Prove that the average of the numbers n sin (n degrees) (n = 2, 4, 6, ... 180) is cot (1 degree)

Question

Hint: Express everything in terms of sin and cos. Can you rearrange the result into a form that allows you to use some of the identities you've learned?

How to start? I know I have to use trigonometric identities but where?
Thanks a lot.

Answer 1

sin0+180sin180 = 0sin0 + 180sin0
2sin2 + 178sin178 = 2sin2 + 178sin2 = 180sin2
...
88sin88 + 102sin102 = 180sin88

so, we have
180(sin0 + sin2 + sin4 + ... + sin88) + 90sin90
= 180(sin0+sin2+...+sin90) - 90
That makes the average
2(sin0+sin2+...+sin90)-1

Gotta think some more. Back later. See what you can figure out.