Top universities for mathematics test prospective students by setting questions in their entrance tests that stretch the most capable students. The example used here has been used in the past by Cambridge University and other top math colleges. The question is an excellent one, that uses De Moivre's Theorem, geometric progressions, trigonometric identities, complex numbers, and algebraic manipulation.
Problem: Find the General Solution to cos(x) + cos(2x) + cos(3x) + ... + cos(nx)
The core of the solution is to recognise that
cos.x + cos.2x + cos.3x + ... + cos.nx
= Real part of (cos.x + i.sin.x + cos.2x + i.sin.2x + cos.3x + i.sin.3x + ... + cos.nx + i.sin.nx)
(where i = √-1)
= Real part of [ (cos.x + i.sin.x) + (cos.x + i.sin.x)^2 + (cos.x + i.sin.x)^3 + ... + (cos.x + i.sin.x)^n ] ... [Eq'n 1]
This is a geometric progression: let Y = (cos.x + i.sin.x), and the series becomes
Real part of (Y + Y^2 + Y^3 + ... + Y^n)
From the rules of a geometric progression, this is
= Real part of [Y.(Y^n - 1) / (Y - 1)]
Now substitute Y = (cos.x + i.sin.x) back in again, and the sum becomes
= Real part of (cos.x + i.sin.x).[ (cos.x + i.sin.x)^n –1] / (cos.x + i.sin.x - 1)
= Real part of (cos.x + i.sin.x).[ (cos.nx + i.sin.nx) –1] / (cos.x + i.sin.x - 1)
Rationalise The Denominator
The denominator is complex. To get rid of the “i” term, multiply numerator and denominator by the complex conjugate of the denominator, (cos.x –1 – i.sin.x), and the summation becomes
=(cos.x + i.sin.x).[ (cos.nx + i.sin.nx) –1].(cos.x -1 + i.sin.x ) / [ (cos.x -1 + i.sin.x ).( cos.x -1 - i.sin.x) ]
For clarity, the denominator will be expanded first:
= (cos.x -1 + i.sin.x ).( cos.x -1 - i.sin.x)
This is simply the difference of two squares:
= (cos.x – 1)^2 – (i.sin.x)^2
= (cos.x)^2 – 2.cos.x. + 1 – (i^2.(sin.x)^2)
= (cos.x)^2 – 2.cos.x. + 1 + (sin.x)^2
But (cos.x)^2 + (sin.x)^2 = 1, so the denominator becomes
= 2 – 2.cos.x
=2.(1 – cos.x)
Evaluate the Numerator
The numerator is
= (cos.x + i.sin.x).[ (cos.nx + i.sin.nx) –1].( cos.x -1 - i.sin.x)
= (cos.x.cos.nx + cos.x.i.sin.nx - cos.x + i.sin.x.cos.nx + i.sin.x.i.sin.nx - i.sin.x)).( cos.x -1 - i.sin.x )
But from the standard trigonometric identities,
(cos.x.cos.nx - sin.x.sin.nx) = cos(nx + x) = cos(n+1).x and
cos.x.i.sin.nx + i.sin.x.cos.nx = sin(nx + x) = sin(n+1).x
So the numerator becomes
= (cos(n+1).x + i.sin(n+1).x - cos.x - i.sin.x). ( cos.x - 1 - i.sin.x )
= cos.x.(cos(n+1).x + i.sin(n+1).x - cos.x - i.sin.x) - (cos(n+1).x + i.sin(n+1).x - cos.x - i.sin.x) - i.sin.x.(cos(n+1).x + i.sin(n+1).x - cos.x - i.sin.x)
= cos.x.cos(n+1).x + i.cos.x.sin(n+1).x - cos².x - i.cos.x.sin.x - cos(n+1).x - i.sin(n+1).x + cos.x + i.sin.x - i.sin.x.cos(n+1).x - i².sin.x.sin(n+1).x + i.sin.x.cos.x + i².sin.x.sin.x
= cos.x.cos(n+1).x + i.cos.x.sin(n+1).x - cos².x - i.cos.x.sin.x - cos(n+1).x - i.sin(n+1).x + cos.x + i.sin.x - i.sin.x.cos(n+1).x + sin.x.sin(n+1).x + i.sin.x.cos.x - sin².x ... [Eq'n 2]
= cos.x + cos.2x + ... cos.nx + i.(sin.x + sin.2x + ... + sin.nx)
Find cos.x + cos.2x + cos.3x + ... + cos.nx
From Equation [1], the sum of the series is the real part of the geometric progression:
Gathering real terms, the real part of equation 2
= cos.x.cos(n+1).x - cos².x - cos(n+1).x + cos.x + sin.x.sin(n+1).x - sin².x
Simplifying,
cos.x.cos(n+1).x + sin.x.sin(n+1).x = cos[(n+1).x - x] = cos.nx, and
-cos²x - sin²x = -(cos²x + sin²x) = -1,
so the real part becomes
= cos.nx - cos.(n+1).x - 1 + cos.x
So the sum of cos.x + cos.2x + ... + cos.nx
= numerator / denominator
= (cos.nx - cos.(n+1).x - 1 + cos.x) / 2.(1 - cos.x)
= (cos.nx - cos.(n+1).x) / (2 - 2.cos.x) - 1 / 2
Q.E.D.
To verify this, choose x = π / 5 and n = 3 as an example:
(cos.3.π / 5 - cos.(4.π / 5)) / (2 - 2.cos.π / 5) - 1 / 2
= 0.809
cos(π / 5) + cos(2.π / 5) + cos(3.π / 5)
= 0.809 + 0.309 - 0.309
= 0.809
Sum of Trigonometric Series Summary
Advanced math students are sometimes asked to sum trigonometric series to enter the best math colleges. The solution to the series cos.x + cos.2x + ... + cos.nx has been shown and verified.
References for Sum of Trigonometric Series
No references have been used in this article, as the solution has been derived from either first principles, or from well-known mathematical formulas. Useful background for this proof may be found in the article "De Moivres Theorem Descripion With Examples and Application", "Trigonometry Sin(a+b) Cos(a+b) Sin(a)+Sin(b) Cos(a)+Cos(b)" and "Geometric Progression Math Formulas - Derivation and Usage".
Martin Bell - Martin holds a B.Sc. degree in chemical engineering, and an M.Sc. degree in electronics and computing. He has spent more than 25 years ...
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