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
设△ABC的外接圆半径为R,由正弦定理可得,sinB=b2R,sinA=a2R,sinC=c2R,所以a(sinB-sinC)+b(sinC-sinA)+c(sinA-sinB)=a(b2R−c2R)