# A pretty difficult trig identity

kbaumen
Posers and Puzzles 25 May '08 09:13
1. 25 May '08 09:13
I had this as homework in my algebra course and it took a fair amount of time for me to find the solution. It's actually quite difficult to notice.

Prove the identity:
cos(2pi/7) + cos(4pi/7) + cos(6pi/7) = -1/2
2. sonhouse
Fast and Curious
25 May '08 20:102 edits
Originally posted by kbaumen
I had this as homework in my algebra course and it took a fair amount of time for me to find the solution. It's actually quite difficult to notice.

Prove the identity:
cos(2pi/7) + cos(4pi/7) + cos(6pi/7) = -1/2
Well for a reality check, I did the calcs on my casio and finally convinced myself, any one of those worked out = 0.999 something and added together, it =2.999 or so. Don't see how that works out to be minus 1/2.
Even if it was supposed to be inverted it would come out to 1/3 or so, a far cry from 1/2.
Are the units in degrees? or maybe radians?
-1/2 radian?
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Back to the cosign: 2Pi is 6.28 and /7= 0.897 and cos (0.8970) = 0.999 any way you look at it and making it 2PI or 4PI or 6PI doesn't change it much, only in the 4th and 5th digit, so can't see adding those three values and coming up with anything like 1/2 or -1/2.
3. 25 May '08 20:372 edits
Originally posted by sonhouse
Well for a reality check, I did the calcs on my casio and finally convinced myself, any one of those worked out = 0.999 something and added together, it =2.999 or so. Don't see how that works out to be minus 1/2.
Even if it was supposed to be inverted it would come out to 1/3 or so, a far cry from 1/2.
Are the units in degrees? or maybe radians?
-1/2 rad igit, so can't see adding those three values and coming up with anything like 1/2 or -1/2.
Everywhere with pi it's radians - 2pi/7 radians, 4pi/7 radians, etc. Isn't that basic notation in trigonometry? To simply write a number without a superscript '0' to point out that those are radians?

-1/2 is a number.

It's a normal problem. I don't think that 120 pupils (ok, 60, some couldn't prove it) and 2 teachers could altogether make a single mistake and prove the unprovable.

EDIT: Btw, I checked it with a calculator (basic Gnome calculator) and it really came out -1/2. So you probably calculated in degrees.
4. 25 May '08 23:322 edits
Originally posted by sonhouse
Back to the cosign: 2Pi is 6.28 and /7= 0.897 and cos (0.8970) = 0.999 any way you look at it and making it 2PI or 4PI or 6PI doesn't change it much, only in the 4th and 5th digit, so can't see adding those three values and coming up with anything like 1/2 or -1/2.
You should try it again, but this time using RADIAN measure (which is what is implied when giving the cos of some factor of PI. (And it works out perfectly using RADIAN measure.)

EDIT: Also, this does not seem like an IDENTITY to me. An identity is something like cos^2(x) + sin^2(x) = 1, where x can be any value. This is a simple, 'type-it-into-your-calculator-and-check-to-see-if-it-is-in-radian-measure' type question, that we math teachers love to give.
5. joe shmo
Strange Egg
25 May '08 23:56
Originally posted by kbaumen
Everywhere with pi it's radians - 2pi/7 radians, 4pi/7 radians, etc. Isn't that basic notation in trigonometry? To simply write a number without a superscript '0' to point out that those are radians?

-1/2 is a number.

It's a normal problem. I don't think that 120 pupils (ok, 60, some couldn't prove it) and 2 teachers could altogether make a single mist ...[text shortened]... sic Gnome calculator) and it really came out -1/2. So you probably calculated in degrees.
I believe you that it is an Identity, but i havent worked with pi/7 in any that I have ever done ( witch isnt many) your probably not going to want decimal equivalents of the radians, since it is such a clean target. i cant think of a way to use cos(t1+/- t2) to find the cosine of a 7th radian......
6. sonhouse
Fast and Curious
26 May '08 00:33
Originally posted by Gastel
You should try it again, but this time using RADIAN measure (which is what is implied when giving the cos of some factor of PI. (And it works out perfectly using RADIAN measure.)

EDIT: Also, this does not seem like an IDENTITY to me. An identity is something like cos^2(x) + sin^2(x) = 1, where x can be any value. This is a simple, 'type-it-into-your-ca ...[text shortened]... nd-check-to-see-if-it-is-in-radian-measure' type question, that we math teachers love to give.
Yep, right of course. I thought that's what my problem was, but I was having trouble getiing my little casio to go into radian mode, finally figured that one out, and it came out at -1/2 (actually -0.5)
The thing that puzzles me is with one radian = 57 odd degrees, 180/PI, why can't I convert the calc using degrees into radians at the end? The numerical answer at around 3, first off is not negative. Oh, I guess the - part comes from the angle being to the left of high noon (the zero degree hack). Is that right? So the -sign doesn't even mean the same thing as an algebaic - sign. I think, correct me if I am in left field here. So anyway, 0.5 radians is about 28 degrees. How do you convert the answer I got doing the calcs with degrees, which came out around 3? Three degrees is more like 0.05 radians not 0.5.
I must be bolluxing the units then. Help!
7. joe shmo
Strange Egg
26 May '08 00:592 edits
Originally posted by sonhouse
Yep, right of course. I thought that's what my problem was, but I was having trouble getiing my little casio to go into radian mode, finally figured that one out, and it came out at -1/2 (actually -0.5)
The thing that puzzles me is with one radian = 57 odd degrees, 180/PI, why can't I convert the calc using degrees into radians at the end? The numerical a Three degrees is more like 0.05 radians not 0.5.
I must be bolluxing the units then. Help!
multiply degrees by Pi/180 to convert from degrees to radians, and vice versa for radians to degrees

You have to think about the Cartesian plane......if the cosine is negative the angle resides in the second or third quadrant then the angle is obtuse either way

if i am understanding the question

I think your calcs are incorrect a cosine of -1/2 belong to an angle of 120 degrees, or 2pi/3 radians and 240 degrees or 4pi/3 radians in the range of ( 0 , 2Pi )

so the whole left side should be able to be manipulated into somthing of the sort using known identities, but it is tough
8. 26 May '08 02:24
It probably has something to do with the real parts of the 7th root of unity.
9. 26 May '08 02:24
Originally posted by kbaumen
I had this as homework in my algebra course and it took a fair amount of time for me to find the solution. It's actually quite difficult to notice.

Prove the identity:
cos(2pi/7) + cos(4pi/7) + cos(6pi/7) = -1/2
I'm aware of a variation of this problem that was included in the 1963 math olympics. So it is indeed a tough problem to solve:

Prove:
cos pi/7 - cos 2pi/7 + cos 3pi/7 = 1/2
10. 26 May '08 08:171 edit
Originally posted by Gastel
EDIT: Also, this does not seem like an IDENTITY to me. An identity is something like cos^2(x) + sin^2(x) = 1, where x can be any value. This is a simple, 'type-it-into-your-calculator-and-check-to-see-if-it-is-in-radian-measure' type question, that we math teachers love to give.
This is an identity. A trigonometric identity is an equality that involves trigonometric functions. What you have provided is is a basic relationship used to prove other identities. Similar to tgx = sinx/cosx.

Now, can anyone prove it? (The one I provided - cos(2pi/7) + ...)
11. 26 May '08 08:49
Again, tgx=sinx/cosx is an identity, x can be anything. The question you've asked is like asking to prove 1+2+3=6. An identity is something like x^2+y^2+2xy=(x+y)^2.
12. 26 May '08 08:561 edit
The problem is to prove LHS=RHS, which can be done..I'll reply with the answer soon..As far as i can remember u have to prove another known identity first and get a 4th degree equation in cos(x) where x = 2n(pi)/7 and n is an integer..so the roots of that equation wil be cos(0),cos(2pi/7),cos(4pi/7) and cos(6pi/7).. then u can directly get the desired equation by getting the value for the summation of the roots..i.e. like in a quadratic equation(ax2+bx+c=0) where alpha+beta = -(b/a)..hope this helps.. ðŸ˜€
13. 26 May '08 09:22
Originally posted by blacknight1985
The problem is to prove LHS=RHS, which can be done..I'll reply with the answer soon..As far as i can remember u have to prove another known identity first and get a 4th degree equation in cos(x) where x = 2n(pi)/7 and n is an integer..so the roots of that equation wil be cos(0),cos(2pi/7),cos(4pi/7) and cos(6pi/7).. then u can directly get the desired ...[text shortened]... ..i.e. like in a quadratic equation(ax2+bx+c=0) where alpha+beta = -(b/a)..hope this helps.. ðŸ˜€
ok here it is..the other identity is when x=2n(pi)/7, cos(3x)=cos(4x)..(u can prove it urself easily) then u expand both sides to get the desired 4th degree equation i've mentioned earlier..then the rest is the same..
14. wolfgang59
howling mad
26 May '08 09:49
Originally posted by kbaumen
This is an identity. A trigonometric identity is an equality that involves trigonometric functions. What you have provided is is a basic relationship used to prove other identities. Similar to tgx = sinx/cosx.

Now, can anyone prove it? (The one I provided - cos(2pi/7) + ...)
What you have here is an equality which needs proving NOT an identity which as previously discussed will involve a variable.

eg x^2 - y^2 = (x-y)(x+y) is an identity

That of course doesnt make it any easier! ðŸ˜ž
15. 26 May '08 12:08
Having tried various different approaches - and proving various identities that weren't asked for (I'm seriously out of practice with these things) - I've got a method that's so simple I suspect there's something wrong with it!

[Will write 2pi/7 as a, to simplify things]

It relies on e^(ia) being a root of x^7 = 1.

But x^7 = (x - 1)(x^6 + x^5 + x^4 + x^3 + x^2 + x + 1)
=> e^(6ia) + e^(5ia) + e^(4ia) + e^(3ia) + e^(2ia) + e^(ia) + 1 = 0

Taking the real part of this:

cos(6a) + cos(5a) + cos(4a) + cos(3a) + cos(2a) + cos(a) + 1 = 0

But note: cos(2pi - x) = cos(x)
so cos(6a) = cos(12pi/7) = cos(2pi/7) = cos(a)
and cos(5a) = cos(2a)
and cos(4a) = cos(3a)

=> 2[cos(a) + cos(2a) + cos(3a)] + 1 = 0

=> cos(2pi/7) + cos(4pi/7) + cos(6pi/7) = -1/2
Ta-da!