△ABC中,sinA=2sincosB

现取(A+B)/2=x,(A-B)/2=y则A=x+yB=x-y于是所证式变为sin(x+y)+sin(x-y)=2sinxcosy这是易证的sin(x+y)=sinxcosy+sinycosxsin

这个式子可以化为：b2-c2=a(√2b-a)b2-c2=√2ab-a2a2+b2-c2=√2abcosC=a2+b2-c2/2ab=√2ab/2ab=√2/2又因为在△ABC中,c在0—180度,所

解析：1）∵(cosA+cosB)sinC=sinA+sinB∴cosA+cosB=(sinA+sinB)/sinC即（b^2+c^2-a^2)/2bc+（a^2+c^2-b^2)/2ac=(a+b)

sin²A=sin²B+sin²C,a/sinA=b/sinB=c/sinC=2R(a/2R)^2=(b/2R)^2+(c/2R)^2a^2=b^2+c^2,ABC是直角

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

sinA=2sinBcosCsin(180-B-C)=2sinBcosCsin(B+C)=2sinBcosCsinBcosC+cosBsinC=2sinBcosCsinBcosC-cosBsinC=0

由和差化积公式：sinA=sin(B+C)=sinBcosC+cosBsinC=2sinBcosC,所以cosBsinC-sinBcosC=0,即sin(B-C)=0.从而B=C,因此三角形ABC是等

sin²A＝sin²B+sin²C根据正弦定理∴a²=b²+c²∴A=90º∵sinA=2sinBsinC∴2sinBsinC=1

因为a/sinA=b/sinB=c/sinC=2Rsin^2A=sin^2B+sin^2C=》a^2=b^2+c^2=>是直角三角形,A=90度=》B+C=90sinA=2sinBcosC=1=》si

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

纠正一下题目：应该是tanC=（sinA+sinB）/（cosA+cosB）因为tanC=(sinA+sinB)/(cosA+cosB),sinC/cosC=(sinA+sinB)/(cosA+cos

sinA=2sinAcosB?改哈题1.1.∵sinA=2sinCcosB∴sinA=sin(B+C)=2sinCcosB即sinBcosC+cosBsinC=2sinCcosB∴sin(B-C)=0

7/4-根号3/2

sin^2A=sin^2B+sin^2C,sinA=2sinBsinC所以sin^2A-sinA=sin^2B+sin^2C-sinA=sin^2B+sin^2C-2sinBsinC即sinA(sin

你的表述出现了一些问题,我想应该是求证：［sin（A/2）］^2＋［sin（B/2）］^2＋［sin（C/2）］^2＝1－2sin（A/2）sin（B/2）sin（C/2）若是这样,则方法如下：在三角

sina+cosa=1/5平方1+2sinacosa=1/25sinacosa=-12/25sina=4/5cosa=-3/5或sina=-4/5舍去cosa=3/5所以sina=4/5cosa=-3

sin²A-sin²(180-A-B)=sinAsinB-sin²Bsin²A-sin²(A+B)=sinAsinB-sin²Bsin&su

证明：（1）左式=sin²A+sin²B-sin²(180-A-B)=sin²A+sin²B-sin²(A+B)=sin²A+si

改了结果相同由正弦定理a/sinA=b/sinB=c/sinC(sinA)^2=(sinB)^2+(sinC)^2等价于a^2=b^2+c^2可知△ABC直角三角形A=π/2sinA=2sinBcos