cosC等于4分之1求sinc怎么求
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由正弦定理可得;sinA:sinB:sinC=a:b:c=2:3:4可设a=2k,b=3k,c=4k(k>0)由余弦定理可得,CosC=a2+b2−c22ab=4k2+9k2−16k22•2k•3k=
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
a:b:c=sinA:sinB:sinC=2:3:4,则设:a=2t、b=3t、c=4t,则:cosC=(a²+b²-c²)/(2ab)=-1/4
令a/sinA=b/sinB=c/sinC=ka:b:c=ksinA:ksinB:ksinC=2:3:4设a=2x,b=3x,c=4xcosC=(a²+b²-c²)/2a
因为sinA:sinB:sinc=2:3:4,根据正弦定理有a:b:c=2:3:4(abc为角ABC所对的角),根据余弦定理又有cosC=(a^2+b^2-c^2)/2ab=(4+9-16)/(2*3
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
方程左边2Sin2c*cosc-sin3c\x05方程右边=(1-Cosc)=2Sin2c*Cosc-Sin(c+2c)\x05=根号3*2sin(c/2)*sin(c/2)=2Sin2c*Cosc-
(cosA-2cosC)/cosB=2c-a/b(cosA-2cosC)/cosB=(2sinC-sinA)/sinB化简得到:2sin(C+B)=sin(A+B)即:sinC/sinA=2因为cos
因为cos2C=2(cosC)的方-1,如果cos2C=1\4,则:1/4=2(cosC)的方-1,解之得:cosC=(根号下10)/4,cosC=-(根号下10)/4,因为cos2C=1/4,所以c
做出来啦!不过这题目有点小问题,只有锐角三角形时此题成立钝角三角形不等式反向若A=120,B=30,C=30直角三角形为等号设q=(A-B)/2sinA+sinB-cosA-cosB=2cos(C/2
∵sinC=2sin0.5C×cos0.5C,cosC=cos0.5C×cos0.5C-sin0.5C×sin0.5C∴2sin0.5C×cos0.5C+cos0.5C×cos0.5C-sin0.5C
正弦定理得:a:b:c=sinA:sinB:sinC=2:3:4设:a=2k,b=3k,c=4kcosC=(a^2+b^2-c^2)/(2ab)=(4k^2+9k^2-16k^2)/(2*2k*3k)
1.cosA-2cosC/cosB=2c-a/b正弦定理(cosA-2cosC)/cosB=(2sinC-sinA)/sinBsinBcosA-2sinBcosC=2cosBsinC-cosBsinA
(1)已知sinA+sinB+sinC=0,cosA+cosB+cosC=0.求cos(B-C)的值.sinA=-(sinB+sinC)cosA=-(cosB+cosC)sinA^2+cosA^2=1
cosa+cosb+cosc=sina+sinb+sinc=0(cosa)^2=(cosb+cosc)^2=(cosb)^2+(cosc)^2+2*cosb*cosc.(1)(sina)^2=(sin
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
(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
(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
(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