IIT JEE , AIEEE Forum - Mathematics , Trignometry

rajanedu

prove that the average of numbers n sin n deg ,

n = 2,4,6,...,180 is cot 1 deg.

gcch29

let c = 2cos2 +4cos4 +........ +180cos180

is = i(2sin2 + 4sin4 +...........+180sin180)

c+is=2e^i.2 +4.e^1.4 +.......... 180e^i.180 e^i

= cos

+isin

let e^i.2=x

c+is=2x + 4x^2 + 6x^2 +............ +180x^180

it is an A.G.P

its sum is

=2x(1-x^180)/[(1-x)^2] + 180(x^180)/(1-x)

substituting back

=2e^1.2(1-e^i*2*180)/[(1-e^i*2)^2] + 180(e^i*2*180) /(1-e^i^2)

since e^i*pi = -1

=180/(1-e^i.2)

=180*[1/1-cos2 -sin2]

=180(1-cos2 +isin2)/(2-2cos2)

therefore s=(180 * sin2)/(2-2cos2)

=90cot1

ave=cot1

rate me for my effotrs

lazycol

e^i= cos

+ isin

;where

is in radians

in d Q

is in degrees,so conversion shud be ther

so first convert

,u get an agp

do sum , finally convert back 2 degrees & then u`ll get d ans