How would you solve this little trig eqn?
February 20, 2011 pm28 5:43 pm
McRib asked me to solve: cos2x + cos4x + cos6x = 0
How would you solve that? Any guess on what I based my quick solution? Other methods?
(Let the domain be anything you like. [0,2π] is fine).
JD2718 
I bit.
But I don’t see a quick solution. (Yet.) I looked up the identity for cos(A+B), and used a few others, and got a cubic in cos2x that wasn’t too hard to factor. Hmm, I can’t find my TI, and Wolfram doesn’t agree with me, maybe I should try again.
I rarely have this sort of insight, but…
here’s the conversation:
4:13 PM McRib: here is something for you to do on the train:
4:14 PM cos2x+cos4x+cos6x=0
never seen anything like this…
just did it… but want to compare the notes
jd2718: domain?
4:15 PM McRib: not given, i assumed 0, 2pi
4:16 PM jd2718: Ok, let’s make that (0,π)
4:17 PM and solve on the unit circle, ok?
4:18 PM Let’s get a vertical vector, and two vectors with opposite x-coordinates
4:19 PM If 4x = π/2, we are good
but let me check for more
4:20 PM If 4x = 3π/2, that looks good
4x = 5π/2, ok this is boring
Am I missing some cases? I will consider for a while
Thanks
I applied
cos (a + b) + cos (a – b) = 2 cos(a)cos(b)
With a=4x and b=2x which got me to
cos(4x)(2cos(2x) + 1) = 0
So either factor must be 0. I also tried solving the cubic but couldn’t factorize it. However I can now, using the solutions I found the other way.
I looked for symmetry on the unit circle, and the answer just fell out. I rarely have that sort of insight.
That was precisely my first instinct too.
JD, unless I’m missing something your symmetry approach, while pretty, missed the x=pi/3 and 5pi/6 solutions? The solution needn’t have a vertical vector at all. And this (missing good solutions) can be the problem with non-algebraic methods, of course.
Yes, I found the pi/4 roots using a geometric approach, but used algebra and the trig identities to get x = pi/3 and x = 2pi/3. (I don’t think x=5pi/6 works…)
For reference:
cos(4x) = 2cos^2(2x)-1
cos(6x) = 4cos^3(2x)-3cos(2x)
and the cubic factors into (2cos(2x)+1)(2cos^2(2x)-1)
You’re quite right – I meant either 2pi/3 or 5pi/3. That’s what I get for doing it in my head I suppose :).
Completely missed! Wow. Thanks.
Jon, that’s awesome because that’s what I did! But, doing so I was missing the case of cos(2x) = cos(4x) = -1/2 and cos(6x) = 1. (I cheated and looked at the graphing calc and cross-referenced it with fergal’s answers. I was beating myself up about it for a couple of days until I saw your post.)
I think that case cos(2x) = cos(4x) = -1/2 and cos(6x) = 1. has something to do with roots of unity but I can’t prove it. In the meanwhile, fergal’s algebraic simplification makes it obvious why algebraically that must be a solution as well.
Don’t they look awfully like roots of unity? But how to make the algebra connect….?
The roots thing is cool.
Roots of unity always sum to zero (because the coefficient of x in the polynomial is the sum of all of the roots and the polynomial has no x term).
So taking the cube roots of unity and then taking root corresponding to 120deg as 2x, you get the second one for 4x and the 3rd one for 6x, so the sum is 0.
So
sum(cos(ix), i=1 to n) = 0
will always have a primitive nth root of unit as a solution. You can switch cos for sin too and I think you can make it cos(aix) for any integer a that’s relatively prime to n because the values will still end up running over all of the roots.
Oh, in fact you can drop the relatively prime condition. If gcd(a, n) = k then you’ll spin over the (n/k)th roots of 1, hitting each one k times which will still sum to 0.
And you don’t need the roots of unity as long as the angles span the appropriate arithmetic sequence, since it just involves the rotation of a regular polygon about the origin:
theta = 2pi/k
cos(phi + theta) + cos(phi + 2*theta) + cos(phi + 3*theta) + … + cos(phi + (k-1)*theta) = 0
I missed it, too.