b-c/a.cosA/2=sin b-c/2
Answers
Answered by
1
= sin B cos(C/2) - sin(C/2) cos B
You could use the half-angle formulas if you like
Since cos(2u) = 1 - 2 cos^2 u = 2 sin^2 u - 1
then sin u = sqrt( [cos (2u) - 1]/2 )
and cos u = sqrt( [1 - cos(2u)]/2 ) if i did that right
so:
= sin B sqrt( [ 1 - cos(C) ] / 2 ) - sqrt( [ cos C - 1 ] / 2 ) cos B
You could use the half-angle formulas if you like
Since cos(2u) = 1 - 2 cos^2 u = 2 sin^2 u - 1
then sin u = sqrt( [cos (2u) - 1]/2 )
and cos u = sqrt( [1 - cos(2u)]/2 ) if i did that right
so:
= sin B sqrt( [ 1 - cos(C) ] / 2 ) - sqrt( [ cos C - 1 ] / 2 ) cos B
Similar questions