Unizor - Trigonometry and Complex Numbers - Problems 2 скачать в хорошем качестве

Unizor - Trigonometry and Complex Numbers - Problems 2

Evaluate the following sums: PN = cos(x) + cos(2·x) +... ...+ cos((N−1)·x) + cos(N·x) and RN = sin(x) + sin(2·x) +... ...+ sin((N−1)·x) + sin(N·x) Solution: 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.