- 09 Apr '07 21:27As 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.
- 10 Apr '07 04:40

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.*Originally posted by sonhouse***Shoudn't that first one be Cos(pi/1)= etc.? otherwise it would just be the Cos(0).** - 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. - 11 Apr '07 21:52The 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. - 15 Apr '07 03:28

The solution is simple...*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.

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 - 18 Apr '07 13:44Cos (pi/7) also came up in this thread: Thread 65139. I got the result to the problem set in that thread (pi/7) by a numerical method, but was left wondering if there was an analytic one as pi/7 is a neat result where 0.4487 isn't. Can anyone see a link between the stuff above and that problem?
- 19 Apr '07 20:53Doesn't look like it. If there were a simple expression for cos(pi/7), it would surely be listed here:

http://mathworld.wolfram.com/TrigonometryAnglesPi7.html

As that page notes, there's definitely no way to do it without using at least cube roots (7 isn't a Fermat prime, so cos(pi/7) isn't constructible). You'll have to settle for what comes out of the cubic formula.