Tuesday, January 6, 2015
Unizor - Trigonometry and Complex Numbers - Problems 2
Evaluate the following sums:
PN = cos(x) + cos(2·x) +...+ cos((N−1)·x) + cos(N·x)
RN = sin(x) + sin(2·x) +...+ sin((N−1)·x) + sin(N·x)
The creative part of this problem is to realize that switching to complex numbers greatly simplifies it, basically converting a trigonometric problem we don't know how to approach to an algebraic problem of summarizing a geometric progression.
Unfortunately, non-creative part of a solution requires some tedious trigonometric transformations (which from pedagogical standpoint play their positive role).
Let's use the Euler's formula
e^(i·φ) = cos(φ)+i·sin(φ).
Consider a series:
SN = e^(i·x) + e^(2·i·x) +...+ e^[(N−1)·i·x]+ e^(N·i·x)
From the Euler's formula for φ=x, φ=2·x,..., φ=(N−1)·x and φ=N·x follows that
SN = PN + i·RN
Since e^(k·i·x) = [e^(i·x)]k, we can transform the above sum SN into
SN = e^(i·x)+[e^(i·x)]^2+...+ [e^(i·x)]^(N−1) + [e^(i·x)]^N
The latter is a sum of a geometric progression.
Without bringing up a formula, which might be hard to remember, we will calculate the sum using the following logic:
SN·e^(i·x) = SN − e^(i·x) + [e^(i·x)](N+1)
Simplifying and resolving it for SN, we obtain:
SN = [e^(i·x) − e^(i·(N+1))·x] / [1−e^(i·x)]
All we have to do now is to convert this complex number into its canonical representation a+b·i and obtain the solutions:
PN = a and RN = b
This is a purely technical exercise with complex numbers.