PLS PROVE

cos 20 cos 100+cos 100 cos 140 - cos 140 cos 200=-3/4

sin@/2 sin 7@/2  +sin 3@/2 sin 11@/2=sin2@ sin5@

cos 20 cos 100+cos 100 cos 140 - cos 140 cos 200=-3/4

1/2[(cos120+cos80)+(cos240+cos40)-(cos340+cos60)]

 

1/2[-1/2 +cos80 - 1/2 + cos40 - cos20 -1/2]

1/2[-3/2 +(cos80+cos40)-cos20]

consider (cos80+cos40)-cos20 = 2cos60cos20-cos20 = 0

 

hence cos 20 cos 100+cos 100 cos 140 - cos 140 cos 200 =

1/2[-3/2] = -3/4

sin@/2 sin 7@/2  +sin 3@/2 sin 11@/2=sin2@ sin5@

let @/2 = t

now, sin@/2 sin 7@/2  +sin 3@/2 sin 11@/2 =sint sin7t + sin3t sin11t

=1/2[(cos6t - cos8t )+(cos8t - cos14t)]

=1/2[ cos6t - cos14t ]

=1/2[ -2 sin10t (-sin4t )]

=sin10t sin4t = sin10@/2  sin4@/2 = sin5@ sin2@

hence proved