cosxcos4xcos8x =1/8: efficient way of solving this?

apple2357

Junior Member
I am trying to find an efficient way of solving
cosxcos4xcos8x=1/8, 0 to 2pi, i am thinking about the product of three trig expressions and what must be the case to equal 1/8.
The alternatives feel too heavy algebraically?

Jomo

Elite Member
I am trying to find an efficient way of solving
cosxcos4xcos8x=1/8, 0 to 2pi, i am thinking about the product of three trig expressions and what must be the case to equal 1/8.
The alternatives feel too heavy algebraically?
What have you tried? Please show us your work. Did you try 1* 1/2* 1/4? How about (1/2)^3... If you want to guess the answer then these are the things you need to try. If this fails then it is time for heavy algebra.

apple2357

Junior Member
What have you tried? Please show us your work. Did you try 1* 1/2* 1/4? How about (1/2)^3... If you want to guess the answer then these are the things you need to try. If this fails then it is time for heavy algebra.
Yes, i was trying different possible products! and wondering if there is some kind of symmetry to exploit.
I have tried sketching it for inspiration.

Dr.Peterson

Elite Member
Yes, i was trying different possible products! and wondering if there is some kind of symmetry to exploit.
I have tried sketching it for inspiration.
After making some attempts at using various identities to simplify the equation, I put it into Wolfram Alpha to see whether it at least ends up with nice solutions. It turns out that there are a couple solutions that you can find by trial and error, among 16 solutions per cycle, the rest of which are horrendous (involving a 12th degree equation). So I doubt that you will be able to manually find all solutions, though you can find a couple.

Where did the problem come from? Were you given any reason to think it could be solved?

apple2357

Junior Member
Yes, a friend who teaches maths sent me this and asked if i can spot a nice trick?

But after musing on it for a while, like you, i haven't made much progress.

I will get back to him in a bit for the 'trick' but wanted to thrash it around on here first!

Jomo

Elite Member
Yes, a friend who teaches maths sent me this and asked if i can spot a nice trick?

But after musing on it for a while, like you, i haven't made much progress.

I will get back to him in a bit for the 'trick' but wanted to thrash it around on here first!
After learning the trick please report back here!

Jomo

Elite Member
I am trying to find an efficient way of solving
cosxcos4xcos8x=1/8, 0 to 2pi, i am thinking about the product of three trig expressions and what must be the case to equal 1/8.
The alternatives feel too heavy algebraically?
Just thinking out loud. How about letting u=4x, 2x=u/2 and 8x = 2u. Will it simplify using the half angle and double angle formulas? I suspect not, since Dr P tried using various identities to simplify the equation with no luck.

apple2357

Junior Member
I have been given a hint:

Try multiplying by sinx!

Dr.Peterson

Elite Member
I have been given a hint:

Try multiplying by sinx!
I can see how that would help (up to a point) for a slightly different equation; but I don't yet see how to use it here.

Just to make sure, did you copy the equation correctly? We have it as

cos(x) cos(4x) cos(8x) = 1/8

apple2357

Junior Member
I can see how that would help (up to a point) for a slightly different equation; but I don't yet see how to use it here.

Just to make sure, did you copy the equation correctly? We have it as

cos(x) cos(4x) cos(8x) = 1/8
yes i believe thats correct. I have just asked him to confirm!

Dr.Peterson

Elite Member
To be more precise, the trick works very nicely for cos(x) cos(2x) cos(4x) = 1/8. I might even have thought of it in that case. (But probably not.)

apple2357

Junior Member
Yes!

My mistake. So sorry!
What a mess up.
The original question is cosx cos2x cos4x = 1/8

Subhotosh Khan

Super Moderator
Staff member
Yes!

My mistake. So sorry!
What a mess up.
The original question is cosx cos2x cos4x = 1/8
Now then follow the hint and show us what do you get.

Jomo

Elite Member
To be more precise, the trick works very nicely for cos(x) cos(2x) cos(4x) = 1/8. I might even have thought of it in that case. (But probably not.)
You are an amazing mathematician.

apple2357

Junior Member
Now then follow the hint and show us what do you get.
so multiply by sinx

sinx cosx cos2x cos4x = 1/8 sinx

1/2 sin2x cos2x cos4x = 1/8 sinx

(1/2)(1/2)sin4x cos4x = 1/8 sinx

(1/8) sin8x = (1/8) sinx

so sin8x = sinx

i think we can do some general solution stuff from here?

Dr.Peterson

Elite Member
so multiply by sinx

sinx cosx cos2x cos4x = 1/8 sinx

1/2 sin2x cos2x cos4x = 1/8 sinx

(1/2)(1/2)sin4x cos4x = 1/8 sinx

(1/8) sin8x = (1/8) sinx

so sin8x = sinx

i think we can do some general solution stuff from here?
Yes; now we ask ourselves, what is the relationship between two angles that have the same sine? There are two main relationships possible.

apple2357

Junior Member
so would this lead to 8x = x+ 360n or 180-x +360n

x = (360/7)n or x = 20 +40n

Is that correct?

Jomo

Elite Member
so would this lead to 8x = x+ 360n or 180-x +360n

x = (360/7)n or x = 20 +40n

Is that correct?
Why not just check these values in the original equation to see if they work? Can you show us your work that lead to 8x = x+ 360n or 180-x +360n

Dr.Peterson

Elite Member
so would this lead to 8x = x+ 360n or 180-x +360n

x = (360/7)n or x = 20 +40n

Is that correct?
These are equivalent to my two sets of solutions (though I used radians, as we normally do in solving trig equations).

Have you tried graphing and comparing what you see to your solutions?