# Trigonometry problem

David113
Posers and Puzzles 07 Apr '07 15:06
1. 07 Apr '07 15:06
Find, without a computer/calculator, the EXACT value of:
cos(pi/7)cos(2pi/7)cos(3pi/7)
2. HandyAndy
Non sum qualis eram
08 Apr '07 00:14
Originally posted by David113
Find, without a computer/calculator, the EXACT value of:
cos(pi/7)cos(2pi/7)cos(3pi/7)
ðŸ˜´ðŸ˜´ðŸ˜´
3. 08 Apr '07 07:29
Wow. Ok. As 7 is not a Fermat prime (i.e. one which can be expressed as 2^(2n)+1), I assume this answer cannot be obtained as a sum or multiple of two fractions or as a root. So I'm stuck. A hint?
4. 08 Apr '07 17:51
Originally posted by itisi
Wow. Ok. As 7 is not a Fermat prime (i.e. one which can be expressed as 2^(2n)+1), I assume this answer cannot be obtained as a sum or multiple of two fractions or as a root. So I'm stuck. A hint?
Hint: the solution is a very simple number.
5. 09 Apr '07 21:27
As I could still not think of a method, I sketched the cos graph, marked off the three points (I got 0.95, 0.7 and 0.2) and multiplied to get 0.133. This is close enough to 1/7 for me to assume (very unmathematically!) that this is the answer as it is in keeping with your hint and seems a reasonable answer to a question with 1/7ths in it.
6. sonhouse
Fast and Curious
10 Apr '07 02:59
The post that was quoted here has been removed
Shoudn't that first one be Cos(pi/1)= etc.? otherwise it would just be the Cos(0).
7. 10 Apr '07 04:40
Originally posted by sonhouse
Shoudn't that first one be Cos(pi/1)= etc.? otherwise it would just be the Cos(0).
Shouldn't that first one be "not equal to" (\neq) since Cos(x) is less than or equal to one? The equation said something like 2 times ... and so that rang as most likely rong (sic). Suck the wrongness.
8. 10 Apr '07 23:24
Originally posted by sonhouse
Shoudn't that first one be Cos(pi/1)= etc.? otherwise it would just be the Cos(0).
It needs to be cos(0), because cos0=1 and all of the other expressions in the list are also equal to 1

Originally posted by dontwantto
Shouldn't that first one be "not equal to" (\neq) since Cos(x) is less than or equal to one? The equation said something like 2 times ... and so that rang as most likely rong (sic). Suck the wrongness.
I would love to know what this means!

I would also love to see an answer to David's original question.
9. 11 Apr '07 21:52
The answer is actually 1/8, and it's not too hard to get, but the usual trick requires a detour through complex analysis.

Easy part first. We can turn this product of cosines into a sum by using an identity: cos A cos B = 1/2 (cos (A+B) + cos (A-B)). Note that if A-B is negative, we can just turn it positive since cos (-x) = cos x. So...

cos pi/7 cos 2pi/7 cos 3pi/7
= 1/2 (cos 3pi/7 + cos pi/7) (cos 3pi/7)
= 1/2 (cos 3pi/7 cos 3pi/7 + cos pi/7 cos 3pi/7)
= 1/2 (1/2 (cos 6pi/7 + cos 0) + 1/2 (cos 4pi/7 + cos 6pi/7))
= 1/4 (cos 2pi/7 + cos 4pi/7 + cos 6pi/7 + 1)

Now we get to the interesting part. A basic rule of complex analysis is that

e^(ix) = cos x + i sin x

Another one is that there are exactly n solutions to the equation x^n = 1, given by e^(i * 2pi (1/n)), e^(i * 2pi (2/n)), ..., up to e^(i * 2pi (n/n)), which is just 1 itself. In particular, there are 7 solutions to x^7 = 1 (we call them seventh roots of unity):

e^(i * 2pi/7), e^(i * 4pi/7), ..., e^(i*12pi/7), e^(i * 14pi/7)=1

These are roots of the degree-seven polynomial x^7 - 1, so they add up to the coefficient of x^6. Since there IS no x^6 term, they must add up to zero. In particular, their real parts add up to zero. And by the first thing I mentioned above, their real parts are as follows:

cos 2pi/7, cos 4pi/7, cos 6pi/7, cos 8pi/7, cos 10pi/7, cos 12pi/7, 1

We now bring one more identity into play: cos (2pi - x) = cos x. So what we have here is

2 cos 2pi/7 + 2 cos 4pi/7 + 2 cos 6pi/7 + 1 = 0

Solving,

cos 2pi/7 + cos 4pi/7 + cos 6pi/7 = -1/2

And we can plug this into that first equation:

cos pi/7 cos 2pi/7 cos 3pi/7 = 1/4 ((-1/2) + 1) = 1/4 (1/2) = 1/8

Voila! There may be a simpler way (probably not), but this is the first thing any complex analysis student will try -- he's seen too many exercises like this.
10. 11 Apr '07 22:01
Oh, by the way, this is where all of RookAttack's equalities come from. It works for any odd number 2n+1:

cos pi/(2n+1) cos 2pi/(2n+1) ... cos n*pi/(2n+1) = (1/2)^n

So his post contained the right answer. But now you know why.
11. 15 Apr '07 03:28
Originally posted by CZeke
The answer is actually 1/8, and it's not too hard to get, but the usual trick requires a detour through complex analysis.

Easy part first. We can turn this product of cosines into a sum by using an identity: cos A cos B = 1/2 (cos (A+B) + cos (A-B)). Note that if A-B is negative, we can just turn it positive since cos (-x) = cos x. So...

cos pi/7 cos ...[text shortened]... e first thing any complex analysis student will try -- he's seen too many exercises like this.
The solution is simple...

cos(pi/7)cos(2pi/7)cos(3pi/7)=
-cos(pi/7)cos(2pi/7)cos(4pi/7)=...

(Now use the identity cosx=sin(2x)/(2sinx))

...=-[sin(2pi/7)/(2sin(pi/7))][sin(4pi/7)/(2sin(2pi/7))][sin(8pi/7)/(2sin(4pi/7))] = -sin(8pi/7)/(8sin(pi/7)) = sin(pi/7)/(8sin(pi/7)) = 1/8
12. 15 Apr '07 08:16
Whoa. That's a lot simpler -- thanks for showing us. It reminds me of telescoping sums.
13. 15 Apr '07 20:15
Very neat. Very impressed.
14. DeepThought