Question

在三角形ABC中，已知AC=2,BC=3,cosA=-4/5，求sinB的值 以及sin(2B+30度)的值

Answer 1

依题意sinA=3/5
由正弦定理 BC/sinA=AC/sinB
3/(3/5)=2/sinB
sinB=2/5
因cosA=-4/5<0,所以A为钝角，那么B为锐角
cosB=√21/5，sin2B=2sinBcocB=4√21/25,cos2B=1-2sin^2B=1-8/25=17/25
sin(2B+30°）=sin2Bcos30°+cos2Bsin30)
=(4√21/25)*(√3/2)+(17/25)*(1/2)
=6√7/25+17/50

Answer 2

sinA=√1-(cosA)^2=3/5
据正弦定理得:sinA/a=sinB/b,得 sinB=2/5,cosB=√1-(sinB)^2=√21/5(锐角取正)
sin(2B)=2sinBcosB=2*(2/5)*√21/5=4√21/25,
cos(2B)=√1-(sin2B)^2=17/25
sin(2B+30°)=sin2Bcos30°+cos2Bsin30°
=(17+12√7)/50