Summation of Trignometric Series

[Himanshu]

Sum the following:

Sin(x) + Sin(x+d) + Sin(x+2d)...+Sin(x+(n-1)d).

I only know that summation of Sin and Cos functions whose arguments are in Arithmetic Progression can be done through telescopic series. But I don't know how to proceed. Please Help!

 

[Count Iblis]

Use the identity: Sin(x) = [Exp(ix) - Exp(-ix)]/(2i)

Then you get two ordinary geometric series.

 

[Architeuthis Dux]

I can suggest using the formula for the sum of a finite geometric series to solve this problem. We can rewrite the series as:

Sin(x) + Sin(x+d) + Sin(x+2d)...+Sin(x+(n-1)d) = Sin(x) + Sin(x)Cos(d) + Sin(x)Cos(2d) + ... + Sin(x)Cos((n-1)d)

This can be simplified to:

Sin(x)(1 + Cos(d) + Cos(2d) + ... + Cos((n-1)d))

Using the formula for the sum of a finite geometric series, we can find the sum of the Cosine terms:

1 + Cos(d) + Cos(2d) + ... + Cos((n-1)d) = (Cos(d)^n - 1)/(Cos(d) - 1)

Substituting this back into our original expression, we get:

Sin(x)(1 + Cos(d) + Cos(2d) + ... + Cos((n-1)d)) = Sin(x)((Cos(d)^n - 1)/(Cos(d) - 1))

This is the sum of the original series, and it can be simplified further if needed. I hope this helps in solving the problem.