Question

Prove that the average of the numbers n sin n° , n =2,4,6....180 , is cot 1°

Answer 1

180/2 = 90.

The total no of terms = 90.

We can write the terms as:

(2 * sin(2) + 4 * sin(4) + 6 sin(6) + ... + 180 * sin(180)) / 90

as we know the relation sin θ = sin (180 - θ)

So, we can write: sin(2) = sin(178), sin(4) = sin(176) , all the way to sin(88) = sin(92)

And we know some of the values as well, like sin 180 is zero, sin90 is one and so on.

(2 * sin(2) + 178 * sin(2) + 4 * sin(4) + 176 * sin(4) + ... + 88 * sin(88) + 92 * sin(92) + 90 * sin(90)) / 90

or, (180 * (sin(2) + sin(4) + sin(6) + ... + sin(88)) + 90) / 90

Now most importantly, we have to know which deduction to be used where and we know sin/cos = tan. and cos/sin = cot.

So, the series can be written as sin(a) + sin(2a) + sin(3a) + ... + sin(na) = (1/2) * cot(a/2) - cos((n + 1/2) * a) / (2 * sin(a/2))

Therefore, sin(2) + sin(4) + ... + sin(88) =>

(1/2) * cot(2/2) - cos((44 + 1/2) * 2) / (2 * sin(2/2)) [ applying the above formula ]

or, (1/2) * (cos(1) - cos(89)) / sin(1)

(180 * (sin(2) + sin(4) + sin(6) + ... + sin(88)) + 90) / 90

or, (180 * (1/2) * (cos(1) - cos(89)) / sin(1) + 90) / 90

or, 90 * ((cos(1) - cos(89)) / sin(1) + 1) / 90

or, (cos(1) - cos(89) + sin(1)) / sin(1)

or, (cos(1) - sin(1) + sin(1)) / sin(1)

or, cos(1)/sin(1)

or, cot(1) = [RHS] Proved.