Question

在△ABC中,内角A,B,C对边的边长分别是a,b,c已知c=根号7,c=60°,COSB=3根号10/10,求sin2

Answer 1

∵cosB=3√10/10,

∴sinB=√10/10,

又∵C=60°,

∴sin(B+C)=sinBcosC + cosBsinC

=√10/10 ·1/2 + 3√10/10·√3/2

=√10/20 ·(1+ 3√3)

∴cos(B+C)=cosBcosC - sinBsinC

=3√10/10 ·1/2 - √10/10·√3/2

=√10/20 ·(3-√3)

所以 sin2A

=sin2(180°-B-C)

=sin[360°-2(B+C)]

=-sin2(B+C)

=-2sin(B+C)cos(B+C)

=-2·√10/20 ·(1+ 3√3)·√10/20 ·(3-√3)

=(3-4√3)/10