sinB=60度,求sinA sinC的取值范围
来源:学生作业帮助网 编辑:作业帮 时间:2024/10/07 21:40:35
sinA+sinC=2sinB2sin[(A+C)/2]·cos[(A-C)/2]=2sinB(和差化积)∵A+B+C=π∴sin(π/2-B/2)·cos(π/6)=sinBcos(B/2)·cos
前提一个公式:(sina)^2+(cosa)^2=1,(sinb)^2+(cosb)^2=1对于sina*sinb=1,由于sin函数的值域[-1,1],则显然sina=sinb=1或sina=sin
(a+b+c)/(sinA+sinB+sinC)=(2RsinA+2RsinB+2RsinC)/(sinA+sinB+sinC)=2R三角形面积S=bc*sinA/2=根号3*c/4=根号3所以c=4
(b*sinB)/c=(sinB*a)/b=[(b*sinA)/a]*(a/b)=sinA=sin60=根号3/2
c^2=a^2+b^2-2ab*cosC=36+16-2*6*4*0.5=28c=2*根号7b/sinB=c/sinC带入数值,可得sinB=根号3/根号7
、根据正弦定理a/sinA=b/sinB=c/sinC得:a=(sinA/sinB)*bc=(sinC/sinB)*b将其带入已知条件a+c=2b中可得sinA+sinC=2sinB根据三角函数和公式
a+c=2b2RsinA+2RsinC=4RsinBsinA+sinC=2sin(A+C)2sin(A+C)/2cos(A-C)/2=4sin(A+C)/2cos(A+C)/2cos(A-C)/2=2
/>C=60度∴B=120°-A且0
SinB+CosB=根2根号2(2/根号2SinB+2/根2CosB)=根号2根号2Sin(B+4/π)=根号2Sin(B+4/π)=1B+4/π=2/π+2kπ,k属于zB=4/π+2kπ,k属于z
sinB=2sinC,∴由正弦定理知:b=2c∵cosA=cos60°=1/2∴(b^2+c^2-a^2)/2bc=1/2即:(4c^2+c^2-a^2)/4c^2=1/2本题应该少1个条件,即a边的
S=1/2bcsinA=1/2*12c*√3/2=18√3c=6a^2=b^2+c^2-2bccosA=144+36-2*12*6*1/2=108a=6√3所以a/sinA=12则a/sinA=b/s
原式=(sina)^2+(sinb)^2+2sinasinb-2sinasinb=(sina+sinb)^2-2sinasinb依据和差化积公式和积化和差公式得={2sin[(a+b)/2]cos[(
B=120°-AsinA+sinB=sinA+sin(120°-A)=3/2*sinA+根号3/2*cosA=根号3*sin(A+30°)
①等腰△ABC内A为顶角,sinB=8/17,得0<B=C<π/4A=π-2B>π/2∴cosB=15/17sinA=sin2B=2sinBcosB=240/289cosA=-161/289tanA=
1/sinB+1/sinA=(sinA+sinB)/(sinA*sinB)=2(sinA+sinB)/[(sinA+sinB)^2-1]设X=sinA+sinB∈(1,根2]则上式=2/(x-1/x)
原式=sin²a+2sinasinb+sin²b+cos²a+2cosacosb+cos²b=(sin²a+cos²a)+2(cosacos
a,b,c成等差数列,2b=a+c,角A-角C=π/32sinB=sinA+sinC,2sin(A+C)/2*cos(A-C)/2=2sinBcosB/2*根号3=2sinB,sinB/2=根号3/4
∵A+B=120º∴C=180º-(A+B)=60º∵c=3根据正弦定理2R=c/sinC=3/(√3/2)=2√3∴sinA=a/(2R)=a/(2√3)sinB=b/
(1)由余弦定理得a²=b²+c²-2bccosA由已知b²+c²=a²+√3bc得a²=b²+c²-√3bc
直角三角形中假设∠A=90°则sinA=sinB得出∠B=90°不能成三角形,∠B同理;故必∠C=90°;另两角都是小于90°的锐角;sinA=sinB得出∠A=∠B=45°即sinAsiinB=√2