Question

在三角形abc中,已知4sin^2(a-b)/2+4sinasinb=2+根号下2求角C

Answer 1

4sin^2(A-B)/2=4[1-cos(a-b)]/2=2[1-cos(A-B)]

所以左边=2-2(cosAcosB+sinAsinB)+4sinAsinB

=2-2cosAcosB+2sinAsinB

=2-2(cosAcosB-sinAsinB)

=2-2cos(A+B)

于是2-2cos(A+B)=2+根号2

所以-cos(A+B)=根号2/2

-cos(A+B)=cos【π-(A+B)】=cosC

所以C=45°