Prove that average of the numbers : 2 sin2° , 4 sin4° , 6 sin6° ..... , 180 sin 180° = cot 1°

we have cos (n-1) - cos (n+1) = 2 sin n sin 1 ,,,1

now 180 sin 180 = 180 sin 0
178 sin 178 = 180 sin 2

so total sum = 180 ( sin 0 + sin 2 + sin 4 + ... + sin 88) + 90 sin 90
= 90 ( 2 sin 0 + 2 sin 2 + 2 sin 4 + 2 sin 88) + 90 sin 90

as there are 90 numbers average
= 2 sin 2 + 2 sin 4 + 2 sin 88 + sin 90 as
= ( cos 1 - cos 3)/ sin 1 + (cos 3 - cos 5)/ sin 1 + ( cos 87- cos 89 / sin 1 + 1
= ( cos 1 - cos 89)/ sin 1 + 1
= cot 1 - sin 1/sin 1 + 1 as cos 89 = sin 1
= cot 1 - 1 + 1
= cot 1