已知A是△ABC的内角,且tanA=-5 4,求sinA,cos
来源:学生作业帮助网 编辑:作业帮 时间:2024/09/20 01:12:19
(sina+cosa)^2=4/91+2sinacosa=4/9sin2a=-5/9180a>90所以是钝角三角形
tanA·tanC=2+√3(sinA·sinC)/(cosA·cosC)=2+√3(cos(A-C)-cos(A+C))/(cos(A+C)+cos(A-C)=2+√3设cos(A-C)=x,因为c
tanA是负数,说明该角是钝角,则用角A补角D(180-A=D)代替A即可.tanD=-tanA,sinD=sinA,cosD=-cosD, 现在角D画图结合勾股定理得知,SIND=3/5,
(1+tanA)(1+tanB)=2,1+tanA+tanB+tanAtanB=2tanA+tanB=1-tanAtanBtan(A+B)=(tanA+tanB)/(1-tanAtanB)=1,A+B
m*n=1*1*cos60=1/2=sinA*sinB-cosA*cosB=-cos(A+B)=cosCC=π/31/2absinc=S=2根号3ab=8c平方=a平方+b平方-2abcosC=a平方
(B+C)/2=(180°-A)/2=90°-A/2,sin[(B+C)/2]=sin(90°-A/2)=cos(A/2)cosA=2[cos(A/2)]^2-1cos2A=2(cosA)^2-1因此
设角平分线为AD则c²+AD²-2cADcosA/2=BD²①b²+AD²-2bADcosA/2=CD²②BD;CD=c:b=2:1所以①÷
解法一:由sinA+cosA=-0.2sin^2A+cos^2A=1得sinA*cosA=-12/2502tan(A/2)/(1+tan^2(A/2))+(1-tan^2(A/2))/(1+tan^2
∵tanA=-5/4,∴tan(180º-A)=5/4∴sinA=sin(180º-A)=5/√(5²+4²)=5/√41=5√41/41cosA=-cos(1
因为0<A<π/4所以cosA>sinA>0sin(π/4+A)=7√2/10sinπ/4cosA+cosπ/4sinA=7√2/10√2/2cosA+√2/2sinA=7√2/10cosA+sinA
B等式两边平方得:1+2sinAcosA=4/9,sinAcosA=-5/18
1、因为根号3b=2asinB,可得到b/sinB=2a/根号3.利用三角形的正玄定理,b/sinB=a/sinA.和前面的等式联立可求得A=60度.2、三角形面积S=1/2乘以bcsinA.可得bc
已知2sinA=√3sinC-sinB,将sinB=sin(A+C)=sinAcosC+cosAsinC代入:2sinA=√3sinC-sinAcosC-cosAsinC;分离A、C:sinA/(√3
sinC=sin(A+B)=sinAcosB+sinBcosA=2cosAsinB+sinBcosA=3cosAsinB∴cosA=sinC/3sinB=c/3b(正弦定理)余弦定理cosA=(c&s
cosC=(a^2+b^2-c^2)/2abcosB=(a^2+c^2-b^2)/2accosA=(c^2+b^2-a^2)/2bc代入a/cosA=b+c/cosB+cosC化简得a^2=b^2+c
(1)m‖n,(a+c)(sinA-sinC)=(a-b)sinB(a+c)(a/2R-c/2R)=(a-b)b/2R(a+c)(a-c)=(a-b)b(a²+b²-c²
(sinA+cosA)^2=1/4sinA^2+cosA^2+2sinAcosA=1/41+2sinAcosA=1/42sinAcosA=sin2A=-3/4cos2A=-根号7/4
sinA+cosA=根号2*(sinAcos45°+cosAsin45°)=根号2*sin(A+45°)=2/3,sin(A+45°)=2/3/根号2=0.47,sin150=0.5,0.47150,
∵tanA,tanB是二次方程x²+mx+m+1的两个实数根∴tanA+tanB=-mtanA*tanB=m+1∴tanC=tan[(180°-(A+B)]=-tan(A+B)=-(tanA
依余弦定理,cosC=(a^2+b^2_c^2)/2ab由已知:a^2+b^2-c^2=-ab,代入上式,得cosC=-1/2,C=120°.由b=2a及正弦定理,得sinB=2sinA.A+B=18