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

apple2357

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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

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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

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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

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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

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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

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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

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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

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I have been given a hint:

Try multiplying by sinx!
 

Dr.Peterson

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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

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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

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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

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Yes!

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

Subhotosh Khan

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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

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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.
 

Denis

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apple2357

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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

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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

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

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

Is that correct?
 

Jomo

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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

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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?
 
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