Question

高一数学题，学霸帮帮忙谢谢！

Answer 1

cos2C=1-2sin^2(C)=cos^2(C)-sin^2(C)=-1/4
2sin^2(C)=5/4 sinC=√5/8
sinA=0.5sinC=1/4√5/2=√5/32
cosB=cos(Pi-A-C)=-cos(A+C)=-(cosAcosC-sinAsinC)=-(√81/256-√25/256)
=-(9/16-5/16)=-4/16=-1/4
a/sinA=b/sinB=c/sinC
2/sinA=c/2sinA c=4
b^2=a^2+c^2-2accosB b^2=4+16+2*2*4*(1/4)=24
b=2√6

Answer 2

（1）cos2C=1-2sin²c=-1/4
∵sinC>0, ∴sinC=√10/4
(2)sinA/a=sinC/c(正弦定理）∴c=4
cosC=√6/4或-√6/4
cosC=(a²+b²-c²)/2ab(余弦定理）
求出b=2√6(cosC=√6/4)或√6（cosC=-√6/4)均符合题意

Answer 3

cos2X=(cosX)^2-(sinX)^2
=2*(cosX)^2-1
=1-2*(sinX)^2
由此公式得到sinC
在列余弦定理公式求解第二道题